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Unit: Matrices

Introduction to matrices.

Representing linear systems of equations with augmented matrices

Elementary matrix row operations

Row-echelon form & Gaussian elimination

Adding & subtracting matrices

Multiplying matrices by scalars

Properties of matrix addition & scalar multiplication

Multiplying matrices by matrices

Properties of matrix multiplication

Matrices as transformations

Determinant of a 2x2 matrix

Introduction to matrix inverses

Finding the inverse of a matrix using its determinant

Practice finding the inverses of 2x2 matrices

Determinants & inverses of large matrices

Solving equations with inverse matrices

Model real-world situations with matrices

About this unit

Free Mathematics Tutorials

Free Mathematics Tutorials

Matrices with examples and questions with solutions.

Examples and questions on matrices along with their solutions are presented .

Definition of a Matrix

Example 1 The following matrix has 3 rows and 6 columns.

Matrix entry (or element)

Square matrix, identity matrix, diagonal matrix, triangular matrix, transpose of a matrix, symmetric matrix, questions on matrices: part a, questions on matrices: part b, solutions to the questions in part a, solutions to the questions in part b, more references and links.

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Math Problems, Tests, Forums

Matrices and Determinants: Problems with Solutions

Multiplication of two 2x2 matrices

Solving Systems of Linear Equations Using Matrices

Hi there! This page is only going to make sense when you know a little about Systems of Linear Equations and Matrices , so please go and learn about those if you don't know them already.

The Example

One of the last examples on Systems of Linear Equations was this one:

Example: Solve

We went on to solve it using "elimination", but we can also solve it using Matrices!

Using Matrices makes life easier because we can use a computer program (such as the Matrix Calculator ) to do all the "number crunching".

But first we need to write the question in Matrix form.

In Matrix Form?

OK. A Matrix is an array of numbers:

Well, think about the equations:

They could be turned into a table of numbers like this:

We could even separate the numbers before and after the "=" into:

Now it looks like we have 2 Matrices.

In fact we have a third one, which is [x y z] :

Why does [x y z] go there? Because when we Multiply Matrices we use the "Dot Product" like this:

Which is the first of our original equations above (you might like to check that). Here it is for the second line.

 Try the third line for yourself.

The Matrix Solution

We can shorten this:

Then (as shown on the Inverse of a Matrix page) the solution is this:

What does that mean?

It means that we can find the X matrix (the values of x, y and z) by multiplying the inverse of the A matrix by the B matrix .

So let's go ahead and do that.

First, we need to find the inverse of the A matrix (assuming it exists!)

Using the Matrix Calculator we get this:

(I left the 1/determinant outside the matrix to make the numbers simpler)

Then multiply A -1 by B (we can use the Matrix Calculator again):

And we are done! The solution is:

x = 5 y = 3 z = −2

Just like on the Systems of Linear Equations page.

Quite neat and elegant, and the human does the thinking while the computer does the calculating.

Just For Fun ... Do It Again!

For fun (and to help you learn), let us do this all again, but put matrix "X" first.

I want to show you this way, because many people think the solution above is so neat it must be the only way.

So we will solve it like this:

And because of the way that matrices are multiplied we need to set up the matrices differently now. The rows and columns have to be switched over ("transposed"):

And XA = B looks like this:

Then (also shown on the Inverse of a Matrix page) the solution is this:

This is what we get for A -1 :

In fact it is just like the Inverse we got before, but Transposed (rows and columns swapped over).

Next we multiply B by A -1 :

And the solution is the same:

x = 5 , y = 3 and z = −2

It didn't look as neat as the previous solution, but it does show us that there is more than one way to set up and solve matrix equations. Just be careful about the rows and columns!

Definition, Formulas, Solved Example Problems - Solved Example Problems on Applications of Matrices: Solving System of Linear Equations | 12th Mathematics : UNIT 1 : Applications of Matrices and Determinants

Chapter: 12th mathematics : unit 1 : applications of matrices and determinants, solved example problems on applications of matrices: solving system of linear equations.

Solution to a System of Linear equations

(i) Matrix Inversion Method

Example 1.22.

Solve the following system of linear equations, using matrix inversion method:

5 x  +  2  y  =  3, 3 x  +  2  y  =  5 .

matrix problem solving questions

Then, applying the formula X = A −1 B , we get

matrix problem solving questions

So the solution is ( x  = −1,  y  = 4).

Example 1.23

Solve the following system of equations, using matrix inversion method:

2 x 1  + 3 x 2  + 3 x 3  = 5,

x 1  – 2 x 2  +  x 3  = -4,

3 x 1  – x 2  – 2 x 3  = 3

The matrix form of the system is AX = B,where

matrix problem solving questions

So, the solution is (  x 1  = 1,  x 2  = 2,  x 3  = −1) .

Example 1.24

matrix problem solving questions

Writing the given system of equations in matrix form, we get

matrix problem solving questions

Hence, the solution is ( x  = 3,  y  = - 2,   z  = −1).

(ii) Cramer’s Rule

Example 1.25.

Solve, by Cramer’s rule, the system of equations

x 1   −   x 2   =  3, 2 x 1   +  3 x 2   +  4 x 3   =  17,  x 2   +  2 x 3   =  7.

First we evaluate the determinants

matrix problem solving questions

So, the solution is  ( x 1  = 2,  x 2  = - 1,   x 3  = 4).

Example 1.26

In a T20 match, Chennai Super Kings needed just 6 runs to win with 1 ball left to go in the last over. The last ball was bowled and the batsman at the crease hit it high up. The ball traversed along a path in a vertical plane and the equation of the path is  y  =   ax 2   +   bx  +   c  with respect to a  xy  -coordinate system in the  vertical  plane  and  the  ball  traversed   through   the   points  (10,8), (20,16), (30,18) , can you conclude that Chennai Super Kings won the  match?

matrix problem solving questions

Justify your answer. (All distances are measured in metres and the meeting point of the plane of  the path with the farthest boundary line is (70, 0).)

The path   y  =  ax 2  +  bx  +  c   passes through the points (10,8), (20,16), (40, 22) . So, we get   the   system   of   equations   100 a   + 10 b   +   c   =   8,   400 a   +   20 b   +   c =   16,1600 a   +   40 b   +   c   =   22.   To   apply   Cramer’s rule, we find

matrix problem solving questions

When  x  = 70, we get  y  = 6. 

So, the ball went by 6 metres high over the boundary line and it is impossible for a fielder  standing even just before theboundary line to jump and catch the ball.

    Hence the ball went for a super six and the Chennai Super Kings won the   match.

(iii) Gaussian Elimination Method

Example 1.27.

Solve the following system of linear equations, by Gaussian elimination method :

4 x  +  3 y  +  6 z  =  25,  x  +  5  y  +  7 z  =  13, 2 x  +  9  y  +   z  =  1.

Transforming the augmented matrix to echelon form, we get

matrix problem solving questions

The equivalent system is written by using the echelon form:

 x + 5y  + 7z = 13 , … (1)

 17y + 22z = 27 , … (2)

 199z = 398 . … (3)

matrix problem solving questions

Substituting z = 2, y = -1  in (1), we get x = 13 - 5  × (−1 ) − 7 × 2 = 4 .

So, the solution is ( x =4, y = - 1, z = 2 ).

Note.  The above method of going from the last equation to the first equation is called the  method of back substitution .

Example 1.28

The  upward  speed   v ( t ) of a rocket  at time t is approximated by v(t) = at 2  + bt + c, 0 ≤  t ≤ 100 where a, b, and c are constants. It has been found that the speed at times t = 3, t = 6 , and t = 9 seconds are respectively, 64, 133, and 208 miles per second respectively. Find the speed at time  t = 15 seconds. (Use Gaussian elimination method.)

matrix problem solving questions

Since  v (3) =64,  v (6) = 133 and  v (9) = 208 , we get the following system of linear equations

 9a +3b + c = 64 ,

 36a + 6b + c = 133,

 81a + 9b + c = 208 .

