Matrices Class 12 MCQ Questions Chapter 3 Question 1.
If [x – 1] \(\left[\begin{array}{cc}
1 & 0 \\
-2 & 0
\end{array}\right]\) = [0,0]
⇒ 0 then x equals
(A) 0
(B) -2
(C) -1
(D) 2
Answer:
(B) -2

Explanation:
[x – 1] \(\left[\begin{array}{cc}
1 & 0 \\
-2 & 0
\end{array}\right]\)
⇒ [x -2 o] = [0 0]
⇒ x – 2 = 0 [By def. of equality)
x = 2

Matrices MCQ Chapter 3 Class 12 Question 2.
if A = [2 -3 4], B = \(\left[\begin{array}{l}
3 \\
2 \\
2
\end{array}\right]\) X = [1 2 3] and Y = \(\left[\begin{array}{l}
2 \\
3 \\
4
\end{array}\right]\),
then AB + XY equals
(A) [28]
(B) [24]
(C) 28
(D) 24
Answer:
(A) [28]

Explanation:
Given, A = [2 -3 4]
B = \(\left[\begin{array}{l}
3 \\
2 \\
2
\end{array}\right]\)

x = [1 2 3],
Y = \(\left[\begin{array}{l}
2\\
3\\
4
\end{array}\right]\)

AB + XY = [2 -3 4] \(\left[\begin{array}{l}
3 \\
2 \\
2
\end{array}\right]\) + [1 2 3] \(\left[\begin{array}{l}
3 \\
2 \\
2
\end{array}\right]\)

= [6 – 6 + 8] + [2 + 6 + 12]
= [8] + [20]
= [28]

Matrix MCQ Chapter 3 Class 12 Question 3.
Suppose P and Q are two different matrices of order 3 x n and n x p, then the order of thc matrix P x Q is?
(A) 3 x p
(B) p x 3
(C) n x n
(D) 3 x 3
Answer:
(A) 3 x p

Matrices Class 12 MCQ Chapter 3 Question 4.
A = \(\left[a_{i}\right]_{m \times n}\) is a square matrix, if
(A) mn
(C) m = n
(D) None of these
Answer:
(C) m = n

Explanation:
It is known that a given matrix is said to be a square matrix if the number of rows is equal to the number of columns.
Therefore,
A = \(\left[a_{i j}\right]_{m \times n}\) is a square matrix, if m = n.

MCQ On Matrices Chapter 3 Question 5.
Which of the given values of x and y make the following pair of matrices equal
A = \(\left[\begin{array}{cc}
3 x+7 & 5 \\
y+1 & 2-3 x
\end{array}\right]\), \(\left[\begin{array}{cc}
0 & y-2 \\
8 & 4
\end{array}\right]\)
[y+l 2-3xj’ [8 4
(A) x = \(\frac{-1}{3}\) y = 7
(B) Not possible to find
(C) y = 7, x = \(\frac{-2}{3}\)
(D) x = \(\frac{-1}{3}\) y = \(\frac{-2}{3}\)
Answer:
(B) Not possible to find

Explanation:
It is given that
\(\left[\begin{array}{cc}
3 x+7 & 5 \\
y+1 & 2-3 x
\end{array}\right]\) = \(\left[\begin{array}{cc}
0 & y-2 \\
8 & 4
\end{array}\right]\)
Equating the corresponding elements, we get
3x + 7 = 0
⇒ x = \(\frac{-7}{3}\)
5 = y – 2
y = 7
y + 1 = 8
y = 7
2 – 3x = 4
x =\(\frac{-2}{3}\)

We find that on comparing the corresponding elements of the two matrices, we get two different values of x, which is not possible. Hence, it is not possible to find the values of x and y for which the given matrices are equal.

MCQ On Matrices Class 12 Chapter 3 Question 6.
The number of all possible matrices of order 3 x 3 with each entry 0 or 1 is:
(A) 27
(B) 18
(C) 81
(D) 512
Answer:
(D) 512

Explanation:
The given matrix of the order 3 X 3 has 9 elements and each of these elements can be either 0 or 1.
Now, each of the 9 elements can be filled in two possible ways.
Therefore,
by the multiplication principle1 the required number of possible matrices 2⁹ = 512.

Matrices MCQ Class 12 Chapter 3 Question 7.
Assume X, Y, Z, Wand P are matrices of order 2 x n, 3 x k, 2 x p. n x 3 and p x k, respectively. The restriction on n, k and p so that PY + WY will be
defined are:
(A) k = 3,p = n
(B) k is arbitrary, p = 2
(C) p is arbitrary,k = 3
(D) k = 2, p = 3
Answer:
(A) k = 3,p = n

Explanation :
Matrices P and Y are of the orders p X k and 3 X k, respectively. Therefore, matrix PY will be defined if k = 3. ConsequentlY, PY will be of the order p X k. Matrices Wand Yare of the orders n x 3 and 3 X k respectively. Since the number of columns in W is equal to the number of rows in Y, matrix WY is well-defined and is of the order n X k. Matrices PY and WY can be added only when their orders are the same. However, J’Y is of the order p X k and WY is of the order n X k. Therefore, we must have p = n. Thus, k = 3 and p = n are the restrictions on n, k, andpsothat PY + Wywilibedefined.

