Important Questions For CBSE Class 12 Maths Chapter 3 Matrices

CBSE Class 12 Maths Chapter 3 Matrices Important Questions

Question 1. The number of all possible matrices of order 2 x 3 with each country 1 or 2 is :

1. 16
2. 6
3. 64
4. 24

Solution: 3. 6

Required number of possible matrices

= (Number of entries)^order

= (2)^2×3 = (2)⁶ = 64

Read and Learn More CBSE Class 12 Maths Important Question and Answers

Important Questions For Class 12 Maths Matrices 

Question 2. If a matrix A is both symmetric and skewed, symmetric. then A is necessarily a

1. Diagonal matrix
2. Zero square matrix
3. Square matrix
4. Identity matrix

Solution: 2. Zero square matrix

Given; A^T = A (symmetric matrix)

and -A^T= A (skew-symmetric matrix)

⇒ 2A = O or A = O

Hence, A is necessarily a zero-square matrix

Question 3. If \(\left[\begin{array}{cc}
3 c+6 & a-d \\
a+d & 2-3 b
\end{array}\right]=\left[\begin{array}{cc}
12 & 2 \\
-8 & -4
\end{array}\right]\), then the value of ab-cd is

1. 4
2. 16
3. -4
4. -16

Solution: 1. 4

Given: \(\left[\begin{array}{cc}
3 c+6 & a-d \\
a+d & 2-3 b
\end{array}\right]=\left[\begin{array}{cc}
12 & 2 \\
-8 & -4
\end{array}\right]\)

On comparing both sides, we get

a – d = 2 and a + d = -8

⇒  2a = -6 Or a=-3

⇒ d = -5

Also; 3c + 6 = 12, 2-3b = -4

⇒ c = 2, b = 2

Hence, ab- cd = (-3)2 – 2(-5) = -6 + 10 = 4

Question 4. For two matrices \(P=\left[\begin{array}{cc}
3 & 4 \\
-1 & 2 \\
0 & 1
\end{array}\right] \text { and } Q^{\top}=\left[\begin{array}{ccc}
-1 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]\), P-Q is

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

Solution: 2. \(\left[\begin{array}{cc}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]\)

⇒ \(Q^{\top}=\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right] \Rightarrow Q=\left[\begin{array}{cc}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right]\)

Hence, \(P-Q=\left[\begin{array}{cc}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]-\left[\begin{array}{cc}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right]=\left[\begin{array}{cc}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]\)

Matrix Multiplication Problems Class 12 Maths 

Question 5. A matrix A = [a_ij]_3×3 is defined by \(a_{i j}=\left\{\begin{array}{ccc}
2 i+3 j & ; & i<j \\
5 & ; & i=j \\
3 i-2 j & ; & i>j
\end{array}\right.\), The number of elements in A, which are more than 5, is

1. 3
2. 4
3. 5
4. 6

Solution: 2. 4

Given; A = [a_ij]_3×3

where \(a_{i j}=\left\{\begin{array}{ccc}
2 i+3 j & ; & i<j \\
5 & ; & i=j \\
3 i-2 j & ; & i>j
\end{array}\right.\)

⇒ A= \(\left[\begin{array}{lll}
5 & 8 & 11 \\
4 & 5 & 13 \\
7 & 5 & 5
\end{array}\right]\)

Hence; required number =4

Question 6. For the matrix X = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\), (X²-X) is:

1. 21
2. 31
3. 1
4. 51

Solution: 1. 21

⇒ \(X^2=\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right] \cdot\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]=\left[\begin{array}{lll}
2 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 2
\end{array}\right]\)

⇒ \(X^2-X=\left[\begin{array}{lll}
2 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 2
\end{array}\right]-\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]=\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{array}\right]=21\)

Question 7. Find the order of matrix A such that \(\left[\begin{array}{cc}2 & -1 \\ 1 & 0 \\ -3 & 4\end{array}\right]\), A = \(\left[\begin{array}{cc}-1 & -8 \\ 1 & -2 \\ 9 & 22\end{array}\right]\)

or,

If B = \(\left[\begin{array}{cc}
1 & -5 \\
0 & -3
\end{array}\right]\) and A+2 B = \(\left[\begin{array}{cc}
0 & 4 \\
-7 & 5
\end{array}\right]\), find matrix A.

Solution:

Let \(\mathrm{B}=\left[\begin{array}{cc}2 & -1 \\ 1 & 0 \\ -3 & 4\end{array}\right]\) and \(\mathrm{C}=\left[\begin{array}{cc}-1 & -8 \\ 1 & -2 \\ 9 & 22\end{array}\right]\)

We have \(\left[\begin{array}{cc}2 & -1 \\ 1 & 0 \\ -3 & 4\end{array}\right]_{3 \times 2} \quad A=\left[\begin{array}{cc}-1 & -8 \\ 1 & -2 \\ 9 & 22\end{array}\right]_{3 \times 2}\)

Let the order of A be m x n

BA will be possible if several columns in matrix B should be equal to several rows in matrix A ⇒ m = 2.

and the order of BA is 3 x n

Since, BA = C.

⇒ Order of BA will be the same as that of matrix C

⇒ 3 x n = 3 x 2

⇒ n = 2

Then, the order of matrix A is 2 x 2

A+2 B \(\left[\begin{array}{cc}
0 & 4 \\
-7 & 5
\end{array}\right]\) (given)

⇒ A = \(\left[\begin{array}{cc}
0 & 4 \\
-7 & 5
\end{array}\right]-2 B=\left[\begin{array}{cc}
0 & 4 \\
-7 & 5
\end{array}\right]-2\left[\begin{array}{cc}
1 & -5 \\
0 & -3
\end{array}\right]=\left[\begin{array}{cc}
0 & 4 \\
-7 & 5
\end{array}\right]-\left[\begin{array}{cc}
2 & -10 \\
0 & -6
\end{array}\right]\)

⇒ A = \(\left[\begin{array}{cc}
-2 & 14 \\
-7 & 11
\end{array}\right]\)

Question 8. If A = \(\left[\begin{array}{lll}
1 & 0 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{l}
2 \\
5 \\
6
\end{array}\right]\), find AB.

Solution:

AB= \(\left[\begin{array}{lll}
1 & 0 & 4
\end{array}\right]_{1 \times 3}\left[\begin{array}{l}
2 \\
5 \\
6
\end{array}\right]_{3 \times 1}\)

= \([1 \times 2+0 \times 5+4 \times 6]_{1 \times 1}=[2+0+24]=[26]\)

Class 12 Matrices Chapter Important Board Questions 

Question 9. Given, a skew-symmetric matrix \(A = \mathrm{A}=\left[\begin{array}{ccc}
0 & \mathrm{a} & 1 \\
-1 & \mathrm{~b} & 1 \\
-1 & \mathrm{c} & 0
\end{array}\right]\), then value of (a+b+c)² is

Solution:

Given \(A = \mathrm{A}=\left[\begin{array}{ccc}
0 & \mathrm{a} & 1 \\
-1 & \mathrm{~b} & 1 \\
-1 & \mathrm{c} & 0
\end{array}\right]\)

A is a skew-symmetric matrix, A^T = -A

⇒ \(A^{\top}=\left[\begin{array}{ccc}
0 & -1 & -1 \\
a & b & c \\
1 & 1 & 0
\end{array}\right] \text { and }-A=\left[\begin{array}{ccc}
0 & -a & -1 \\
1 & -b & -1 \\
1 & -c & 0
\end{array}\right]\)

So, a = 1, b = 0 and c = -1

Now (a + b + c)² = (1 +0-1)² = 0

Question 10. If the matrices A = \(\left[\begin{array}{ccc}
0 & \mathrm{a} & -3 \\
2 & 0 & -\mathrm{t} \\
\mathrm{b} & 1 & 0
\end{array}\right]\) is skew symmetric, find the values of ‘a’ and ‘b’.

Solution:

A = \(\left[\begin{array}{ccc}
0 & \mathrm{a} & -3 \\
2 & 0 & -\mathrm{t} \\
\mathrm{b} & 1 & 0
\end{array}\right]\) (given)

A is skew-symmetric ⇒ A^T = -A

⇒ \(\left[\begin{array}{ccc}
0 & 2 & b \\
a & 0 & 1 \\
-3 & -1 & 0
\end{array}\right]=\left[\begin{array}{ccc}
0 & -a & 3 \\
-2 & 0 & 1 \\
-b & -1 & 0
\end{array}\right]\)

⇒ a=-2, b=3

Question 11. If the matrix A = \(\left[\begin{array}{cc}
0 & 6-5 x \\
x^2 & x+3
\end{array}\right]\) is symmetric, find the values of x?

Solution:

A = \(\left[\begin{array}{cc}
0 & 6-5 x \\
x^2 & x+3
\end{array}\right]\) (given)

For a symmetric matrix, A = A¹

⇒ \(\left[\begin{array}{cc}
0 & 6-5 x \\
x^2 & x+3
\end{array}\right]=\left[\begin{array}{cc}
0 & x^2 \\
6-5 x & x+3
\end{array}\right]\)….(1)

∴ 6-5x = x² [from (1)]

⇒ x² + 5x – 6 = 0

⇒ (x + 6) (x -1) = 0

⇒ x = – 6, 1

Question 12. If A = \(\left[\begin{array}{cc}
-3 & 2 \\
1 & -1
\end{array}\right] \text { and } I=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\), find scalar k such that A² + I = KA

Solution:

Given, A = \(\left[\begin{array}{cc}
-3 & 2 \\
1 & -1
\end{array}\right] \text { and } I=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)

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

(because \(A^2+I=K A\))

⇒ \(\left[\begin{array}{cc}
11 & -8 \\
-4 & 3
\end{array}\right]+\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right]\) = \(\left[\begin{array}{cc}
-3 \mathrm{~K} & 2 \mathrm{~K} \\
\mathrm{~K} & -\mathrm{K}
\end{array}\right]\)

∴ \(\left[\begin{array}{cc}
12 & -8 \\
-4 & 4
\end{array}\right]\) = \(\left[\begin{array}{cc}
-3 \mathrm{~K} & 2 \mathrm{~K} \\
\mathrm{~K} & -\mathrm{K}
\end{array}\right]\)

⇒ K = -4 (on comparing both sides)

Application Of Matrices Class 12 Maths Questions 

Question 13. If A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\) , show that (A-2I)(A-3I) = O.

Solution:

A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\) (given)

⇒ A-2I = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]-2\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
2 & 2 \\
-1 & -1
\end{array}\right]\)

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

⇒ (A-2I)(A-3I) = \(\left[\begin{array}{cc}
2 & 2 \\
-1 & -1
\end{array}\right]\left[\begin{array}{cc}
1 & 2 \\
-1 & -2
\end{array}\right]=\left[\begin{array}{cc}
2-2 & 4-4 \\
-1+1 & -2+2
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=O\)