Chapter 3 Matrices | class 12th | Important Question for Maths

Matrices Class 12 Important Questions with Solutions Previous Year Questions

Matrix and Operations on Matrices

Question 1.
If 3A – B = [5101] and B = [4235] then find the value of matrix A. (Delhi 2019)
Answer:

Question 2.
Find the value of x – y, if (Delhi 2019)
2[103x]+[y102]=[5168]
Answer:
Given that,

Here, both matrices are equal, so we equate the corresponding elements,
2 + y = 5 and 2x + 2 = 8
⇒ y = 3 and 2x = 6 ⇒ x = 3
Therefore, x – y = 3 – 3 = 0

Question 3.
If A is a square matrix such that A² = I, then find the simplified value of (A – I)³ + (A + I)³ – 7A. (Delhi 2016)
Answer:
Given, A² = 7 ……. (i)
Now, (A – I)³ + (A + I)³ – 7A
= (A³ – 3A²I + 3AI² – I) + (A³ + 3A²I + 3AI² + I³) – 7A
= A³ – 3A² + 3AI – I + A³ + 3A² + 3AI + I – 7A
[∵ A²I = A² and I³ = I³ = I]
= 2A³ + 6AI – 7A = 2A² A + 6A – 7A [∵ AI = A]
= 2IA – A [from Eq. (1)]
= 2A – A = A [∵ IA = A]

Question 4.
Write the number of all possible matrices of order 2 × 2 with each entry 1, 2 or 3. (All India 2016)
Answer:
We know that, a matrix of order 2 × 2 has 4 entries. Since, each entry has 3 choices, namely 1, 2 or 3, therefore number of required matrices
3⁴ = 3 × 3 × 3 × 3 = 81.

Question 5.
If [2 1 3] ⎡⎣⎢−1−10011−101⎤⎦⎥⎡⎣⎢10−1⎤⎦⎥ = A, then write the order of matrix A. (Foregin 2016)
Answer:

= [- 3 – I] = [- 4]_{1 × 1}
∴ Order of matrix A is 1 × 1.

Question 6.
Write the element a of a 3 × 3 matrix A = [a_ij], whose elements are given by a_ij = |i−j|2 (Delhi 2015)
Answer:
Given, A = [a_ij]_{3 × 3}
where, a_ij = |i−j|2
Now, a₂₃ = |2−3|2=|−1|2=12
[put i = 2 and j = 3]

Question 7.
If [2x 3] [1−320] [x3] = 0, find x. (Delhi 2015C)
Answer:
Given, matrix equation is

⇒ [2x² – 9x + 12x] = [0]
⇒ 2x² + 3x = 0
⇒ x(2x + 3) = 0
∴ x = 0 or x = – 3/2

Question 8.
If 2[354x]+[10y1]=[71005], then find (x – y). (Delhi 2014)
Answer:

On equating the corresponding elements, we get
8 + y = 0 and 2x + 1 = 5
⇒ y = – 8 and x = 5−12 = 2
∴ x – y = 2 – (-8) = 10

Question 9.
Solve the following matrix equation for x.
[x 1] [1−200] = 0 (Delhi 2014)
Answer:
Given, [x1][1−200] = 0
By using matrix multiplication, we get
[x – 2 0] = [0 0]
On equating the corresponding elements, we get
x – 2 = 0
⇒ x = 2

Question 10.
If A is a square matrix such that A² = A, then write the value of 7A — (I + A)³, where I is an identity matrix. (All India 2014)
Answer:
Given, A² = A
Now, 7A – (I + A)³ = 7A – [I³ + A³ + 3I4(I + A)]
[∵ (x + y)³ = x³ + y³ + 3xy (x + y)]
= 7A – [I + A².A + 3A(I + A)]
[∵ I³ = I and IA = A]
= 7A – (I + A . A + 3AI + 3A²)
[∵ A² = A]
= 7A – (I + A + 3A + 3A)
[∵AI = A and A² = AI]
= 7A – (I + 7A) = – 1