Matrices MCQ Questions Class 12 Chapter 3 Question 8.
Assume X, Y, Z, W and P are matrices of order 2 X n, 3 X k, 2 X p, n X 3 and p X k, respectively. 1f n = p, then the order of the matrix 7X – 5Z is:
(A) p x 2
(B) 2 x n
(C) n x 3
(D) p x n
Answer:
(B) 2 x n

Explanation:
Matrix X is of the order 2 X n. Therefore, matrix 7X is also of the same order. Matrix Z is of the order 2 X p, i.e., 2 X n [Since n = p] Therefore, matrix 5Z is also of the same order. Now, both the matrices 7X and 5Z are of the order 2 X n. Thus, matrix 7X – 5Z is well- defined and is of the order 2 X n.

Matrix MCQ Questions Chapter 3 Question 9.
If A = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\), then A + A’ = I, then the value of α is:

(A) \(\frac {π}{6}\)
(B) \(\frac {π}{3}\)
(C) π
(D) \(\frac {3π}{2}\)
Answer:
(B) \(\frac {π}{3}\)

Explanation:
Given that, A = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)
Also A + A’ = I
⇒ \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]+\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2 \cos \alpha & 0 \\
0 & 2 \cos \alpha
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Equating corresponding entries, we have
⇒ 2 cos α = 1
⇒ cos α = \(\frac {1}{2}\)
⇒ cos α = ⇒ cos \(\frac {π}{3}\)
∴ α = \(\frac {π}{3}\)

Matrix Class 12 MCQ Maths Question 10.
Matrices A and B will be inverse of each other only if
(A) AB = BA
(B) AB =BA0
(C)AB = 0, BA = 1
(D)AB = BA =1
Answer:
(D)AB = BA =1

Explanation:
We know that if A is a square matrix of order m, and if there exists another square matrix B of the same order m, such that AB = BA = I, then B is said to be the inverse of A. In this case, it is clear that A is the inverse of B. Thus, matrices A and B will be the inverse of each other only if AB = BA = 1.

Class 12 Maths Chapter 3 MCQ Question 11.
The matrix P = \(\left[\begin{array}{lll}
0 & 0 & 4 \\
0 & 4 & 0 \\
4 & 0 & 0
\end{array}\right]\) is a
(A) square matrix
(B) diagonal matrix
(C) unit matrix
(D) None of these
Answer:
(A) square matrix

Explanation:
We know that, in a square matrix number of rows is equal to the number of columns. So, the matrix P = \(\left[\begin{array}{lll}
0 & 0 & 4 \\
0 & 4 & 0 \\
4 & 0 & 0
\end{array}\right]\) is a square matrix.

MCQ On Matrices Class 12 Pdf Chapter 3 Question 12.
If A and Bare symmetric matrices of same order, then AB – BA is a:
(A) Skew-symmetric matrix
(B) Symmetric matrix
(C) Zero matrix
(D) Identity matrix
Answer:
(A) Skew-symmetric matrix

Explanation :
Given that, A and B are symmetric matrices.
⇒ A = A’ and B = B’
Now, (AB – BA)’ = (AB)’ – (BA)’ ………..(i)
(AB – BA)’ = B’A’ – A’B’
(AB-BA)’ = BA – AB [From Equestion (1)]
(AB-BA)’ = -(AB – BA)
= (AB – BA) is a skew-symmetric matrix.

Class 12 Matrices MCQ Chapter 3 Question 13.
1f the matrix A is both symmetric and skew- symmetric1 then
(A) A is a diagonal matrix
(B) A is a zero matrix
(C) A is a square matrix
(D) None of these
Answer:
(B) A is a zero matrix

Explanation:
If A is both symmetric and skew symmetric then we have,
A’ = A and A’ = -A
⇒ A = – A
⇒ A + A = O
⇒ 2A = O
⇒ A = O
Therefore, A is a zero matrix.

MCQ On Matrices Class 12 Pdf Download Chapter 3 Question 14.
The matrix \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 4
\end{array}\right]\) is a
(A) identity matrix
(B) symmetric matrix
(C) skew-symmetric matrix
(D) None of these
Answer:
(B) symmetric matrix

Explanation:
A = \(=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 4
\end{array}\right]\)
∴ A’ = \(=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 4
\end{array}\right]\) = A
So, the given matrix is a symmetric matrix. [Since, in a square matrix A, if A’ = A, then A is called symmetric matrix.]