We solve the above system of linear equations by Gaussian elimination method.

Reducing the augmented matrix to an equivalent row-echelon form by using elementary row  operations, we get

matrix problem solving questions

Writing the equivalent equations from the row-echelon matrix, we get

9a + 3b + c =  64, 2b + c = 41, c= 1.

By back substitution, we get 

matrix problem solving questions

So, we get v (t) = 1/3 t 2    + 20t + 1.

Hence,  v (15) = 1/3 (225) + 20(15) + 1 = 75 + 300 + 1 = 376

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Matrix Multiplication Questions

Matrix multiplication questions with solutions are given here for practice. These matrix multiplication questions will not only help students with the preparation for board exams but also with many competitive exams as well.

Also check: Important questions for class 12 Matrices .

In linear algebra, a matrix is an arrangement of elements in the form of rows and columns. An m × n matrix has m rows and n columns. If a matrix has an equal number of rows and columns, then the matrix is called a square matrix. When we perform multiplication operations with matrices, it is called matrix multiplication.

Know more about the properties of matrix multiplication .

Matrix Multiplication Questions with Solutions

Let us solve some questions to practise matrix multiplications.

Question 1:

Find the product of the following matrices:

\(\begin{array}{l}A=\begin{bmatrix}1 & 2 & 3 \\0 & 2 & 1 \\ 1& 2 & 5 \\\end{bmatrix}\:\:and\:\:B=\begin{bmatrix}1 & 0 \\0 & 1 \\2 & 1 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}A\times B=\begin{bmatrix}1 & 2 & 3 \\0 & 2 & 1 \\ 1& 2 & 5 \\\end{bmatrix}\times \begin{bmatrix}1 & 0 \\0 & 1 \\2 & 1 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}=\begin{bmatrix}1+0+6 &0+2+3 \\0+0+2 &0+2+1 \\1+0+10 &0+2+5 \\\end{bmatrix} = \begin{bmatrix}7 &5 \\ 2& 3\\ 11& 7 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}\therefore A\times B =\begin{bmatrix}7 &5 \\ 2& 3\\ 11& 7 \\\end{bmatrix}\end{array} \)

Question 2:

Prove that for the matrices A and B, (A + B) 2 ≠ A 2 + 2AB + B 2 where

\(\begin{array}{l}A=\begin{bmatrix}1 & 0 \\2 & 3 \\\end{bmatrix}\:\:and\:\:B=\begin{bmatrix}1 & 1\\-1 & 2 \end{bmatrix}\end{array} \)

\(\begin{array}{l}A+B=\begin{bmatrix}1 & 0 \\2 & 3 \\\end{bmatrix}+\begin{bmatrix}1 & 1\\-1 & 2 \end{bmatrix}=\begin{bmatrix}1 & 1 \\1 & 5 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}(A+B)^{2}=\begin{bmatrix}1 & 1 \\1 & 5 \\\end{bmatrix}\times \begin{bmatrix}1 & 1 \\1 & 5 \\\end{bmatrix}=\begin{bmatrix}2 & 6 \\6 & 26 \\\end{bmatrix}\:\:\:\:….(i)\end{array} \)

\(\begin{array}{l}A^{2}=\begin{bmatrix}1 & 0 \\2 & 3 \\\end{bmatrix}\times \begin{bmatrix}1 & 0 \\2 & 3 \\\end{bmatrix}= \begin{bmatrix}1 & 0 \\8 & 9 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}B^{2}=\begin{bmatrix}1 & 1\\-1 & 2 \end{bmatrix}\times \begin{bmatrix}1 & 1\\-1 & 2 \end{bmatrix}=\begin{bmatrix}0 & 3 \\-3 & 3 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}A\times B=\begin{bmatrix}1 & 0 \\2 & 3 \\\end{bmatrix}\times\begin{bmatrix}1 & 1\\-1 & 2 \end{bmatrix}=\begin{bmatrix}1 & 1 \\-1 & 8 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}2AB=2\begin{bmatrix}1 & 1 \\-1 & 8 \\\end{bmatrix}=\begin{bmatrix}2 & 2 \\-2 & 16 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}A^{2}+2AB+B^{2}=\begin{bmatrix}1 & 0 \\8 & 9 \\\end{bmatrix}+\begin{bmatrix}0 & 3 \\-3 & 3 \\\end{bmatrix}+\begin{bmatrix}2 & 2 \\-2 & 16 \\\end{bmatrix}=\begin{bmatrix}3 &5 \\3 & 28\\\end{bmatrix}\:\:….(ii)\end{array} \)

From (i) and (ii), it is proved that (A + B) 2 ≠ A 2 + 2AB + B 2 .

Question 3:

For given matrices A and B, find AB and BA, also prove that AB ≠ BA.

\(\begin{array}{l}A=\begin{bmatrix}4 & 2 & -1 \\3 & 2 & 0 \\\end{bmatrix}\:\:and\:\:B=\begin{bmatrix}0 & 1 \\2 & 3 \\7 & 4 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}AB=\begin{bmatrix}4 & 2 & -1 \\3 & 2 & 0 \\\end{bmatrix}\begin{bmatrix}0 & 1 \\2 & 3 \\7 & 4 \\\end{bmatrix}=\begin{bmatrix}-3 & 6 \\4 & 9 \\\end{bmatrix}\:\:\:…(i)\end{array} \)

\(\begin{array}{l}BA=\begin{bmatrix}0 & 1 \\2 & 3 \\7 & 4 \\\end{bmatrix}\begin{bmatrix}4 & 2 & -1 \\3 & 2 & 0 \\\end{bmatrix}=\begin{bmatrix}3 & 2 & 0 \\17 & 10 & -2 \\40 & 22 & -7 \\\end{bmatrix}\:\:\:….(ii)\end{array} \)

From (i) and (ii) AB ≠ BA.

Question 4:

Let X, Y, Z, W and S be matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. Find the values of n, k, and p so that SY + WY is well defined.

Order of S = p × k

Order of Y = 3 × k

If the matrices S and Y are multiplication conformable, then k = 3.

Order of WY = n × k

Order of SY = p × k

If the matrices SY and WY are addition conformable, then p = n.

Question 5:

Verify the associative property for multiplication of matrices:

\(\begin{array}{l}A=\begin{bmatrix}1 & 2 \\-2 & 3 \\\end{bmatrix},B=\begin{bmatrix}2 & 1 \\2 & 3 \\\end{bmatrix}\:and\:C=\begin{bmatrix}-3 & 1 \\2 & 0 \\\end{bmatrix}\end{array} \)

We need to prove that A(BC) = (AB)C.

\(\begin{array}{l}BC=\begin{bmatrix}2 & 1 \\2 & 3 \\\end{bmatrix}\begin{bmatrix}-3 & 1 \\2 & 0 \\\end{bmatrix}=\begin{bmatrix}-4 & 2 \\0 & 2 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}A(BC)=\begin{bmatrix}1 & 2 \\-2 & 3 \\\end{bmatrix}\begin{bmatrix}-4 & 2 \\0 & 2 \\\end{bmatrix}=\begin{bmatrix}-4 & 6 \\8 & 2 \\\end{bmatrix}\:\:\:…(i)\end{array} \)

\(\begin{array}{l}AB=\begin{bmatrix}1 & 2 \\-2 & 3 \\\end{bmatrix}\begin{bmatrix}2 & 1 \\2 & 3 \\\end{bmatrix}=\begin{bmatrix}6 & 7 \\2 & 7 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}(AB)C=\begin{bmatrix}6 & 7 \\2 & 7 \\\end{bmatrix}\begin{bmatrix}-3 & 1 \\2 & 0 \\\end{bmatrix}=\begin{bmatrix}-4 & 6 \\8 & 2 \\\end{bmatrix}\:\:\:….(ii)\end{array} \)

From (i) and (ii) we get that A(BC) = (AB)C.

