Matrices

Download CBSE Class 12th Standard CBSE Maths question papers, sample papers, important questions, and previous year solved papers in PDF format. Get free study materials, NCERT solutions, and exam preparation resources for Class 12th Standard CBSE Maths

Questions:

Generic 2

13 x 2 = 26

Write the order of matrix B if any matrix of order m x n such that AB and BA both are defined

It is possible to have the product of two matrices to be the null matrix.
While neither of them is the null matrix? if it is so, give an example

if \(\left[ \begin{matrix} a+b & 2 \\ 5 & ab \end{matrix} \right] =\left[ \begin{matrix} 6 & 2 \\ 5 & 8 \end{matrix} \right] \)find the relation between a and b

Let A = [a_ij]be a matric of order 2 x 3 and a_ij = \(\frac { i-j }{ i+j } \), write the value of a₂₃

If A and B are invertible matrices of the same order, then (AB)^–1 = B^–1 A^–1.

Find the value of X and Y if
\(X+Y=\left[ \begin{matrix} 2 & 3 \\ 5 & 1 \end{matrix} \right] ,X-Y=\left[ \begin{matrix} 6 & 5 \\ 7 & 3 \end{matrix} \right] \)

Solve the matrix equation \(\left[ \begin{matrix} { x }^{ 2 } \\ { y }^{ 2 } \end{matrix} \right] -3\left[ \begin{matrix} x \\ 2y \end{matrix} \right] =\left[ \begin{matrix} -2 \\ -9 \end{matrix} \right] \)

If \(A=\left[ \begin{matrix} 1 & 4 \\ 3 & 2 \\ 2 & 1 \end{matrix} \right] B=\left[ \begin{matrix} 5 & 2 \\ -1 & 0 \\ 1 & 1 \end{matrix} \right] \), then find the matrix X for which A + B - X = 0.

If \(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ 2 & 5 \end{bmatrix}=\begin{bmatrix} 7 & 11 \\ k & 23 \end{bmatrix}\), find the value of k.

10.

If \(A=\left[ { a }_{ ij } \right] =\left[ \begin{matrix} 2 & 3 & -5 \\ 1 & 4 & 9 \\ 0 & 7 & -2 \end{matrix} \right] andb=\left[ { b }_{ ij } \right] =\left[ \begin{matrix} 2 & 1 & -1 \\ -3 & 4 & 4 \\ 1 & 5 & 2 \end{matrix} \right] ,then\quad find\quad { a }_{ 22 }+{ b }_{ 21 }.\)

11.

If \(\left[ \begin{matrix} y & +2x & 5 \\ & -x & 3 \end{matrix} \right] =\begin{bmatrix} 7 & 5 \\ -2 & 3 \end{bmatrix}\), find the value of y.

12.

If matrix A = \([\begin{matrix} 1 & 2 & 3 \end{matrix}]\) write AA' , where A' is the transpose of matrix A.

13.

If \(\left[ \begin{matrix} x & +3y & y \\ 7 & -x & 4 \end{matrix} \right] \)=\(\begin{bmatrix} 4 & -1 \\ 0 & 4 \end{bmatrix}\), find the values of x and y.

Generic 5

8 x 5 = 40

14.

If \(A=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right] \) Prove that , A =\(\left[ \begin{matrix} { 3 }^{ x-1 } & { 3 }^{ x-1 } & { 3 }^{ x-1 } \\ { 3 }^{ x-1 } & { 3 }^{ x-1 } & { 3 }^{ x-1 } \\ { 3 }^{ x-1 } & { 3 }^{ x-1 } & { 3 }^{ x-1 } \end{matrix} \right] \) for every positive integer n.

15.

Let f(x) = x² - 5x+6 find f(A) If, A =\(\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{matrix} \right] \)

16.

Find the values of x and y for which the following matrix equation A-3B = C is satisfied, where \(A=\left[ \begin{matrix} { x }^{ 2 } \\ { y }^{ 2 } \end{matrix} \right] \), \(B=\left[ \begin{matrix} x \\ 2y \end{matrix} \right] \),\(c=\left[ \begin{matrix} -2 \\ 9 \end{matrix} \right] \)

17.