Matrix MCQ Class 12 Chapter 3 Question 15.
The matrix \(\left[\begin{array}{ccc}
0 & -5 & 8 \\
5 & 0 & 12 \\
-8 & -12 & 0
\end{array}\right]\) is a
(A) diagonal matrix
(B) symmetric matrix
(C) skew symmetric matrix
(D) scalar matrix
Answer:
(C) skew symmetric matrix

Explanation:
We know that, in a square matrix, if b_ij = O when i ≠ j then it is said to be a diagonal matrix. Here, b₁₂, b₁₃…. ≠ 0 so the given matrix is not a diagonal matrix.
Now, B = \(\left[\begin{array}{ccc}
0 & -5 & 8 \\
5 & 0 & 12 \\
-8 & -12 & 0
\end{array}\right]\)
B’ = \(\left[\begin{array}{ccc}
0 & 5 & -8 \\
-5 & 0 & -12 \\
8 & 12 & 0
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
0 & -5 & 8 \\
5 & 0 & 12 \\
-8 & -12 & 0
\end{array}\right]\)
= -B
So, the given matrix is a skew – symmetric matrix, since we know that in a square matrix B, if B’ = -B, then it is called skew-symmetric matrix.

MCQ Of Matrix Class 12 Chapter 3 Question 16.
If A is a matrix of order n x n and Bisamatrixsuch that AB’ and BA are both defined, then the order of matrix B is …….
(A) m x m
(B) n x n
(C) n x m
(D) m x n
Answer:
(D) m x n

Explanation:
Let, A = \(\left[a_{i j}\right]\)_{m x n} and B = \(\left[b_{i j}\right]\)_{p x q}
∴ B = \(\left[b_{i j}\right]\)_{p x q}
Now,AB’ is deflned, son = q and B’A is also defined, so p = m
∴ Order of B’ = \(\left[b_{i j}\right]\)_{n x m}
And order of B = \(\left[b_{i j}\right]\)_{m x n}

Assertion And Reason Based MCQs (1 Mark each)

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is True

MCQ On Matrix Maths Chapter 3 Question 1.
Assertion (A): If A is a square matrix such that A2 = A, then (I + A)² – 3A = I
Reason (R): AI = IA = A
Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
AI = IA = A is true.
Hence R is true.
Given A² = A,
(I + A)² – 3A = I² + 2IA + A² – 3A = I + 2A + A – 3A = I
Hence A is true.
R is the correct explanation for A.

Matrices MCQ With Answers Pdf Chapter 3 Question 2.
Assertion (A): \(\left[\begin{array}{lll}
7 & 0 & 0 \\
0 & 7 & 0 \\
0 & 0 & 7
\end{array}\right]\) is a scalar matrix.
Reason (R): If all the elements of the principal diagonal are equal, it is called a scalar matrix.
Answer:
(C) A is true but R is false

Explanation:
In a scalar matrix the diagonal elements are equal and the non-diagonal elements are zero. Hence R is false.
A is true since the diagonal elements are equal and the non-diagonal elements are zero.

MCQ Of Chapter 3 Maths Class 12 Question 3.
Assertion (A): (A + B)² ≠ A² + 2AB + B².
Reason (R): Generally AB ≠ BA
Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
For two matrices A and B, generally AB ≠ BA. i.e., matrix multiplication is not commutative.
∴ R is true
(A + B)² = (A + B)(A + B) ?
= A² + AB + BA + B²
≠ A² + 2AB + B²
∴ A is true
R is the correct explanation for A.

Question 4.
A and B are two matrices such that both AB and BA are defined.
Assertion (A): (A + B)(A – B) = A² – B²
Reason(R): (A + B)(A – B) = A² – AB + BA – B²
Answer:
(D) A is false and R is True

Explanation:
For two matrices A and B, even if both AB and BA are defined, generally AB ≠ BA.
(A + B)(A – B) = A² – AB + BA – B².
Since AB ≠ BA, (A + B) (A – B) ≠ A² – B².
Hence R is true and A is false.

Question 5.
Let A and B be two symmetric matrices of order 3. Assertion (A): A(BA) and (AB)A are symmetric matrices.
Reason (R): AB is symmetric matrix if matrix multiplication of A with B is commutative.
Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
Generally (AB)’ = B’ A’
If AB = BA, then (AB)’ = (BA)’ =A’ B’ = AB Since (AB)’ = AB, AB is a symmetric matrix. Hence R is true.
A(BA) = (AB)A = ABA (ABA)’ = A’ B’ A’ = ABA.
A(BA) and (AB)A are symmetric matrices.
Hence A is true.
But R is not the correct explanation for A.

Question 6.
Assertion (A): If the matrix P = \(\left[\begin{array}{ccc}
0 & 2 b & -2 \\
3 & 1 & 3 \\
3 a & 3 & 3
\end{array}\right]\) is a symmetric matrix1 then a = \(\frac {-2}{3}\) and b = \(\frac {3}{2}\)
Reason (R): If P is a symmetric matrix, then P’ = P.
Answer:
(C) A is true but R is false

Explanation:
If P is a symmetric matrix, then P’ = P.
Hence R is false.
As P is a symmetric matrix,P’ = P
∴ \(\left[\begin{array}{ccc}
0 & 3 & 3 a \\
2 b & 1 & 3 \\
-2 & 3 & -1
\end{array}\right]\) = \(\left[\begin{array}{ccc}
0 & 2 b & -2 \\
3 & 1 & 3 \\
3 a & 3 & -1
\end{array}\right]\)
∴ By equality of matrices, a = \(\frac {-2}{3}\) and b = \(\frac {-3}{2}\)
Hence A is true.

Question 7.
Assertion (A): If A is a symmetric matrix, then B’AB is also symmetric.
Reason (R): (ABC)’ = C’B’A’
Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
For three matrices A, B and C, if ABC is defined then (ABC)’ = C’B’A’.
Hence R is true.
Given that A is symmetric
⇒ A’ = A (B’AB)’ = B’A'(B’)’ = B’AB.
Hence A is true.
R is the correct explanation for A.

Question 8.
Assertion (A): If A and B are symmetric matrices, then AB – BA is a skew symmetric matrix
then AB – BA is a skew symmetric mathx
Reason (R): (AB)’ = B’A’
Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
(AB)’ = B’A’
⇒R is true.
Given that A and B are symmetric matrices.
∴ A’ = A and B’ = B
(AB – BA)’ = (AB)’ – (BA)’
= B’A’-A’B’ = BA – AB
Since (AB – BA)’ = -(AB – BA),
AB – BA is skew symmetric.
Hence A is true.
R is the correct explanation for A.

Case-Based MCQs

Attempt any four sub-parts from each question. Each sub-part carries 1 mark.

I. Read the following text and answer the following questions on the basis of the same:

A manufacture produces three stationery products Pencil, Eraser and Sharpener which he sells in two markets. Annual sales are indicated below

If the unit Sale price of Pencil, Eraser and Sharpener are ₹2.50, ₹1.50 and ₹1.00 respectively, and unit cost of the above three commodities are ₹2.00, ? 1.00 and ₹ 0.50 respectively, then,

Question 1.
Total revenue of market A
(A) ₹ 64,000
(B) ₹ 60,400
(C) ₹ 46,000
(D) ₹ 40600
Answer:
(C) ₹ 46,000

Explanation:
Total revenue of
=[10,000 2,000 18,000] \(\left[\begin{array}{l}
2.50 \\
1.50 \\
1.00
\end{array}\right]\)
= [2.50 x 10,000 + 1.50 x 2,000 + 1.00 x 18,0001
= [46,000]

Question 2.
Total revenue of market B
(A) ₹ 35,000
(B) ₹ 53,000
(C) ₹ 50,300
(D) ₹ 30,500
Answer:
(B) ₹ 53,000

Explanation:
Total revenue of market B
= [6,000 20,000 8,000] \(\left[\begin{array}{l}
2.50 \\
1.50 \\
1.00
\end{array}\right]\)
= [2.50 x 6,000 + 1.50 x 20,000 + 1.00 x 8,000]
= [53,000]

Question 3.
Cost incurred in market A
(A) ₹ 23,000
(B) ₹ 20,300
(C) ₹ 32,000
(D) ₹ 30,200
Answer:
(C) ₹ 32,000

Explanation:
Cost incurred in market A
= [10,000 2,000 18,000] \(\left[\begin{array}{l}
2.00 \\
1.00 \\
0.50
\end{array}\right]\)
= [2.00 x 10,000+1.00 x 2,000+ 0.50 x 18,000]
= [31,000]

Question 4.
Profits in market A and B respectively are
(A) (₹ 15,000, ₹ 17,000)
(B) (₹ 17,000, ₹ 15,000)
(c) (₹ 51,000, ₹ 71,000)
(D) (₹ 10,000, ₹ 20,000)
Answer:
(A) (₹ 15,000, ₹ 17,000)

Explanation:
Cost incurred in market B
=[6,000 20,000 8,000] \(\left[\begin{array}{l}
2.00 \\
1.00 \\
0.50
\end{array}\right]\)
0.50
=[2.00 x 6,000 + 1.00 x 20,000 +0.50 x 8,000]
= [36,000]
Profit of market A & B = total revenue of A and B – Cost increased in market A and B
\(\left[\begin{array}{l}
A \\
B
\end{array}\right]\) = \(\left[\begin{array}{l}
46,000 \\
50,000
\end{array}\right]\) – \(\left[\begin{array}{l}
31,000 \\
36,000
\end{array}\right]\)
= \(\left[\begin{array}{l}
15,000 \\
17,000
\end{array}\right]\)
i.e., (₹ 15,000, ₹ 17,000)

Question 5.
Gross profit in both markets
(A) ₹ 23,000
(B) ₹ 20,300
(C) ₹ 32,000
(D) ₹ 30,200
Answer:
(C) ₹ 32,000

Explanation:
Gross profit in both markets = Profit in A + Profit in B
= 15,000 + 17,000
= ₹ 32,000

II. Read the following text and answer the following questions on the basis of the same:
Amit, Biraj and Chirag were given the task of creating a square matrix of order 2.
Below are the matrices created by them. A, B, C are the matrices created by Amit, Biraj and Chirag respectively.
A = \(\left[\begin{array}{rr}
1 & 2 \\
-1 & 3
\end{array}\right]\)
B = \(\left[\begin{array}{ll}
4 & 0 \\
1 & 5
\end{array}\right]\)
C = \(\left[\begin{array}{rr}
2 & 0 \\
1 & -2
\end{array}\right]\)
if a = 4 and b = -2

Question 1.
Sum of the matrices A, B and C, A+(B + C) is
(A) \(\left[\begin{array}{ll}
1 & 6 \\
2 & 7
\end{array}\right]\)
(B) \(\left[\begin{array}{ll}
6 & 1 \\
7 & 2
\end{array}\right]\)
(C) \(\left[\begin{array}{ll}
7 & 2 \\
1 & 6
\end{array}\right]\)
(D) \(\left[\begin{array}{ll}
2 & 1 \\
7 & 6
\end{array}\right]\)
Answer:
(C) \(\left[\begin{array}{ll}
7 & 2 \\
1 & 6
\end{array}\right]\)

Explanation:
A + (B + C) = \(\left[\begin{array}{ll}
1 & 2 \\
-1 & 3
\end{array}\right]\) + \(\left(\left[\begin{array}{ll}
4 & 0 \\
1 & 5
\end{array}\right]+\left[\begin{array}{ll}
2 & 0 \\
1 & -2
\end{array}\right]\right)\)
= \(\left[\begin{array}{ll}
1 & 2 \\
-1 & 3
\end{array}\right]+\left[\begin{array}{ll}
6 & 0 \\
2 & 3
\end{array}\right]\)
= \(\left[\begin{array}{ll}
7 & 2 \\
1 & 6
\end{array}\right]\)

Question 2.
(A^T)^T is equal to
(A) \(\left[\begin{array}{rr}
1 & 2 \\
-1 & 3
\end{array}\right]\)
(B) \(\left[\begin{array}{rr}
2 & 1 \\
3 & -1
\end{array}\right]\)
(C) \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\)
(D) \(\left[\begin{array}{rr}
2 & 3 \\
-1 & 1
\end{array}\right]\)
Answer:
(C) \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\)

Explanation:
(A^T) = \(\left[\begin{array}{ll}
1 & -1 \\
2 & 3
\end{array}\right]\)
(A^T)^T = \(\left[\begin{array}{ll}
1 & 2 \\
-1 & 3
\end{array}\right]\)

Question 3.
(bA)^T is equal to
(A) \(\left[\begin{array}{rr}
-2 & -4 \\
2 & -6
\end{array}\right]\)
(B) \(\left[\begin{array}{rr}
-2 & 2 \\
-4 & -6
\end{array}\right]\)
(C) \(\left[\begin{array}{rr}
-2 & 2 \\
-6 & -4
\end{array}\right]\)
(D) \(\left[\begin{array}{rr}
-6 & -2 \\
2 & 4
\end{array}\right]\)
Answer:
(B) \(\left[\begin{array}{rr}
-2 & 2 \\
-4 & -6
\end{array}\right]\)

Explanation:
bA = -2A = \(=\left[\begin{array}{ll}
-2 & -4 \\
2 & -6
\end{array}\right]\)
(bA )^T = \(\left[\begin{array}{ll}
-2 & 2 \\
-4 & -6
\end{array}\right]\)

Question 4.
AC – BC is Equal to
(A) \(\left[\begin{array}{rr}
-4 & -6 \\
-4 & 4
\end{array}\right]\)
(B) \(\left[\begin{array}{rr}
-4 & -4 \\
4 & -6
\end{array}\right]\)
(C) \(\left[\begin{array}{rr}
-4 & -4 \\
-6 & 4
\end{array}\right]\)
(D) \(\left[\begin{array}{rr}
-6 & 4 \\
-4 & -4
\end{array}\right]\)
Answer:
(C) \(\left[\begin{array}{rr}
-4 & -4 \\
-6 & 4
\end{array}\right]\)