Question 11.
If [x−y2x−yzw]=[−1045], then find the value of x + y. (All India 2014)
Answer:
Given
[x−y2x−yzw]=[−1045]
On equating the corresponding elements, we get
x – y = – 1 …… (i)
and 2x – y = 0 …… (ii)
On solving the Eqs.(i) and (ii), we get
x = 1 and y = 2
∴ x + y = 1 + 2 = 3

Question 12.
If [a+483b−6]=[2a+28b+2a−8b], then write the value of a – 2b. (Foreign 2014)
Answer:
Given,
[a+483b−6]=[2a+28b+2a−8b]
On equating the corresponding elements, we get
∴ a + 4 = 2a+ 2 ……… (i)
3b = b + 2 ……. (ii)
and – 6 = a – 8b ……… (iii)
On solving the Eqs. (i), (ii) and (iii), we get
a = 2 and b = 1
Now, a – 2b= 2 – 2(1) = 2 – 2 = 0

Question 13.
If [x⋅yz+64x+y]=[80w6], then write the value of (x + y + z). (Delhi 2014C)
Answer:
Given,
[x⋅yz+64x+y]=[80w6]
On equating the corresponding elements, we get
∴ x – y = 8 ……. (i)
Z + 6 = 0
⇒ z = – 6 ……… (ii)
and x + y = 6 ……… (iii)
Now, on adding Eqs. (ii) and (iii), we get
x + y + z = 6 + (- 6) = 0

Question 14.
The elements a of a 3 × 3 matrix are given by a_ij = 12|- 3i + j|. Write the value of element a₃₂. (All India 2014C)
Answer:
72

Question 15.
If [2x 4][x−8] = 0, then find the positive value of x. (All India 2014C)
Answer:
Given,
[2x 4][x−8] = 0
On equating the corresponding elements, we get
⇒ 2x² – 32 = 0
⇒ 2x² = 32
⇒ x² =16
⇒ x = ±4
∴ Positive value of x is 4.

Question 16.
If 2[103x]+[y102]=[5168] then find the value of(x + y). (Delhi 2013C; All India 2012)
Answer:
8

Question 17.
Find the value of a, if (Delhi 2013)
[a−b2a−b2a+c3c+d]=[−10513]
Answer:
1

Question 18.
If [9−2−1143] = A + [1024−19], then find the matrix A. (Delhi 2013)
Answer:
Given, matrix equation can be rewritten as

NOTE: Two matrices can be subtracted only when their orders are same.

Question 19.
If matrix A = [1−1−11] and A² = kA. then write the value of k. (All India 2013)
Answer:

Question 20.
If matrix A = [2−2−22] and A² = pA. then write the value of p. (All India 2013)
Answer:
19

Question 21.
If matrix A = [3−3−33] and A² = λA, then write the value of λ. (All India 2013)
Answer:
19

Question 22.
Simplify
cos θ [cosθ−sinθsinθcosθ] + sin θ [sinθcosθ−cosθsinθ] (Delhi 2012)
Answer:
First, multiply each element of the first matrix by cos θ and second matrix by sin θ and then use the matrix addition.

Question 23.
If [2537][1−2−34]=[−4−96x], write the value of x. (Delhi 2012)
Answer:
Given matrix equation is

On equating the corresponding elements, we get
x = 13

Question 24.
Find the value of y – x from following equation.
2[x75y−3]+[31−42]=[715614] (All India 2012)
Answer:
7

Question 25.
If x[23]+y[−11]=[105], then write the value of x. (Foreign 2012)
Answer:

On equating the corresponding elements, we get
2x – y = 10 …… (i)
and 3x + y = 5 …… (ii)
On adding Eqs. (i) and (ii), we get
5x = 15
∴ x = 3

Question 26.
If 3A – B = [5101] and B = [4235], then find the matrix A. (Delhi 2012C)
Answer:

Question 27.
Write the value of x – y + z from following equation. (Foregin 2011)
⎡⎣⎢x+y+zx+zy+z⎤⎦⎥=⎡⎣⎢957⎤⎦⎥
Answer:
Given matrix equation is
⎡⎣⎢x+y+zx+zy+z⎤⎦⎥=⎡⎣⎢957⎤⎦⎥
On equating the corresponding elements, we get
x + y+ z = 9 …… (i)
x + z = 5 …….. (ii)
and y + z = 7 ……. (iii)
On putting the value of x + z from Eq. (ii) in Eq. (i), we get
y + 5 = 9 ⇒ y = 4
On putting y = 4 in Eq. (iii), we get z = 3
Again, putting z = 3 in Eq. (ii), we get x = 2
∴ x – y + z = 2 – 4 + 3 = 1