Find matrix A such that 2A - 3B + 5C = 0 Where \(B=\left[ \begin{matrix} -2 & 2 & 0 \\ 3 & 1 & 4 \end{matrix} \right] \)\(C=\left[ \begin{matrix} 2 & 0 & -2 \\ 7 & 1 & 6 \end{matrix} \right] \)

18.

If \(A=\left( \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right) \), then show that A³ - 4A² - 3A + 11I = 0

19.

A shopkeeper has 3 varieties of pen 'A' , 'B' and 'C'. Meenu purchase 1 pen of each variety for a total of Rs. 21. Jeevan purchase 4 pens of 'A' variety, 3 pen say 'B' variety and 2 pens of 'C' variety for Rs. 60. While Shikha purchased 6 pens of 'A' variety, 2 pens of 'B' variety and 3 pens of 'C' variety for Rs. 70. Using matrix method, find cost of each variety of pen.

20.

Obtain the inverse of the following matrix, using elementary operations : \(A=\left[ \begin{matrix} 3 & 0 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{matrix} \right] \)

21.

If \(A=\left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right] \), verify that \({ A }^{ 2 }-4A-5l=0\)

Generic 3

4 x 3 = 12

22.

Show that: \(\left[ \left( \begin{matrix} 1 & \omega & { \omega }^{ 2 } \\ \omega & { \omega }^{ 2 } & 1 \\ { \omega }^{ 2 } & 1 & \omega \end{matrix} \right) +\left( \begin{matrix} \omega & { \omega }^{ 2 } & 1 \\ { \omega }^{ 2 } & 1 & \omega \\ \omega & { \omega }^{ 2 } & 1 \end{matrix} \right) \right] \left[ \begin{matrix} 1 \\ \omega \\ { \omega }^{ 2 } \end{matrix} \right] =\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] ,\) where \(\omega \) is a cube root of unity.

23.

If \(A=\left[ \begin{matrix} 3 & 1 & -1 \\ 0 & 1 & 2 \end{matrix} \right] \) , then show that \(AA\prime \) is a symmetric matrix.

24.

If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

25.

If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?

Visit QB365-SLMS - https://www.qb365.in/slms/

Answers Key:

Generic 2

13 x 2 = 26

n x m

Under what conditions is the matrix equation
A² -B² = (A-B)(A+B)is true
\(A=\left[ \begin{matrix} 0 & 2 \\ 0 & 0 \end{matrix} \right] B=\left[ \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right] AB=\left[ \begin{matrix} 0 & 2 \\ 0 & 0 \end{matrix} \right] \)

a = 2b {a=4, b=2}

-1/5

From the definition of inverse of a matrix, we have
(AB) (AB)^–1 = 1
or A^–1(AB) (AB)^–1 = A^–1I (Pre multiplying both sides by A^–1)
or (A^–1A) B (AB)^–1 = A–1 (Since A^–1 I = A^–1)
or IB (AB)^–1 = A^–1
or B (AB)^–1 = A^–1
or B^–1 B (AB)^–1 = B^–1 A^–1
or I (AB)^–1 = B^–1 A^–1
Hence (AB)^–1 = B^–1 A^–1

We have, \(X+Y=\left[ \begin{matrix} 2 & 3 \\ 5 & 1 \end{matrix} \right] ,X-Y=\left[ \begin{matrix} 6 & 5 \\ 7 & 3 \end{matrix} \right] \)
\(\left( X+Y \right) +\left( X-Y \right) =\left[ \begin{matrix} 2 & 3 \\ 5 & 1 \end{matrix} \right] +\left[ \begin{matrix} 6 & 5 \\ 7 & 3 \end{matrix} \right] \)
\(2X=\left[ \begin{matrix} 8 & 8 \\ 12 & 4 \end{matrix} \right] \)
\(\Rightarrow X=\left[ \begin{matrix} 4 & 4 \\ 6 & 2 \end{matrix} \right] \)
\(\therefore \ X=\left[ \begin{matrix} 4 & 4 \\ 6 & 2 \end{matrix} \right] \)
and \(Y=\left[ \begin{matrix} 2 & 3 \\ 5 & 1 \end{matrix} \right] -\left[ \begin{matrix} 4 & 4 \\ 6 & 2 \end{matrix} \right] \)
\(=\left[ \begin{matrix} -2 & -1 \\ -1 & -1 \end{matrix} \right] \)