Explanation:
AC = \(\left[\begin{array}{ll}
1 & 2 \\
-1 & 3
\end{array}\right]\left[\begin{array}{ll}
2 & 0 \\
1 & -2
\end{array}\right]\)
= \(\left[\begin{array}{ll}
4 & -4 \\
1 & -6
\end{array}\right]\)
BC = \(\left[\begin{array}{ll}
4 & 0 \\
1 & 5
\end{array}\right]\left[\begin{array}{ll}
2 & 0 \\
1 & -2
\end{array}\right]\)
= \(\left[\begin{array}{ll}
8 & 0 \\
7 & -10
\end{array}\right]\)
AC – BC = \(\left[\begin{array}{ll}
4 & -4 \\
1 & -6
\end{array}\right]-\left[\begin{array}{ll}
8 & 0 \\
7 & -10
\end{array}\right]\)
= \(\left[\begin{array}{ll}
-4 & -4 \\
-6 & 4
\end{array}\right]\)

Question 5.
(a + b)B is equal to
(A) \(\left[\begin{array}{rr}
0 & 8 \\
10 & 2
\end{array}\right]\)
(B) \(\left[\begin{array}{rr}
2 & 10 \\
8 & 0
\end{array}\right]\)
(C) \(\left[\begin{array}{rr}
8 & 0 \\
2 & 10
\end{array}\right]\)
(D) \(\left[\begin{array}{rr}
2 & 0 \\
8 & 10
\end{array}\right]\)
Answer:
(C) \(\left[\begin{array}{rr}
8 & 0 \\
2 & 10
\end{array}\right]\)

Explanation:
(a + b)B = (4 – 2)\(\left[\begin{array}{ll}
4 & 0 \\
1 & 5
\end{array}\right]\)
= \(\left[\begin{array}{ll}
8 & 0 \\
2 & 10
\end{array}\right]\)

III. Read the following text and answer the following questions on the basis of the same:
Three schools DPS, CVC and KVS decided to organize a fair for collecting money for helping the flood victims. They sold handmade fans, mats and plates from recycled material at a cost of f25, 100 and 50 each respectively. The numbers of articles sold are given as

Question 1.
What is the total money (in Rupees) collected by the school DPS?
(A) 700
(B) 7,000
(C) 6,125
(D) 7,875
Answer:
(B) 7,000

Explanation:
The funds collected by the schools can be obtained by matrix multiplication:
\(\left[\begin{array}{lll}
40 & 50 & 20 \\
25 & 40 & 30 \\
25 & 50 & 40
\end{array}\right]\left[\begin{array}{l}
25 \\
100 \\
50
\end{array}\right]=\left[\begin{array}{l}
7000 \\
6125 \\
7875
\end{array}\right]\)
Funds collected by school DPS = 7000
Funds collected by school, CVC = 6125
Funds collected by school KVS = 7875

Question 2.
What is the total amount of money (in ₹) collected by schools CVC and KVS?
(A) 14,000
(B) 15,725
(C) 21,000
(D) 13,125
Answer:
(A) 14,000

Explanation:
Total amount of money collected by school
= 6125 + 7875
= 14000

Question 3.
What is the total amount of money collected by all three schools DPS, CVC and KVS?
(A) ₹ 15,775
(B) ₹ 14,0O0
(C) ₹ 21,O00
(D) ₹ 17,125
Answer:
(C) ₹ 21,O00

Explanation:
Total amount of money collected by all school DI’S, CVC and KVS
= 7000 + 7875 + 6125
= 21000

Question 4.
if the number of handmade fans and plates are interchanged for all the schools, then what is the total money collected by all schools?
(A) ₹ 18,000
(B) ₹ 6,750
(C) ₹ 5,000
(D) ₹ 21,250
Answer:
(D) ₹ 21,250

Question 5.
How many articles (in total) are sold by three schools?
(A) 230
(B) 130
(C) 430
(D) 330
Answer:
(D) 330

IV. Read the following text and answer the following questions on the basis of the same:

On her birthday, Seema decided to donate some money to children of an orphanage home. If there were 8 children less, everyone would have got ₹ 10 more. However, if there were 16 children more, everyone would have got ₹ 10 less. Let the number of children be x and the amount distributed by Seema for one child be y (in ₹).

Question 1.
The equations in terms of x and y are

Explanation:
Given equations are;
(A) 5x – 4 y = 40
5x – 8y = -80

(B) 5x – 4y = 40
5x – 8y = 80

(C) 5 x- 4y =4 0
5x + 8y = -80

(D) 5x + 4y = 40
5x -8y = -80
Answer:
(A) 5x – 4 y = 40
5x – 8y = -80

Explanation:
According to question,
(x – 8)(y + 10) = xy
xy + 10x – 8y – 80 = xy
5x – 4y = 40 ………(i)
and (x + 16)(y – 10) = xy
xy – 10x + 16y – 160 = xy
5x – 8y = -80 ……….(ii)

Question 2.
Which of the following matrix equations represent the information given above?
(A) \(\left[\begin{array}{ll}
5 & 4 \\
5 & 8
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{r}
40 \\
-80
\end{array}\right]\)
(B) \(\left[\begin{array}{ll}
5 & -4 \\
5 & -8
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
40 \\
80
\end{array}\right]\)
(C) \(\left[\begin{array}{ll}
5 & -4 \\
5 & -8
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{r}
40 \\
-80
\end{array}\right]\)
(D) \(\left[\begin{array}{rr}
5 & 4 \\
5 & -8
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{r}
40 \\
-80
\end{array}\right]\)
Answer:
(C) \(\left[\begin{array}{ll}
5 & -4 \\
5 & -8
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{r}
40 \\
-80
\end{array}\right]\)

Explanation:
Give equestion are
5 x – 4y = 4 0
5x + 8y = -80
\(\left[\begin{array}{ll}
5 & -4 \\
5 & -8
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
40 \\
-80
\end{array}\right]\)

Question 3.
The number of children who were given some money by Seema, is
(A) 30
(B) 40
(C) 23
(D) 32
Answer:
(D) 32

Explanation:
Since,
\(\left[\begin{array}{ll}
5 & -4 \\
5 & -8
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
40 \\
-80
\end{array}\right]\)
Let A = \(\left[\begin{array}{ll}
5 & -4 \\
5 & -8
\end{array}\right]\), x = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\)
B = \(\left[\begin{array}{c}
40 \\
-80
\end{array}\right]\)
∴ AX = B
X = A²B
|A| = 5(-8) – (-4) x 5 = -40 + 20 = -20
adj (A) = \(\left[\begin{array}{cc}
-8 & -5 \\
4 & 5
\end{array}\right]^{T}\)
= \(\left[\begin{array}{ll}
-8 & 4 \\
-5 & 5
\end{array}\right]\)

A^-1 = \(\frac{1}{-20}\left[\begin{array}{ll}
-8 & 4 \\
-5 & 5
\end{array}\right]\)
= \(=\left[\begin{array}{cc}
\frac{2}{5} & \frac{-1}{5} \\
\frac{1}{4} & \frac{-1}{4}
\end{array}\right]\)
X = A^-1B
\(\left[\begin{array}{cc}
\frac{2}{5} & \frac{-1}{5} \\
\frac{1}{4} & \frac{-1}{4}
\end{array}\right]\left[\begin{array}{l}
40 \\
-80
\end{array}\right]\)
\(\left[\begin{array}{l}
\frac{2}{5} \times 40-\frac{1}{5} \times(-80) \\
\frac{1}{4} \times 40-\frac{1}{4} \times(-80)
\end{array}\right]\)
\(\left[\begin{array}{l}
16+16 \\
10+20
\end{array}\right]\)
\(\left[\begin{array}{l}
32 \\
30
\end{array}\right]\)
∴ \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
32 \\
30
\end{array}\right]\)
x = 32
y = 30
Hence, 32 children were given some money by Seema.

Question 4.
How much amount is given to each child by Seema?
(A) ₹ 32
(B) ₹ 30
(C) ₹ 62
(D) ₹ 26
Answer:
(B) ₹ 30

Explanation:
₹ 30 is given to each child by Seema [∴y = 30]

Question 5.
How much amount Seema spends in distributing the money to all the students of the Orphanage?
(A) ₹ 609
(B) ₹ 960
(C) ₹ 906
(D) ₹ 690
Answer:
(B) ₹ 960

Explanation:
Total amount Seema spends in distributing the money to all the students of the orphanage
= x x y = 32 x 30 = ₹ 960

V. Read the following text and answer the following questions on the basis of the same:

Two farmers Ramakishan and Gurucharan Singh cultivate only three varieties of rice namely Basmati, Permal and Naura. The sale (in ?) of these varieties of rice by both the farmers in the month of September and October are given by the following matrices A and B

September sales (in Rupees)

October sales (in Rupees)

Question 1.
The total sales in September and October for each farmer in each variety can be represented as ………….
(A) A + B
(B) A – B
(C) A > B
(D) A < B
Answer:
(A) A + B

Explanation:
Combined sales in September and October for each farmer in each variety is given by
A + B =

Question 2.
What is the value of A₂₃
(A) 10,000
(B) 20,00
(C) 30,000
(D) 40,00
Answer:
(A) 10,000

Explanation:
A₂₃ = 10,000

Question 3.
The decrease in sales from September to October is given by ………..
(A) A + B
(B) A – B
(C) A > B
(D) A < B
Answer:
(B) A – B

Explanation:
Change in sales from September to October is given by A – B =

Question 4.
If Ramkishan receives 2% profit on gross sales, compute his profit for each variety sold in October.
(A) ₹ 100, ₹ 200 and ₹ 120
(B) ₹ 100, ₹ 200 and ₹ 130
(C) ₹ 100, ₹ 220 and ₹ 120
(D) ₹ 110, ₹ 200 and ₹120
Answer:
(A) ₹ 100, ₹ 200 and ₹ 120

Explanation:
2% of B = \(\frac{2}{100} \times \mathrm{B}\)
= 0.02 x B = 0.02

Thus, in October Ramkishan receives ₹ 100, ₹ 200 and ₹ 120 as profit in the sale of each variety of rice, respectively.

Question 5.
If Gurucharan receives 2% profit on gross sales, compute his profit for each variety sold in September.
(A) ₹ 100, ₹ 200, ₹ 120
(B) ₹ 1000, ₹ 600, ₹ 200
(C) ₹ 400, ₹ 200, ₹ 120
(D) ₹ 1200, ₹ 200, ₹ 120
Answer:
(B) ₹ 1000, ₹ 600, ₹ 200

Explanation:
2% of A = \(\frac {2}{100}\) x A
= 0.02 x A
0.02

Thus, in September Gurucharan receives ₹ 1000, ₹ 600 and ₹ 200 as Profit in the sale of each variety of rice, respectively.

VI. Read the following text and answer the following questions on the basis of the same:

There are three families A, B and C. The number of members in these families are given in the table below.

The daily expenses of each man, woman and child are respectively ₹ 200, ₹ 100 and ₹ 50.

Question 1.
The total daily expense of family A is …………..
(A) 850
(B) 900
(C) 1,200
(D) 2,950
Answer:
(A) 850

Explanation:
Expense of family A
= [3 2 1] \(\left[\begin{array}{c}
200 \\
100 \\
50
\end{array}\right]\) = [850]

Question 2.
The totaL daily expense of family C is ……………
(A) 850
(B) 900
(C) 1,200
(D) 2,950
Answer:
(C) 1,200

Explanation:
Expense of family C
= [4 3 2] \(\left[\begin{array}{c}
200 \\
100 \\
50
\end{array}\right]\) = [1200]

Question 4.
The family with highest expense is ……………
(A) A
(B) B
(C) C
(D) All have same expense
Answer:
(D) All have same expense

Explanation:
Most expensive family is C with I an expense of ₹ 1200.

Question 5.
The combined expense of men in family A and children in family C is …………..
(A) 600
(B) 700
(C) 800
(D) 900
Answer:
(B) 700

Explanation:
[3 2] \(\left[\begin{array}{c}
200 \\
50
\end{array}\right]\) = [700]

VII. Read the following text and answer the following questions on the basis of the same:

Three schools SNT, SNP and TKM organised a mela for collecting funds for helping the rehabilitation of flood victims. They sold hand-made fans, mats and plates from recycled material at a cost of ₹ 25, ₹ 100 and ₹ 50 each. The number of articles sold are given below.

Question 1.
Funds collected by SNT is …………..
(A) ₹ 7000
(C) ₹ 7875
(B) ₹ 6125
(D) ₹ 21,000
Answer:
(A) ₹ 7000

Explanation:
Funds raised by SNT = ₹ 7000
= [40 50 20] \(\left[\begin{array}{c}
25 \\
100 \\
50
\end{array}\right]\) =[7,000]

Question 2.
Funds collected by SNP is ………….
(A) ₹ 7000
(B) ₹ 6125
(C) ₹ 7875
(D) ₹ 21,000
Answer:
(B) ₹ 6125

Explanation:
= [25 40 30] \(\left[\begin{array}{c}
25 \\
100 \\
50
\end{array}\right]\) = [6,125]

Question 3.
The total fund raised by all the three schools together is …………
(A) ₹ 7000
(C) ₹ 7875
(B) ₹ 6125
(D) ₹ 21,000
Answer:
(D) ₹ 21,000

Explanation:
[25 40 30] \(\left[\begin{array}{c}
25 \\
100 \\
50
\end{array}\right]\) = [7,875]
Fund raised by TKM = 7,875
Total a amount raised = 7000 + 6,125 + 7,875 = ₹ 21,000

Question 4.
The total fund raised by seLling fans is ………….
(A) ₹ 4000
(B) ₹ 2,5oo
(C) ₹ 2,ooo
(D) ₹ 35,000
Answer:
(C) ₹ 2,ooo

Explanation:
[40 + 25 + 35][ 25] = [2,500]
Fund raised by selling fans = ₹ 2,500

Question 5.
TKM collected ₹ ……………. by selling plates.
(A) 4000
(B) 2000
(C) 2500
(D) 3500
Answer:
(B) 2000

Explanation:
[40] [50] = [2000]
TKM collected ₹ 2000 by selling plates.

MCQ Questions for Class 12 Maths with Answers