Question 6:

Show that the product of the two matrices

\(\begin{array}{l}\begin{bmatrix}cos^{2}\theta & cos\:\theta sin\:\theta\\cos\:\theta sin\:\theta & sin^{2}\theta \\\end{bmatrix}\:\:and\:\:\begin{bmatrix}cos^{2}\phi & cos\:\phi sin\:\phi\\cos\:\phi sin\:\phi & sin^{2}\phi \\\end{bmatrix}\end{array} \)

It is a null matrix when the difference between both angles is an odd multiple of 𝜋/2.

The product of both the matrix is given by

\(\begin{array}{l}\begin{bmatrix}cos^{2}\theta & cos\:\theta sin\:\theta\\cos\:\theta sin\:\theta & sin^{2}\theta \\\end{bmatrix}\begin{bmatrix}cos^{2}\phi & cos\:\phi sin\:\phi\\cos\:\phi sin\:\phi & sin^{2}\phi \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}=\begin{bmatrix}cos^{2}\theta cos^{2}\phi +cos \theta \:cos \phi \:sin \theta \:sin \phi & cos^{2}\theta \:cos \phi\: sin \phi+cos \theta \:sin\theta \:sin^{2}\phi \\cos^{2}\phi \:cos \theta\: sin \theta+cos \phi \:sin\phi\:sin^{2}\theta &sin^{2}\theta sin^{2}\phi +cos \theta \:cos \phi \:sin \theta \:sin \phi \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}=\begin{bmatrix}cos\theta\:cos\phi\:cos(\theta-\phi) & cos\theta\:sin\phi\:cos(\theta-\phi) \\sin\theta\:cos\phi\:cos(\theta-\phi) &sin\theta\:sin\phi\:cos(\theta-\phi) \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}=\begin{bmatrix}0 & 0 \\0 & 0 \\\end{bmatrix}\:\:\:\:\left [ as \:\:\theta -\phi = (2n+1)\frac{\pi}{2},n\epsilon \mathbb{Z},then\:\:cos[(2n+1)\frac{\pi}{2}] =0\right ]\end{array} \)

Question 7:

Let the given matrices A, B and C be

\(\begin{array}{l}A=\begin{bmatrix}1 & 0 \\2 & 4 \\\end{bmatrix},B=\begin{bmatrix}2 & 4 \\1 & 0 \\\end{bmatrix}\:\:and\:\:\:C=\begin{bmatrix}-1 & 5 \\1 & -1 \\\end{bmatrix}\end{array} \)

Find a matrix D such that CD – AB = 0

\(\begin{array}{l}Let,\:\:D=\begin{bmatrix}a & b \\c & d \\\end{bmatrix}\end{array} \)

Given, CD – AB = 0 ⇒ CD = AB

\(\begin{array}{l}AB=\begin{bmatrix}1 & 0 \\2 & 4 \\\end{bmatrix}\begin{bmatrix}2 & 4 \\1 & 0 \\\end{bmatrix}=\begin{bmatrix}2 & 4 \\8 & 8 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}CD=\begin{bmatrix}-1 & 5 \\1 & -1 \\\end{bmatrix}\begin{bmatrix}a & b \\c & d \\\end{bmatrix}=\begin{bmatrix}-a+5c & -b+5d \\-a+c & -b+d \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}\begin{bmatrix}-a+5c & -b+5d \\-a+c & -b+d \\\end{bmatrix}=\begin{bmatrix}2 & 4 \\8 & 8 \\\end{bmatrix}\end{array} \)

We get the following equations,

– a + 5c = 2 ….(i)

– a + c = 8 ….(ii)

– b + 5d = 4 ….(iii)

– b + d = 8 ….(iv)

Solving (i) and (ii) we get, a = – 19/2 and c = –3/2

Solving (iii) and (iv) we get, b = –9 and d = –1.

Therefore, we get the matrix

\(\begin{array}{l}D = \begin{bmatrix}\frac{-19}{2} & -9 \\ \frac{-3}{2}&-1 \\\end{bmatrix}\end{array} \)

Question 8:

Show that A 2 – 3I = 2A where

\(\begin{array}{l}A=\begin{bmatrix}1 & 2 \\2 & 1 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}Now,\:\:A^{2}=A.A=\begin{bmatrix}1 & 2 \\2 & 1 \\\end{bmatrix}.\begin{bmatrix}1 & 2 \\2 & 1 \\\end{bmatrix}=\begin{bmatrix}5 & 4 \\4 & 5 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}3I=3\begin{bmatrix}1& 0 \\0 & 1 \\\end{bmatrix}=\begin{bmatrix}3& 0 \\0 & 3 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}A^{2}-3I=\begin{bmatrix}5 & 4 \\4 & 5 \\\end{bmatrix}\begin{bmatrix}3& 0 \\0 & 3 \\\end{bmatrix}=\begin{bmatrix}2 & 4 \\4 & 2 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}Also,\:\:\:\:2A=2\begin{bmatrix}1 & 2 \\2 & 1 \\\end{bmatrix}=\begin{bmatrix}2 & 4 \\4 & 2 \\\end{bmatrix}\end{array} \)

Hence, A 2 – 3I = 2A.

Question 9:

Let f(x) = 2x 2 – 3x, find det [f(A)] if

\(\begin{array}{l}A=\begin{bmatrix}-2 & 1 \\0 & 3 \\\end{bmatrix}\end{array} \)

Given, f(x) = 2x 2 – 3x, then f(A) = 2A 2 – 3A

\(\begin{array}{l}2A^{2}-3A=2\begin{bmatrix}-2 & 1 \\0 & 3 \\\end{bmatrix}\begin{bmatrix}-2 & 1 \\0 & 3 \\\end{bmatrix}-3\begin{bmatrix}-2 & 1 \\0 & 3 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}=2\begin{bmatrix}4 & 1 \\0 & 9 \\\end{bmatrix}-\begin{bmatrix}-6 & 3 \\0 & 9 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}=\begin{bmatrix}8 & 2 \\0 & 18 \\\end{bmatrix}-\begin{bmatrix}-6 & 3 \\0 & 9 \\\end{bmatrix}=\begin{bmatrix}14 & -1 \\0 & 9 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}\Rightarrow f(A)=\begin{bmatrix}14 & -1 \\0 & 9 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l} |f(A)|=\begin{vmatrix}14 & -1 \\0 & 9 \\\end{vmatrix}=14 \times 9 -0 = 126\end{array} \)

Question 10:

Show that f(x).f(y) = f(x + y) where

\(\begin{array}{l}f(\theta)=\begin{bmatrix} cos\:\theta& -sin\:\theta& 0\\sin\:\theta & cos\:\theta & 0\\ 0&0 & 1 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}f(x)=\begin{bmatrix} cos\:x& -sin\:x& 0\\sin\:x& cos\:x & 0\\ 0&0 & 1 \\\end{bmatrix},f(y)=\begin{bmatrix} cos\:y& -sin\:y& 0\\sin\:y& cos\:y & 0\\ 0&0 & 1 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}and\:\:\:f(x+y)=\begin{bmatrix} cos\:(x+y)& -sin\:(x+y)& 0\\sin\:(x+y)& cos\:(x+y) & 0\\ 0&0 & 1 \\\end{bmatrix}\end{array} \)

\(\begin{array}{l}f(x).f(y)=\begin{bmatrix} cos\:x& -sin\:x& 0\\sin\:x& cos\:x & 0\\ 0&0 & 1 \\\end{bmatrix}\begin{bmatrix} cos\:y& -sin\:y& 0\\sin\:y& cos\:y & 0\\ 0&0 & 1 \\\end{bmatrix}=\begin{bmatrix}cos(x+y) & -sin(x+y) & 0 \\ sin(x+y)& cos(x+y) & 0 \\ 0& 0 & 1\\\end{bmatrix}=f(x+y)\end{array} \)

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matrix problem solving questions

Practice Questions on Matrix Multiplication

1. Find the product of the following matrices:

\(\begin{array}{l}A=\begin{bmatrix}1 & -2 & 3 \\3 & 2 & -1 \\\end{bmatrix}\:\:and\:\:B=\begin{bmatrix}2 & 3 \\-1 & 2 \\4 & -5 \\\end{bmatrix}\end{array} \)

2. Let X, Y, Z, W and S are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. Find the order of matrix 7X – 5Z, if n = p.