We have, \(\left[ \begin{matrix} { x }^{ 2 } \\ { y }^{ 2 } \end{matrix} \right] -3\left[ \begin{matrix} x \\ 2y \end{matrix} \right] =\left[ \begin{matrix} -2 \\ -9 \end{matrix} \right] \)
\(\Rightarrow\) x² - 3x = - 2 and y² - 6y = - 9
\(\Rightarrow\) x² - 3x + 2 = 0 and y² - 6y + 9 = 0
\(\Rightarrow\) x² - 2x - x + 2 = 0 and y² - 3y - 3y + 9 = 0
\(\Rightarrow\) x(x - 2) - 1(x - 2) = 0 and y(y - 3) - 3(y - 3) = 0
\(\Rightarrow\) (x - 2)(x - 1) = 0 and (y - 3)(y - 3) = 0
\(\therefore\) x = 1, 2 and y = 3, 3

We have A + B - X = 0
By adding X on both the sides,
A + B - X + X = 0 + X
\(\Rightarrow\) A + B = X
\(\Rightarrow X=\left[ \begin{matrix} 1 & 4 \\ 3 & 2 \\ 2 & 1 \end{matrix} \right] +\left[ \begin{matrix} 5 & 2 \\ -1 & 0 \\ 1 & 1 \end{matrix} \right] \)
\(\Rightarrow X=\left[ \begin{matrix} 6 & 6 \\ 2 & 2 \\ 3 & 2 \end{matrix} \right] \)

\(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ 2 & 5 \end{bmatrix}=\left[ \begin{matrix} 3\quad + & 4 \\ 9\quad + & 8 \end{matrix}\quad \begin{matrix} 1\quad + & 10 \\ 3\quad + & 20 \end{matrix} \right] =\begin{bmatrix} 7 & 11 \\ k & 23 \end{bmatrix} \)
\(\Rightarrow k=9+8=17\)

10.

\({ a }_{ 22 }+{ b }_{ 21 }=4 - 3 = 1\)

11.

\(y+2x=7,-x=-2\Rightarrow x=2,y=3\)

12.

\(AA'= \left[ \begin{matrix} 1 & 2 & 3 \end{matrix} \right] \left[ \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right] =\left[ 1+4+9 \right] =\left[ 14 \right]\)

13.

\(\left[ \begin{matrix} x & +3y & y \\ 7 & -x & 4 \end{matrix} \right] =\begin{bmatrix} 4 & -1 \\ 0 & 4 \end{bmatrix}\)
\(\Rightarrow x+3y=4;y=-1;7-x=0 \)
\(\Rightarrow x=7,y=-1\)

Generic 5

8 x 5 = 40

14.

\({ A }^{ -1 }=\frac { 1 }{ 8 } \left[ \begin{matrix} 5 & -1 \\ -7 & 3 \end{matrix} \right] \)

15.

\(f(A)\left[ \begin{matrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{matrix} \right] \)

16.

x = 1, 2 y = 3\(\pm\)3\(\sqrt { 2 } \)

17.

\(A=\left[ \begin{matrix} -8 & 3 & 5 \\ -13 & -1 & -9 \end{matrix} \right] \)

18.

Getting \({ A }^{ 2 }=\left( \begin{matrix} 9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9 \end{matrix} \right) \)
and \({ A }^{ 3 }=\left( \begin{matrix} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{matrix} \right) \)
\(\therefore\) A³ - 4A² - 3A + 11I
\(=\left( \begin{matrix} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{matrix} \right) -\left( \begin{matrix} 36 & 28 & 20 \\ 4 & 16 & 4 \\ 32 & 36 & 36 \end{matrix} \right) -\left( \begin{matrix} 3 & 9 & 6 \\ 6 & 0 & -3 \\ 3 & 6 & 9 \end{matrix} \right) +\left( \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) =0\)
Alternative Method :
\(A=\left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right] \)
\({ A }^{ 2 }=\left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right] \left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9 \end{matrix} \right] \)
\({ A }^{ 3 }=\left[ \begin{matrix} 9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9 \end{matrix} \right] \left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{matrix} \right] \)
\(\therefore\) A³ - 4A² - 4A + 11A
\(=\left[ \begin{matrix} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{matrix} \right] -4\left[ \begin{matrix} 9 & 7 & 5 \\ 1 & 4 & 1 \\ 8 & 9 & 9 \end{matrix} \right] -3\left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 0 & -1 \\ 1 & 2 & 3 \end{matrix} \right] +\left[ \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 28 & 37 & 26 \\ 10 & 5 & 1 \\ 35 & 42 & 34 \end{matrix} \right] -\left[ \begin{matrix} 36 & 28 & 20 \\ 4 & 16 & 4 \\ 32 & 36 & 36 \end{matrix} \right] -\left[ \begin{matrix} 3 & 9 & 6 \\ 6 & 0 & -3 \\ 3 & 6 & 9 \end{matrix} \right] +\left[ \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)
= 0

19.

Let the cost of 'A' variety pen be Rs. x, the cost of 'B' variety pen be Rs. y and the cost of 'C' variety pen be Rs. z.
According to the question,
x + y + z = 21
4x + 3y + 2z = 60
and 6x + 2y + 3z = 70
It can be written as,
\(\left[ \begin{matrix} 1 & 1 & 1 \\ 4 & 3 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 21 \\ 60 \\ 70 \end{matrix} \right] \)
i.e., AX = B,
where \(A=\left[ \begin{matrix} 1 & 1 & 1 \\ 4 & 3 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \)
\(X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)
and \(B=\left[ \begin{matrix} 21 \\ 60 \\ 70 \end{matrix} \right] \)
\(\left| A \right| =\left| \begin{matrix} 1 & 1 & 1 \\ 4 & 3 & 2 \\ 6 & 2 & 3 \end{matrix} \right| \)
= 1(9 - 4) - 1(12 - 12) + 1(8 - 18)
= 5 - 10 = - 5 \(\neq \) 0.
\(\Rightarrow\)  A^-1 exists.
C₁₁ = (- 1)² (9 - 4) = 5
C₁₂ = (- 1)³ (12-12) = 0
C₁₃ = (- 1)⁴ (8 - 18) = - 10
C₂₁ = - (3 - 2) = - 1
C₂₂ = - 3
C₂₃ = - (2 - 6) = 4
C₃₁ = (2 - 3) = - 1
C₃₂ = - (2 - 4) = 2
C₃₃ = 3 - 4 = - 1
\(adjA=\left[ \begin{matrix} 5 & -1 & -1 \\ 0 & -3 & 2 \\ -10 & 4 & -1 \end{matrix} \right] \)
\({ A }^{ -1 }=\frac { 1 }{ \left| A \right| } adjA\)
Also AX = B
Premultiplying by A^-1
\(\Rightarrow\)   X = A^-1B
\(\Rightarrow\)   X = \(\frac { 1 }{ \left| A \right| } \left( adjA \right) B\)
\(=\frac { 1 }{ \left| A \right| } \left[ \begin{matrix} 5 & -1 & -1 \\ 0 & -3 & 2 \\ -10 & 4 & -1 \end{matrix} \right] \left[ \begin{matrix} 21 \\ 60 \\ 70 \end{matrix} \right] \)
\(=\frac { -1 }{ 5 } \left[ \begin{matrix} 105-60-70 \\ -180+140 \\ -210+240-70 \end{matrix} \right] \)
\(=\frac { -1 }{ 5 } \left[ \begin{matrix} -25 \\ -40 \\ -40 \end{matrix} \right] \)
\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 8 \\ 8 \end{matrix} \right] \)
\(\therefore\)   x = 5, y = 8, z = 8
\(\therefore\)  cost of 'A' variety pen = Rs. 5
'B' variety pen = Rs. 8
'C' variety pen = Rs. 8

20.

\(\text { Let } A=I A \Rightarrow\left[\begin{array}{rrr} 3 & 0 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A\)
\(\Rightarrow\left[\begin{array}{rrr} 1 & -3 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{array}\right]=\left[\begin{array}{crr} 1 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A\)
\(\text { [by performing } \left.R_{1} \rightarrow R_{1}-R_{2}\right]\)
\(\Rightarrow\left[\begin{array}{rrr} 1 & -3 & -1 \\ 0 & 9 & 2 \\ 0 & 4 & 1 \end{array}\right]=\left[\begin{array}{rrr} 1 & -1 & 0 \\ -2 & 3 & 0 \\ 0 & 0 & 1 \end{array}\right] A\)
\(\text { [by performing } \left.R_{2} \rightarrow R_{2}-2 R_{1}\right]\)
\(\Rightarrow\left[\begin{array}{rrr} 1 & -3 & -1 \\ 0 & -1 & 0 \\ 0 & 4 & 1 \end{array}\right]=\left[\begin{array}{rrr} 1 & -1 & 0 \\ -2 & 3 & -2 \\ 0 & 0 & 1 \end{array}\right] A\)
\(\text { [by performing } \left.R_{2} \rightarrow R_{2}-2 R_{3}\right]\)
\(\Rightarrow\left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{rrr} -5 & 8 & -6 \\ -2 & 3 & -2 \\ 8 & -12 & 9 \end{array}\right] A\)
\(\text { [by performing } \left.R_{1} \rightarrow R_{1}+3 R_{2} \text { and } R_{3} \rightarrow R_{3}-4 R_{2}\right]\)
\(\Rightarrow\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{rrr} 3 & -4 & 3 \\ -2 & 3 & -2 \\ 8 & -12 & 9 \end{array}\right] A\)
\(\text { [by performing } \left.R_{1} \rightarrow R_{1}+R_{3}\right]\)
\(\text { Hence }A^{-1}=\left[\begin{array}{rrr} 3 & -4 & 3 \\ -2 & 3 & -2 \\ 8 & -12 & 9 \end{array}\right]\)

21.

\(A^{2}-4 A-5 I\)
\(=\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]-\left[\begin{array}{lll} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{array}\right]-\left[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]\)
\(=\left[\begin{array}{ccc} 1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1 \end{array}\right]\)
\(-\left[\begin{array}{lll} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{array}\right]-\left[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]\)
\(=\left[\begin{array}{lll} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{array}\right]-\left[\begin{array}{lll} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{array}\right]-\left[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]\)
\(=\left[\begin{array}{lll} 9-4-5 & 8-8-0 & 8-8-0 \\ 8-8-0 & 9-4-5 & 8-8-0 \\ 8-8-0 & 8-8-0 & 9-4-5 \end{array}\right]\)
\(=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)
\(\text { Hence, } A^{2}-4 A-5 I=O\)

Generic 3

4 x 3 = 12

22.

\(\left[ \left( \begin{matrix} 1 & \omega & { \omega }^{ 2 } \\ \omega & { \omega }^{ 2 } & 1 \\ { \omega }^{ 2 } & 1 & \omega \end{matrix} \right) +\left( \begin{matrix} \omega & { \omega }^{ 2 } & 1 \\ { \omega }^{ 2 } & 1 & \omega \\ \omega & { \omega }^{ 2 } & 1 \end{matrix} \right) \right] \left[ \begin{matrix} 1 \\ \omega \\ { \omega }^{ 2 } \end{matrix} \right] =\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\( \left[ \left( \begin{matrix} 1 & \omega & { \omega }^{ 2 } \\ \omega & { \omega }^{ 2 } & 1 \\ { \omega }^{ 2 } & 1 & \omega \end{matrix} \right) +\left( \begin{matrix} \omega & { \omega }^{ 2 } & 1 \\ { \omega }^{ 2 } & 1 & \omega \\ \omega & { \omega }^{ 2 } & 1 \end{matrix} \right) \right] \left[ \begin{matrix} 1 \\ \omega \\ { \omega }^{ 2 } \end{matrix} \right] =\left( \begin{matrix} 1+\omega & \omega +{ \omega }^{ 2 } & { \omega }^{ 2 }+1 \\ \omega +{ \omega }^{ 2 } & { \omega }^{ 2 }+1 & 1+\omega \\ { \omega }^{ 2 }+\omega & 1+{ \omega }^{ 2 } & \omega +1 \end{matrix} \right) \left[ \begin{matrix} 1 \\ \omega \\ { \omega }^{ 2 } \end{matrix} \right] \)
\(=\left( \begin{matrix} -{ \omega }^{ 2 } & -1 & -\omega \\ -1 & -\omega & -{ \omega }^{ 2 } \\ -1 & -\omega & -{ \omega }^{ 2 } \end{matrix} \right) \left[ \begin{matrix} 1 \\ \omega \\ { \omega }^{ 2 } \end{matrix} \right] \quad \quad \left[ \because \quad 1+\omega +{ \omega }^{ 2 }=0 \right] \)
\(=\left( \begin{matrix} -{ \omega }^{ 2 } & -\omega & -{ \omega }^{ 3 } \\ -1 & -{ \omega }^{ 2 } & -{ \omega }^{ 4 } \\ -1 & -{ \omega }^{ 2 } & -{ \omega }^{ 4 } \end{matrix} \right) =\left[ \begin{matrix} -{ \omega }^{ 2 } & -\omega & -1 \\ -1 & -{ \omega }^{ 2 } & -{ \omega } \\ -1 & -{ \omega }^{ 2 } & -{ \omega } \end{matrix} \right] \quad \left[ \because { \omega }^{ 3 }=1\quad \& \quad { \omega }^{ 4 }=\omega \quad \right] \)
\(=\left[ \begin{matrix} -0 \\ -0 \\ -0 \end{matrix} \right] =\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
which is true.

23.

We have: \(A=\left[ \begin{matrix} 3 & 1 & -1 \\ 0 & 1 & 2 \end{matrix} \right] .\)
\(\therefore \ A\prime =\left[ \begin{matrix} 3 & 0 \\ 1 & 1 \\ -1 & 2 \end{matrix} \right]\)
\(\therefore \ AA\prime =\left[ \begin{matrix} 3 & 1 & -1 \\ 0 & 1 & 2 \end{matrix} \right] \left[ \begin{matrix} 3 & 0 \\ 1 & 1 \\ -1 & 2 \end{matrix} \right] \)
\(=\begin{bmatrix} 9+1+1 & 0+1-2 \\ 0+1-2 & 0+1+4 \end{bmatrix}=\begin{bmatrix} 11 & -1 \\ -1 & 5 \end{bmatrix}.\)
Which is a symmetric matrix.

24.

We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 18 elements, we have to find all the ordered pairs of natural numbers whose product is 18.
The ordered pairs are: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6), and (6, 3)
Hence, the possible orders of a matrix having 18 elements are:
1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, and 6 × 3
(1, 5) and (5, 1) are the ordered pairs of natural numbers whose product is 5.
Hence, the possible orders of a matrix having 5 elements are 1 × 5 and 5 × 1.

25.

We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 24 elements, we have to find all the ordered pairs of natural numbers whose product is 24.
The ordered pairs are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), and (6, 4)
Hence, the possible orders of a matrix having 24 elements are:
1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, and 6 × 4
(1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13.
Hence, the possible orders of a matrix having 13 elements are 1 × 13 and 13 × 1.