Question 28.
Write the order of product matrix (Foreign 2011)
⎡⎣⎢123⎤⎦⎥[234]
Answer:
Use the fact that if a matrix A has order m × n and other matrix B has order n × z, then the matrix AB has order m × z.
Let A = ⎡⎣⎢123⎤⎦⎥ and B = [2, 3, 4]
Here, order of matrix A = 3 × 1
and order of matrix B = 1 × 3
∴ Order of product matrix AB = 3 × 3

Question 29.
If a matrix has 5 elements, then write all possible orders it can have. (All India 2011)
Answer:
Use the result that if a matrix has order m × n, then total number of elements in that matrix is mn.
Given, a matrix has 5 elements. So, possible order of this matrix are 5 × 1 and 1 × 5.

Question 30.
For a 2 × 2 matrix, A = [a_ij] whose elements are given by a_ij = i/j, write the value of a₁₂. (Delhi 2011)
Answer:
12

Question 31.
If [x2x+yx−y7] = [3817], then find the value of y. (Delhi 2011C)
Answer:
Given,
[x2x+yx−y7] = [3817]
On equating the corresponding elements, we get
x = 3 and x – y = 1
⇒ y = x – 1 = 3 – 1 = 2

Question 32.
From the following matrix equation, find the value of x. (Foreign 2010)
[x+y−543y]=[3−546]
Answer:
Given
[x+y−543y]=[3−546]
On equating the corresponding elements, we get
x + y = 3 … (i)
and 3y = 6 …. (ii)
From Eq. (ii), we get
y = 2
On substituting y = 2 in Eq. (i), we get
x + 2 = 3
⇒ x = 1

Question 33.
Find x from the matrix equation (Foreign 2010)
[1435][x2]=[56]
Answer:
First, determine the multiplication of matrices in . LHS and then equate the corresponding elements of both sides.
Given matrix equation is

On equating the corresponding elements, we get
x + 6 ⇒ 5 x = – 1

Question 34
If [324x][x1]=[1915], then find the value of x. (Foreign 2010)
Answer:
5

Question 35.
If A = \left[\begin{array}{rr} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right] then for what value of α, A is an identity matrix? (Delhi 2010)
Answer:
First, put the given matrix A equal to an identity matrix and then equate the corresponding elements to get the value of α.

On equating the element a₁₁ of both matrices, we get
cos α = 1
⇒ cos α = cos 0° [∵ cos 0° = 1]
∴ α = 0
Hence, for α = 0, A is an identity matrix.
[∵ sin 0 = 0]

Question 36.
If \left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\left[\begin{array}{ll} 3 & 1 \\ 2 & 5 \end{array}\right]=\left[\begin{array}{ll} 7 & 11 \\ k & 23 \end{array}\right], then write the value of k, (Delhi 2010)
Answer:
17

Question 37.
If A is a matrix of order 3 × 4 and B is a matrix of order 4 × 3, then find order of matrix (AB). (DelhI 2010C)
Answer:
3 × 3

Question 38.
If \left[\begin{array}{cc} x+y & 1 \\ 2 y & 5 \end{array}\right]=\left[\begin{array}{ll} 7 & 1 \\ 4 & 5 \end{array}\right],then find the value of x. (Delhi 2010C)
Answer:
x = 5

Question 39.
If \left[\begin{array}{cr} 2 x+y & 3 y \\ 0 & 4 \end{array}\right]=\left[\begin{array}{ll} 6 & 0 \\ 0 & 4 \end{array}\right], then find the value of x. (All India 2010C)
Answer:
x = 3

Question 40.
If \left[\begin{array}{cr} 3 y-x & -2 x \\ 3 & 7 \end{array}\right]=\left[\begin{array}{rr} 5 & -2 \\ 3 & 7 \end{array}\right], then find the value of y. (All India 2010C)
Answer:
y = 2