3. Verify the distributive property for multiplication over addition for the matrices:

4. For the given matrices A and B, verify (A + B) 2 = A 2 + B 2 , where

\(\begin{array}{l}A=\begin{bmatrix}7 & 2 \\0 & 1 \\\end{bmatrix}\:\:\:and \:\:\:B=\begin{bmatrix}2 & 3 \\0 & -1 \\\end{bmatrix}\end{array} \)

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How to Solve Matrices

Last Updated: October 7, 2022 References

This article was co-authored by wikiHow Staff . Our trained team of editors and researchers validate articles for accuracy and comprehensiveness. wikiHow's Content Management Team carefully monitors the work from our editorial staff to ensure that each article is backed by trusted research and meets our high quality standards. There are 10 references cited in this article, which can be found at the bottom of the page. This article has been viewed 135,108 times. Learn more...

A matrix is a very useful way of representing numbers in a block format, [1] X Research source which you can then use to solve a system of linear equations. If you only have two variables, you will probably use a different method. See Solve a System of Two Linear Equations and Solve Systems of Equations for examples of these other methods. But when you have three or more variables, a matrix is ideal. By using repeated combinations of multiplication and addition, you can systematically reach a solution.

Setting up the Matrix for Solving

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Learning the Operations for Solving a System with a Matrix

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Putting the Steps Together to Solve the System

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Verifying Your Solution

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By properly setting up a matrix, you can use them to solve a system of linear equations. Start by writing out your equations and then transfer the numbers from them into a matrix by copying the coefficients and results into a single row. Stack the rows one on top of each other to form a block-looking format. Add a large square bracket around your full matrix and use the abbreviation “R” for the rows and “C” for the columns. This allows you to refer to a specific position in the matrix with a combination of R and C, such as R4C1. To solve the matrix, you can use different operations. For instance, you could use row-addition or row-subtraction, which allows you to add or subtract any two rows of the matrix. To learn about other ways to create a solution matrix, keep reading! Did this summary help you? Yes No

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Matrices is a plural form of a matrix, which is a rectangular array or a table where numbers or elements are arranged in rows and columns. They can have any number of columns and rows. Different operations can be performed on matrices such as addition, scalar multiplication, multiplication, transposition, etc.

There are certain rules to be followed while performing these matrix operations like they can be added or subtracted if only they have the same number of rows and columns whereas they can be multiplied if only columns in first and rows in second are exactly the same. Let us understand the different types of matrices and these rules in detail.

What are Matrices?

Matrices , the plural form of a matrix, are the arrangements of numbers, variables, symbols, or expressions in a rectangular table that contains various numbers of rows and columns. They are rectangular-shaped arrays, for which different operations like addition, multiplication, and transposition are defined. The numbers or entries in the matrix are known as its elements. Horizontal entries of matrices are called rows and vertical entries are known as columns.

Matrix Definition

A matrix is a rectangular array of numbers, variables, symbols, or expressions that are defined for the operations like subtraction, addition, and multiplications. The size of a matrix (which is known as the order of the matrix ) is determined by the number of rows and columns in the matrix. The order of a matrix with 6 rows and 4 columns is represented as a 6 × 4 and is read as 6 by 4. For example, the given matrix B is a 3 × 4 matrix and is written as \([{B}]_{3 \times 4}\):

\(B = \left[\begin{array}{ccc} 2 & -1 & 3 & 5 \\ 0 & 5 & 2 & 7\\ 1 & -1 & -2 & 9 \end{array}\right]\)

Matrix definition

Notation of Matrices

If a matrix has m rows and n columns, then it will have m × n elements . A matrix is represented by the uppercase letter, in this case, 'A', and the elements in the matrix are represented by the lower case letter and two subscripts representing the position of the element in the number of row and column in the same order, in this case, '\(a_{ij}\)', where i is the number of rows, and j is the number of columns. For example, in the given matrix A, element in the 3rd row and 2nd column would be \(a_{32}\), can be verified in the matrix given below:

\(A = \left[\begin{array}{ccc} a_{11} & a_{12} & a_{13} .. .& a_{1n} \\ a_{21} & a_{22} & a_{23} ... & a_{2n} \\ a_{31} & a_{32} & a_{33} ...& a_{3n} \\ : & : & : & : \\ a_{m1} & a_{m2} & a_{m3} ...& a_{mn} \end{array}\right] \)

Calculate Matrices

We can solve matrices by performing operations on them like addition, subtraction, multiplication, and so on. Calculating matrices depends upon the number of rows and columns. For addition and subtraction, the number of rows and columns must be the same whereas, for multiplication, number of columns in the first and the number of rows in the second matrix must be equal. The basic operations that can be performed on matrices are:

Addition of Matrices

Subtraction of matrices, scalar multiplication, multiplication of matrices.

The addition of matrices can only be possible if the number of rows and columns of both the matrices are the same. While adding 2 matrices, we add the corresponding elments. i.e., (A + B) = [a\(_{ij}\)] + [b\(_{ij}\)] = [a\(_{ij}\) + b\(_{ij}\)], where i and j are the number of rows and columns respectively. For example: \(\begin{bmatrix} 2 & {-1}\\ \\ 0 & 5\end{bmatrix} + \begin{bmatrix} 0 & 2 \\ \\ 1 & -2 \end{bmatrix}\\ = \begin{bmatrix} 2+0 & {-1} +2 \\ \\ 0+1 & 5+(-2) \end{bmatrix}\\ = \begin{bmatrix} 2 & 1 \\ \\1 & 3 \end{bmatrix} \)

Matrices subtraction is also possible only if the number of rows and columns of both the matrices are the same. While subtracting 2 matrices, we subtract the corresponding elements. i.e., (A - B) = [a\(_{ij}\)] - [b\(_{ij}\)] = [a\(_{ij}\) - b\(_{ij}\)], where i and j are the row number and column number respectively. For example: \( \begin{bmatrix} 2 & {-1}\\ \\ 0 & 5 \end{bmatrix} -\begin{bmatrix} 0 & 2 \\ \\1 & -2 \end{bmatrix} \\ = \begin{bmatrix} 2-0 & {-1} -2\\ \\ 0-1 & 5-(-2) \end{bmatrix} \\ = \begin{bmatrix} 2 & -3\\ \\ -1 & 7 \end{bmatrix} \)

The product of a matrix A with any number 'c' is obtained by multiplying every entry of the matrix A by c, is called scalar multiplication . i.e., (cA)\(_{ij}\) = c(A\(_{ij}\))

Properties of scalar multiplication in matrices

The different properties of matrices for scalar multiplication of any scalars K and l, with matrices A and B are given as,

Matrices multiplication is defined only if the number of columns in the first matrix and rows in the second matrix are equal. To understand how matrices are multiplied, let us first consider a row vector \(R=\left[ {{r}_{1}}\ {{r}_{2}}...{{r}_{n}} \right]\) and a column vector \(C=\left[ \begin{align} \; \ {{c}_{1}} \;\\ \; \ {{c}_{2}} \; \\ \; \ \ \vdots \; \ \\ \; \ {{c}_{n}} \;\ \\ \end{align} \right]\). Then the product of R and C can be defined as

\(RC=\left[ {{r}_{1}}\ \ {{r}_{2}}\ \ ...\ {{r}_{n}} \right]\ \left[ \begin{align} & \ {{c}_{1}} \\ & \ {{c}_{2}} \\ & \ \ \vdots \ \\ & \ {{c}_{n}}\ \\ \end{align} \right]\ \\ =[{{r}_{1}}{{c}_{1}}+{{r}_{2}}{{c}_{2}}+...+{{r}_{n}}{{c}_{n}}]\). For example,

\(\left[ 1\ \ 3\ \ 2 \right]\ \ \left[ \begin{align} & \ \ 2 \\ & -1 \\ & \ \ 4 \\ \end{align} \right]=[7]\)

Now, we will discuss matrix multiplication. It will soon become evident that to multiply 2 matrices A and B and to find AB , the number of columns in A should equal the number of rows in B .

Let A be of order m × n and B be of order n × p . The matrix AB will be of order m × p and will be obtained by multiplying each row vector of A successively with column vectors in B . Let us understand this using a concrete example:\(A=\left[ \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\\end{matrix} \right]B=\left[ \begin{matrix} {{\alpha }_{1}} & {{\beta }_{1}} \\ {{\alpha }_{1}} & {{\beta }_{2}} \\ {{\alpha }_{3}} & {{\beta }_{3}} \\\end{matrix} \right]\)

To obtain the element \(a_{11}\) of AB, we multiply \(R_1\) of A with \(C_1\) of B :

Multiplication of Matrices

To obtain the element \(a_{12}\) of AB, we multiply \(R_1\) of A with \(C_2\) of B:

Matrix Multiplication

To obtain the element \({{a}_{21}}\) of AB, we multiply \(R_2\) of A with \(C_1\) of B:

calculating matrices

Proceeding this way, we obtain all the elements of AB.

Let us generalize this: if A is or order m × n, and B of order n × p, then to obtain the element \( a_{ij}\) in AB, we multiply \(R_i\) in A with \(C_j\) in B:

solve matrices

Properties of Matrix Multiplication

There are different properties associated with the multiplication of matrices. For any three matrices A, B, and C:

Transpose of Matrix

The transpose of a matrix is done when we replace the rows of a matrix to the columns and columns to the rows. Interchanging of rows and columns is known as the transpose of matrices. In the matrix given below, we have row elements as row-1: 2, -3, -4, and row-2: -1, 7, -7. On transposing, we will get the elements in column-1: 2, -3, -4, and column-2: -1, 7, -7, we can check that in the image given below:

Transpose of a Matrix

Properties of transposition in matrices

There are various properties associated with transposition. For matrices A and B, given as,

Apart from these operations, we have several other operations on matrices like finding its trace, determinant, minors and cofactors , adjoint, inverse, etc. Let us learn each of these in detail in the upcoming sections.

Trace of a Matrix

The trace of any matrix A, Tr(A) is defined as the sum of its diagonal elements. Some properties of trace of matrices are,

Determinant of Matrices

The determinant of a matrix is a number defined only for square matrices. It is used in the analysis of linear equations and their solution. The determinant formula helps calculate the determinant of a matrix using the elements of the matrix. Determinant of a matrix is equal to the summation of the product of the elements of a particular row or column with their respective cofactors. Determinant of a matrix A is denoted as |A|. Let say we want to find the determinant of the matrix \(A = \left[\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right] \)

Then determinant formula of matrix A:

\(a_{11}(-1)^{1 + 1} \!\!\left|\begin{matrix}a_{22}\!\!\!&a_{23}\\a_{32}\!\!\!&a_{33}\end{matrix}\right| \!\!+\!\! a_{12}(-1)^{1 + 2} \!\!\left|\begin{matrix}a_{21}\!\!\!&a_{23}\\a_{31}\!\!\!&a_{33}\end{matrix}\right| \!\!+\!\! a_{13}(-1)^{1 + 3} \!\!\left|\begin{matrix}a_{21}\!\!\!&a_{22}\\a_{31}\!\!\!&a_{32}\end{matrix}\right|\)

Minor of Matrix

Minor for a particular element in the matrices is defined as the determinant of the matrix that is obtained when the row and column of the matrix in which that particular element lies are deleted, and the minor of the element \(a_{ij}\) is denoted as \(M_{ij}\). For example, for the given matrix, minor of \( a_{12}\) of the matrix \(A = \left[\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right] \) is:

\(M_{12} = \left|\begin{array}{ccc} a_{21} & a_{23} \\ \\ a_{31} & a_{33} \end{array}\right|\)

Similarly, we can find all the minors of the matrix and will get a minor matrix M of the given matrix A as:

\(M = \left[\begin{array}{ccc} M_{11} & M_{12} & M_{13} \\ M_{21} & M_{22} & M_{23} \\ M_{31} & M_{32} & M_{33} \end{array}\right]\)

Cofactor of Matrix

Cofactor of an element in the matrix A is obtained when the minor \(M_{ij}\) of the matrix is multiplied with (-1) i+j . The cofactor of a matrix is denoted as \(C_{ij}\). If the minor of a matrix is \(M_{ij}\), then the cofactor of the matrix would be:

\(C_{ij} = (-1)^{i+j} M_{ij}\)

On finding all the cofactors of the matrix, we will get a cofactor matrix C of the given matrix A:

\(C = \left[\begin{array}{ccc} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{array}\right] \)

Note: Be extra cautious about the negative sign while calculating the cofactor of the matrix.

Adjoint of Matrices

The adjoint of matrices is calculated by finding the transpose of the cofactors of the elements of the given matrices. To find the adjoint of a matrix, we have to calculate the cofactors of the elements of the matrix and then transpose the cofactor matrix to get the adjoint of the given matrix. The adjoint of matrix A is denoted by adj(A). Let us understand this with an example: We have a matrix \(A = \left[\begin{array}{ccc} 2 & -1 & 3 \\ 0 & 5 & 2 \\ 1 & -1 & -2 \end{array}\right] \)

Then the minor matrix M of the given matrix would be:

\(M = \left[\begin{array}{ccc} -8 & -2 & -5 \\ 5 & -7 & -1 \\ -17 & 4 & 10 \end{array}\right] \)

We will get the cofactor matrix C of the given matrix A as:

\(C = \left[\begin{array}{ccc} -8 & 2 & -5 \\ -5 & -7 & 1 \\ -17 & -4 & 10 \end{array}\right] \)

Then the transpose of the cofactor matrix will give the adjoint of the given matrix:

adj(A) = C T = \(\left[\begin{array}{ccc} -8 & -5 & -17 \\ 2 & -7 & -4 \\ -5 & 1 & 10 \end{array}\right] \)

Inverse of Matrices

The inverse of any matrix is denoted as the matrix raised to the power (-1), i.e. for any matrix "A", the inverse matrix is denoted as A -1 . The inverse of a square matrix, A is A -1 only when: A × A -1 = A -1 × A = I . There is a possibility that sometimes the inverse of a matrix does not exist if the determinant of the matrix is equal to zero(|A| = 0). The inverse of a matrix is shown by A -1 . Matrices inverse is calculated by using the following formula:

A -1 = (1/|A|)(Adj A)

The inverse of a 2 × 2 matrix \(A = \left[\begin{array}{ccc} a_{11} & a_{12} \\ \\ a_{21} & a_{22} \end{array}\right] \) is calculated by: A -1 = \(\dfrac{1}{a_{11}a_{22} - a_{12}a_{21}}\left(\begin{matrix}a_{22}&-a_{12}\\ \\-a_{21}&a_{11}\end{matrix}\right)\)

Let us find the inverse of the 3 × 3 matrix we have used in the previous section: \(A = \left[\begin{array}{ccc} 2 & -1 & 3 \\ 0 & 5 & 2 \\ 1 & -1 & -2 \end{array}\right] \)

Since adj(A) = \(\left[\begin{array}{ccc} -8 & -5 & -17 \\ 2 & -7 & -4 \\ -5 & 1 & 10 \end{array}\right] \)

And on calculating the determinant , we will get |A| = -33

Therefore, A -1 = (1/-33) × \(\left[\begin{array}{ccc} -8 & -5 & -17 \\ 2 & -7 & -4 \\ -5 & 1 & 10 \end{array}\right] \)

Hence, A -1 = \(\left[\begin{array}{ccc} 0.24 & 0.15 & 0.51 \\ -0.06 & 0.21 & 0.12 \\ 0.15 & -0.03 & -0.39 \end{array}\right] \)

Types of Matrices

There are various types of matrices based on the number of elements and the arrangement of elements in them.

Row matrix: A row matrix is a matrix having a single row is called a row matrix. Example: [1, −2, 4].

Column matrix: A column matrix is a matrix having a single column is called a column matrix. Example: [−1, 2, 5] T .

Square matrix: A matrix having equal number of rows and columns is called a square matrix . For example: \(B= \left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 &6& 1 \end{array}\right] \)

Rectangular Matrix: A matrix having unequal number of rows and columns is called a rectangular matrix . For example: \(B= \left[\begin{array}{ccc} 1 & 2 & 3 \\ \\ 0 & 1 & 4 \end{array}\right] \)

Diagonal matrices: A matrix with all non-diagonal elements to be zeros is known as a diagonal matrix . Example: \(A = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 &0 & 3 \end{array}\right] \)

Identity matrices: A diagonal matrix having all the diagonal elements equal to 1 is called an identity matrix . Example: \(B= \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 &0 & 1 \end{array}\right] \)

Symmetric and skew-symmetric matrices: Symmetric matrices: A square matrix D of size n×n is considered to be symmetric if and only if D T = D. For example, D = \(\left[\begin{array}{lll} 2 & 3 & 6 \\ 3 & 4 & 5 \\ 6 & 5 & 9 \end{array}\right] \) is a symmetric matrix because

D T = \(\left[\begin{array}{lll} 2 & 3 & 6 \\ 3 & 4 & 5 \\ 6 & 5 & 9 \end{array}\right]\) = D

Skew-symmetric matrices-A square matrix F of size n×n is considered to be skew-symmetric if and only if F T = - F.

\(F = \left[\begin{array}{ccc} 0 & 3 \\ \\ -3 & 0 \end{array}\right]\) is a skew-symmetric matrix because

Invertible Matrix: Any square matrix A is called invertible matrix , if there exists another matrix B, such that, AB = BA = \(I_n\), where \(I_n\) is an identity matrix with n × n.

Orthogonal Matrix: Any square matrix A is orthogonal if its transpose is equal to its inverse. i.e., A T = A -1

Solving a System of Equations Using Matrices

While solving the system of equations using matrices, we have three matrices A, B, and X where A is known as the coefficient matrix, B is known as the constant matrix, and X contains all the variables of the equations which is known as a variable matrix. Matrix A is of the order m × n, while B is the column matrix of the order m × 1. The product of matrix A and matrix X results in matrix B; hence, X is a column matrix as well of the order n × 1.

The matrices are arranged as:

A • X = B

Let's understand how to solve a system of equations using matrices with the help of an example. We have a set of two equations as given below. The equations are:

x + y = 8 2x + 3y = 10

Arrange all the coefficients, variables, and constants in the matrix in such a way that whenever we find the product of the matrices, the result obtained must result in the equation. Then the matrix equation is, AX = B where:

\(A = \begin{bmatrix} 1 & 1\\ \\ 2 & 3\\ \end{bmatrix}\)

\(X = \begin{bmatrix} x\\ \\ y\\ \end{bmatrix}\)

\(B = \begin{bmatrix} 8\\ \\ 10\\ \end{bmatrix}\)

To solve the equations, we need to find matrix X. It can be found by multiplying the inverse of matrix A with B, which is given as \( X = (A^{-1})B\). To find the determinant of matrix A, we will follow the below steps:

\( |A| = \begin{vmatrix} 1 & 1\\ \\ 2 & 3\\ \end{vmatrix}\)

Hence, |A| = 3 - 2 = 1 \(\because\) \(|A| \neq 0\), it is possible to find the inverse of matrix A.

Now, by using the formula for finding the inverse of 2x2 matrix (which is mentioned in previous sections),

\(A^{-1} = \begin{bmatrix} 3 & -1\\ \\ -2 & 1\\ \end{bmatrix}\)

Now to find the matrix X, we'll multiply \(A^{-1}\) and B. We get,

\(\begin{bmatrix} 3 & -1 \\ \\ -2 & 1 \end{bmatrix} % \begin{bmatrix} 8 \\ \\ 10 \end{bmatrix} \ = \begin{bmatrix} 14 \\ \\ -6 \end{bmatrix} \)

Hence, the value of matrix X is,

\(X = \begin{bmatrix} 14\\ \\ -6\\ \end{bmatrix}\)

Rank of a Matrix

The rank of a matrix A is defined as the maximum number of linearly independent row(or column) vectors of the matrix. That means the rank of a matrix will always be less than or equal to the number of its rows or columns. The rank of a null matrix is zero since it has no independent row or column vectors.

Eigen Values and Eigen Vectors of Matrices

If A is any square matrix of order 'n', a matrix of A - λI can be formed, where I is a unit matrix of order n, such that the number λ, called the eigenvalue and a non-zero vector v, called the eigenvector , satisfy the equation, Av = λv. λ is an eigenvalue of an n×n-matrix A if and only if A − λI n is not invertible, which is equivalent to Det(A - λI) = 0.

Matrices Formulas

There are different formulas associated with matrix operations depending upon the type of matrix. Some of the matrices formulas are listed below:

Important Notes on Matrices:

☛ Related Topics:

Solved Examples on Matrices

Example 1: Let \(A=\left[ \begin{matrix} 1 & 2\\ \\ 3 & 1 \\\end{matrix} \right],\ B=\left[ \begin{matrix} 1 & 4\\ \\ 3 & -1 \\\end{matrix} \right]\). Calculate A + B.

Solution: Here, matrix A = \(\left[ \begin{matrix} 1 & 2\\ \\ 3 & 1 \\\end{matrix} \right]\) matrix B = \(\left[ \begin{matrix} 1 & 4\\ \\ 3 & -1 \\\end{matrix} \right]\)

Using addition of matrices property, A + B = \(\left[ \begin{matrix} 1 & 2 \\ \\ 3 & 1 \\\end{matrix} \right]\) + \(\left[ \begin{matrix} 1 & 4\\ \\ 3 & -1 \\\end{matrix} \right]\) = \(\left[ \begin{matrix} 2 & 6\\ \\ 6 & 0 \end{matrix} \right]\)

Answer: Sum of matrices A and B, A + B = \(\left[ \begin{matrix} 2 & 6\\ \\ 6 & 0 \end{matrix} \right]\)

Example 2: Find the inverse of a matrix A =\(\left[\begin{matrix}1 & -2\\ \\2 & -3 \end{matrix}\right]\).

The given matrix is A = \(\left[\begin{matrix}1 & -2\\ \\2 & -3 \end{matrix}\right]\).

Using the formula of matrix inverse: A -1 = \(\dfrac{1}{a_{11}a_{22} - a_{12}a_{21}}\left[\begin{matrix}a_{22}&-a_{12}\\ \\-a_{21}&a_{11}\end{matrix}\right]\)

Using the inverse of matrix formula we can calculate A -1 as follows.

A -1 = \(\dfrac{1}{(1× -3) - (-2 × 2)}\left[\begin{matrix}-3&2\\ \\-2&1\end{matrix}\right]\)

= \(\dfrac{1}{-3 +4}\left[\begin{matrix}-3&2\\ \\-2&1\end{matrix}\right]\)

= \(\left[\begin{matrix}-3&2\\ \\-2&1\end{matrix}\right]\)

Answer: Therefore A -1 = \(\left[\begin{matrix}-3&2\\ \\-2&1\end{matrix}\right]\).

Example 3: Prove that the product of the matrices A = \(\left[\begin{array}{rr}1 & 2 & -1\\ 3 & 2 & 0\\ -4 & 0 & 2\end{array}\right]\) and the identity matrix of order 3x3 is the matrix itself.

The identity matrix of order 3x3 is, I = \(\left[\begin{array}{rr}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right]\).

Now, AI = \(\left[\begin{array}{rr}1 & 2 & -1\\ 3 & 2 & 0\\ -4 & 0 & 2\end{array}\right]\) \(\left[\begin{array}{rr}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right]\)

= \(\left[\begin{array}{rr}1+0+0 & 0+2+0 & 0+0-1 \\ 3+0+0 & 0+2+0 & 0+0+0 \\ -4+0+0& 0+0+0& 0+0+2\end{array}\right]\)

= \(\left[\begin{array}{rr}1 & 2 & -1\\ 3 & 2 & 0\\ -4 & 0 & 2\end{array}\right]\)

Answer: We have proved that AI = A.

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Practice Questions on Matrices

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FAQs on Matrices

What is the meaning of matrix in math.

A matrix in is an arrangement of numbers, variables, symbols, or expressions in the rectangular table which contains various numbers of rows and columns, for which the operations like addition, multiplication, transposition, etc are defined.

How to Solve Matrices?

We can solve matrices by performing matrix operations on them like addition, subtraction, multiplication, and so on. We have to take care of the orders while solving matrices.

How to Solve Systems of Equations with Matrices?

To solve the system of equations with matrices, we will follow the steps given below.

What is 3×3 Inverse Matrix Formula?

The inverse matrix formula for a 3×3 matrix is, A -1 = adj(A)/|A|; |A| ≠ 0 where A = square matrix, adj(A) = adjoint of square matrix, and A -1 = inverse matrix of A.

What is the Special Feature Of the Determinant Formula For Matrices?

The determinant of a matrix is defined only for square matrices, and this property of the determinant formula makes it unique. Also, the determinant value can be calculated by using the elements of any row or any column.

How To Calculate the Determinant of a 2×2 Matrix Using Determinant Formula?

The determinant formula for 2x2 matrix, \(A =\begin{pmatrix}a &b\\ \\c&d\end{pmatrix}\) is given by the formula |A| = ad - bc.

What is the Condition for Matrix Multiplication to be Possible?

Matrix multiplication is possible only if the matrices are compatible i.e., matrix multiplication is valid only if the number of columns of the first matrix is equal to the number of rows of the second matrix.

What Are Properties of Transposition of Matrices?

For given 2 matrices, A and B, the properties of transposition of matrices can be explained as given below,

What is the Formula for Inverse of Matrices?

The inverse matrix formula is used to determine the inverse matrix for any given matrix. The inverse of a square matrix, A is A -1 . The inverse matrix formula can be given as, A -1 = adj(A)/|A|; |A| ≠ 0, where A is a square matrix. Also for a matrix and its inverse we have A × A -1 = A -1 × A = I.

How To Use Inverse of Matrix Formula?

The inverse matrix formula can be used following the given steps:

What are the Different Types of a Matrix?

There are different types of matrices depending upon the properties of their properties. Some of them are given as,

What are the Properties of Scalar Multiplication in Matrices?

Given the matrices A and B (both of the same order) and scalars K and l, the different properties associated with the multiplication of matrices can be given as,

What is a Matrix Polynomial?

Given a polynomial of the form, f(x) = a 0 x n + a 1 x n-1 + a 2 x n-1 + . . . + a n-1 x + a n , and A as a square matrix of order n. Then, f(A) = a 0 A n + a 1 A n-1 + a 2 A n-2 + . . . + a n-1 A + a n A + a n is called the matrix polynomial.

What is the Echelon Form of Matrices?

A matrix A = (a\(_{ij}\)\(_{m\times n}\) is said to be of echelon form if it is in either upper triangular or lower triangular form. To convert a matrix into the echelon form, we apply elementary row operations .

How to Express a Matrix as a Sum of Symmetric and Non-Symmetric Matrix?

Any square matrix A can be written as, A = P + Q, where P and Q are symmetric and skew-symmetric matrices respectively, such that, P = (A + A T )/2 and Q = (A - A T )/2.

Top 20 Problem Solving Interview Questions (Example Answers Included)

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matrix problem solving questions

By Mike Simpson

When candidates prepare for interviews, they usually focus on highlighting their leadership, communication, teamwork, and similar crucial soft skills . However, not everyone gets ready for problem-solving interview questions. And that can be a big mistake.

Problem-solving is relevant to nearly any job on the planet. Yes, it’s more prevalent in certain industries, but it’s helpful almost everywhere.

Regardless of the role you want to land, you may be asked to provide problem-solving examples or describe how you would deal with specific situations. That’s why being ready to showcase your problem-solving skills is so vital.

If you aren’t sure who to tackle problem-solving questions, don’t worry, we have your back. Come with us as we explore this exciting part of the interview process, as well as some problem-solving interview questions and example answers.

What Is Problem-Solving?

When you’re trying to land a position, there’s a good chance you’ll face some problem-solving interview questions. But what exactly is problem-solving? And why is it so important to hiring managers?

Well, the good folks at Merriam-Webster define problem-solving as “the process or act of finding a solution to a problem.” While that may seem like common sense, there’s a critical part to that definition that should catch your eye.

What part is that? The word “process.”

In the end, problem-solving is an activity. It’s your ability to take appropriate steps to find answers, determine how to proceed, or otherwise overcome the challenge.

Being great at it usually means having a range of helpful problem-solving skills and traits. Research, diligence, patience, attention-to-detail , collaboration… they can all play a role. So can analytical thinking , creativity, and open-mindedness.

But why do hiring managers worry about your problem-solving skills? Well, mainly, because every job comes with its fair share of problems.

While problem-solving is relevant to scientific, technical, legal, medical, and a whole slew of other careers. It helps you overcome challenges and deal with the unexpected. It plays a role in troubleshooting and innovation. That’s why it matters to hiring managers.

How to Answer Problem-Solving Interview Questions

Okay, before we get to our examples, let’s take a quick second to talk about strategy. Knowing how to answer problem-solving interview questions is crucial. Why? Because the hiring manager might ask you something that you don’t anticipate.

Problem-solving interview questions are all about seeing how you think. As a result, they can be a bit… unconventional.

These aren’t your run-of-the-mill job interview questions . Instead, they are tricky behavioral interview questions . After all, the goal is to find out how you approach problem-solving, so most are going to feature scenarios, brainteasers, or something similar.

So, having a great strategy means knowing how to deal with behavioral questions. Luckily, there are a couple of tools that can help.

First, when it comes to the classic approach to behavioral interview questions, look no further than the STAR Method . With the STAR method, you learn how to turn your answers into captivating stories. This makes your responses tons more engaging, ensuring you keep the hiring manager’s attention from beginning to end.

Now, should you stop with the STAR Method? Of course not. If you want to take your answers to the next level, spend some time with the Tailoring Method , too.

With the Tailoring Method, it’s all about relevance. So, if you get a chance to choose an example that demonstrates your problem-solving skills, this is really the way to go.

We also wanted to let you know that we created an amazing free cheat sheet that will give you word-for-word answers for some of the toughest interview questions you are going to face in your upcoming interview. After all, hiring managers will often ask you more generalized interview questions!

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Top 3 Problem-Solving-Based Interview Questions

Alright, here is what you’ve been waiting for: the problem-solving questions and sample answers.

While many questions in this category are job-specific, these tend to apply to nearly any job. That means there’s a good chance you’ll come across them at some point in your career, making them a great starting point when you’re practicing for an interview.

So, let’s dive in, shall we? Here’s a look at the top three problem-solving interview questions and example responses.

1. Can you tell me about a time when you had to solve a challenging problem?

In the land of problem-solving questions, this one might be your best-case scenario. It lets you choose your own problem-solving examples to highlight, putting you in complete control.

When you choose an example, go with one that is relevant to what you’ll face in the role. The closer the match, the better the answer is in the eyes of the hiring manager.

EXAMPLE ANSWER:

“While working as a mobile telecom support specialist for a large organization, we had to transition our MDM service from one vendor to another within 45 days. This personally physically handling 500 devices within the agency. Devices had to be gathered from the headquarters and satellite offices, which were located all across the state, something that was challenging even without the tight deadline. I approached the situation by identifying the location assignment of all personnel within the organization, enabling me to estimate transit times for receiving the devices. Next, I timed out how many devices I could personally update in a day. Together, this allowed me to create a general timeline. After that, I coordinated with each location, both expressing the urgency of adhering to deadlines and scheduling bulk shipping options. While there were occasional bouts of resistance, I worked with location leaders to calm concerns and facilitate action. While performing all of the updates was daunting, my approach to organizing the event made it a success. Ultimately, the entire transition was finished five days before the deadline, exceeding the expectations of many.”

2. Describe a time where you made a mistake. What did you do to fix it?

While this might not look like it’s based on problem-solving on the surface, it actually is. When you make a mistake, it creates a challenge, one you have to work your way through. At a minimum, it’s an opportunity to highlight problem-solving skills, even if you don’t address the topic directly.

When you choose an example, you want to go with a situation where the end was positive. However, the issue still has to be significant, causing something negative to happen in the moment that you, ideally, overcame.

“When I first began in a supervisory role, I had trouble setting down my individual contributor hat. I tried to keep up with my past duties while also taking on the responsibilities of my new role. As a result, I began rushing and introduced an error into the code of the software my team was updating. The error led to a memory leak. We became aware of the issue when the performance was hindered, though we didn’t immediately know the cause. I dove back into the code, reviewing recent changes, and, ultimately, determined the issue was a mistake on my end. When I made that discovery, I took several steps. First, I let my team know that the error was mine and let them know its nature. Second, I worked with my team to correct the issue, resolving the memory leak. Finally, I took this as a lesson about delegation. I began assigning work to my team more effectively, a move that allowed me to excel as a manager and help them thrive as contributors. It was a crucial learning moment, one that I have valued every day since.”

3. If you identify a potential risk in a project, what steps do you take to prevent it?

Yes, this is also a problem-solving question. The difference is, with this one, it’s not about fixing an issue; it’s about stopping it from happening. Still, you use problem-solving skills along the way, so it falls in this question category.

If you can, use an example of a moment when you mitigated risk in the past. If you haven’t had that opportunity, approach it theoretically, discussing the steps you would take to prevent an issue from developing.

“If I identify a potential risk in a project, my first step is to assess the various factors that could lead to a poor outcome. Prevention requires analysis. Ensuring I fully understand what can trigger the undesired event creates the right foundation, allowing me to figure out how to reduce the likelihood of those events occurring. Once I have the right level of understanding, I come up with a mitigation plan. Exactly what this includes varies depending on the nature of the issue, though it usually involves various steps and checks designed to monitor the project as it progresses to spot paths that may make the problem more likely to happen. I find this approach effective as it combines knowledge and ongoing vigilance. That way, if the project begins to head into risky territory, I can correct its trajectory.”

17 More Problem-Solving-Based Interview Questions

In the world of problem-solving questions, some apply to a wide range of jobs, while others are more niche. For example, customer service reps and IT helpdesk professionals both encounter challenges, but not usually the same kind.

As a result, some of the questions in this list may be more relevant to certain careers than others. However, they all give you insights into what this kind of question looks like, making them worth reviewing.

Here are 17 more problem-solving interview questions you might face off against during your job search:

Putting It All Together

At this point, you should have a solid idea of how to approach problem-solving interview questions. Use the tips above to your advantage. That way, you can thrive during your next interview.

FREE : Job Interview Questions & Answers PDF Cheat Sheet!

Download our " Job Interview Questions & Answers PDF Cheat Sheet " that gives you word-for-word sample answers to some of the most common interview questions including:

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matrix problem solving questions

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Decision Matrix Resources

Decision Matrix Related Topics

What is a Decision Matrix?

Quality Glossary Definition: Decision matrix

Also called: Pugh matrix, decision grid, selection matrix or grid, problem matrix, problem selection matrix, opportunity analysis, solution matrix, criteria rating form, criteria-based matrix

A decision matrix evaluates and prioritizes a list of options and is a decision-making tool . The team first establishes a list of weighted criteria and then evaluates each option against those criteria. This is a variation of the L-shaped matrix.

When to Use a Decision Matrix

Typical situations are:

Decision Matrix Procedure

Method 1: Establish a rating scale for each criterion. Some options are:

Method 2: For each criterion, rank-order all options according to how well each meets the criterion. Number them with 1 being the option that is least desirable according to that criterion.

Method 3 (Pugh matrix): Establish a baseline, which may be one of the alternatives or the current product or service. For each criterion, rate each other alternative in comparison to the baseline, using scores of worse (-1), same (0), or better (+1). Finer rating scales can be used, such as 2, 1, 0, -1, -2 for a five-point scale or 3, 2, 1, 0, -1, -2, -3 for a seven-point scale. Again, be sure that positive numbers reflect desirable ratings.

Decision Matrix Example

Figure 1 shows a decision matrix used by the customer service team at the Parisian Experience restaurant to decide which aspect of the overall problem of "long wait time" to tackle first. The problems they identified are customers waiting for the host, the waiter, the food, and the check.

The criteria they identified are "Customer pain" (how much does this negatively affect the customer?), "Ease to solve," "Effect on other systems," and "Speed to solve." Originally, the criteria "Ease to solve" was written as "Difficulty to solve," but that wording reversed the rating scale. With the current wording, a high rating on each criterion defines a state that would encourage selecting the problem: high customer pain, very easy to solve, high effect on other systems, and quick solution.

Decision Matrix Example

Figure 1: Decision Matrix Example

"Customer pain" has been weighted with 5 points, showing that the team considers it by far the most important criterion, compared to 1 or 2 points for the others.

The team chose a rating scale of high = 3, medium = 2, and low = 1 and used it for the problem. "Customers wait for food." In this example, the customer pain is medium (2), because the restaurant ambiance is nice. This problem would not be easy to solve (low ease = 1), as it involves both waiters and kitchen staff. The effect on other systems is medium (2), because waiters have to make several trips to the kitchen. The problem will take a while to solve (low speed = 1), as the kitchen is cramped and inflexible.

Each rating is multiplied by the weight for that criterion. For example, "Customer pain" (weight of 5) for "Customers wait for host" rates high (3) for a score of 15. The scores are added across the rows to obtain a total for each problem. "Customers wait for host" has the highest score at 28. Since the next highest score is 18, the host problem probably should be addressed first.

Decision Matrix Considerations

Additional considerations

Adapted from The Quality Toolbox, Second Edition, ASQ Quality Press.

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