Question 41.
If A = \left[\begin{array}{rr} 4 & 2 \\ -1 & 1 \end{array}\right], show that (A – 2I) (A – 3I) = 0. (All IndIa 2019C)
Answer:

Question 42.
Find a matrix A such that 2A – 3B + 5C = 0, where B = \left[\begin{array}{rrr} -2 & 2 & 0 \\ 3 & 1 & 4 \end{array}\right] and C = \left[\begin{array}{rrr} 2 & 0 & -2 \\ 7 & 1 & 6 \end{array}\right]. (Delhi 2019)
Answer:
Given, 2A – 3B + 5C = 0 ⇒ 2A = 3B – 5C

Question 43.
If A = \left[\begin{array}{rrr} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right] then find the values of (A² – 5A). (Delhi 2019)
Answer:

Question 44.
Find matrix A such that (All India 2017)
\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]
Answer:
Let the order of A is m × n ∴ m = 2, n = 2

On equating corresponding elements both sides, we get
2x – s = – 1, x = 1, y = – 2 and 2y – t = – 8
At x = 1, 2x – s = – 1 ⇒ 2 × 1 – s = – 1
⇒ – s = – 1 – 2 = s = 3 and at y = – 2, 2y – t = – 8,
⇒ 2 × (- 2) – t = – 8
⇒ – 4 – t = – 8
⇒ t = 4
On putting x = 1, y = – 2, s = 3 and t = 4 in Eq. (i),
we get A = \left[\begin{array}{cc} 1 & -2 \\ 3 & 4 \end{array}\right]

Question 45.
Let A = \left[\begin{array}{cc} 2 & -1 \\ 3 & 4 \end{array}\right], B = \left[\begin{array}{ll} 5 & 2 \\ 7 & 4 \end{array}\right], C = \left[\begin{array}{ll} 2 & 5 \\ 3 & 8 \end{array}\right], find a matrix D such that CD – AB = 0. (Delhi 2017)
Answer:

Question 46.
If A = \left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right], then find A² – 5A + 4I and hence find a matrix X such that A² – 5A + 4I + X = 0. (Delhi 2015)
Answer:

Question 47.
If A = \left[\begin{array}{ll}1 & -1 \\ 2 & -1\end{array}\right], B = \left[\begin{array}{cc} a & 1 \\ b & -1 \end{array}\right] and (A + B)² = A² + B², then find the values of a and b. (Foreign 2015)
Answer:

On equating the corresponding elements, we get
a² + 2a + I = a² + b – 1 ⇒ 2a – b = – 2
a – 1 = 0 ⇒ a = 1 ….. (ii)
2a – b + ab – 2 = ab – b
⇒ 2a – 2 = 0 ⇒ a = 1 ….. (iii)
and b = 4 …….(iv)
Since, a = 1 and b = 4 also satisfy Eq. (1). therefore
a = 1 and b = 4

Question 48.
If A = \left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right], then find value of A² – 3A + 2I. (All India 2010)
Answer:
\left[\begin{array}{ccc} 1 & -1 & -1 \\ 3 & -3 & -4 \\ -3 & 2 & 0 \end{array}\right]

Question 49.
If A = \left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right] and A³ – 6A² + 7A + kI₃ = 0, find the value of k. (All India 2016)
Answer:

Transpose of a Matrix, Symmetric and Skew-Symmetric Matrices

Question 1.
If A = \left[\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & x \\ -2 & 2 & -1 \end{array}\right] is a matric satisfying AA’ = 9I, find x. (CBSE 2018C)
Answer:

On equating the corresponding elements, we get
4 + 2x = 0 and 5 + x² = 9
⇒ x = – 2 and x² = 4
⇒ x = – 2 and x = ±2
∴ The value of x is -2

Question 2.
If the matrix A = \left[\begin{array}{ccc} 0 & a & -3 \\ 2 & 0 & -1 \\ b & 1 & 0 \end{array}\right] is skew-symmetric, find the values of ‘a’ and ‘b’. (CBSE 2018)
Answer:

Question 3.
Matrix A = \left[\begin{array}{ccc} 0 & 2 b & -2 \\ 3 & 1 & 3 \\ 3 a & 3 & -1 \end{array}\right] is given to be symmetric, find the values of a and b.
Answer:

Question 4.
If A = \left[\begin{array}{cc} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right], then find a satisfying 0 < α < \frac{\pi}{2} when A + A^T = √2 I₂;
where A^T is transpose of A. (All India 2016)
Answer:

Question 5.
If A = \left(\begin{array}{ll} 3 & 5 \\ 7 & 9 \end{array}\right) is written as A = P + Q, where P is a symmetric matrix and Q is skew-symmetric matrix, then write the matrix P. (Foreign 2016)
Answer:
We have A = \left(\begin{array}{ll} 3 & 5 \\ 7 & 9 \end{array}\right) and A = P + Q. where P is symmetric matrix and Q is skew-symmetric matrix.

Question 6.
Write 2 × 2 matrix which is both symmetric and skew-symmetric. (Delhi 2014C)
Answer:
A null matrix of order 2 × 2, i.e. \left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] is both symmetric and skew-symmetric.

Question 7.
For what value of x, is the matrix A = \left[\begin{array}{rrr} 0 & 1 & -2 \\ -1 & 0 & 3 \\ x & -3 & 0 \end{array}\right] a skew-symmetric matrix? (All India 2013)
Answer:
If A is a skew-symmetric matrix, then A = – A^T, where A^T is transpose of matrix A.

Question 8.
If A^T = \left[\begin{array}{rr} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{array}\right] and B = \left[\begin{array}{rrr} -1 & 2 & 1 \\ 1 & 2 & 3 \end{array}\right], then find A^T – B^T (All India 2012)
Answer:
First, find the transpose of matrix 6 and then subtract the corresponding elements of both matrices A^T and B^T.

Question 9
If A = \left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right], then find A + A’. (All India 2010C)
Answer:

Question 10.
If \left(\begin{array}{cc} 2 x+y & 3 y \\ 0 & 4 \end{array}\right)=\left(\begin{array}{ll} 6 & 0 \\ 6 & 4 \end{array}\right), then find the value of x. (All India 2010)
Answer:

We know that, if two matrices are equal, then their corresponding elements are equal.
∴ 2x + y = 6 ….. (i)
and 3y = 6 ….. (ii)
From Eq. (ii), we get
y = 2
On substituting y = 2 in Eq. (i), we get
2x+ 2 = 6
⇒ x +1 = 3
∴ x = 2

Question 11.
Show that all the diagonal elements of a skew-symmetric matrix are zero. (Delhi 2017)
Answer:
Let A = [a_ij] be a skew-symmetric matrix.
Then, a_ij = – a_ij for all i, j
Now, put i = j, we get
⇒ a_ii = – a_ii for all values of i
⇒ 2 a_ii = 0
⇒ a_ii = 0 for all values of i
∴ a₁₁ = a₂₂ = a₃₃ = ….. = a_nn = 0
Hence, all the diagonal elements of a skew symmetric matrix are zero.
Hence proved.

Question 12.
Express the matrix A = \left[\begin{array}{ccc} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{array}\right] as the sum of a symmetric and skew-symmetric matrix. (All India 2015C)
Answer:
Any square matrix A can be expressed as the sum of a symmetric matrix and skew-symmetric matrix, i.e.
A = \frac{A+A^{\prime}}{2}+\frac{A-A^{\prime}}{2}, where \frac{A+A^{\prime}}{2} and \frac{A-A^{\prime}}{2}
are symmetric and skew-symmetric matrices, respectively.


Thus, matrix A is expressed as the sum of symmetric matrix and skew-symmetric matrix.

Question 13.
For the following matrices A and B, verify that [AB]’ = B’A’;
A = \left[\begin{array}{r} 1 \\ -4 \\ 3 \end{array}\right], B = [-1 2 1]. (All India 2010).
Answer:

Question 14
Express the following matrix as a sum of a symmetric and a skew-symmetric matrices and verify your result: (All India 2010)
\left[\begin{array}{rrr} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{array}\right]
Answer: