Matrices MCQ AP Inter 2nd Year Maths Chapter 3

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Practice AP Inter 2nd Year Maths Study Material Chapter 3 Matrices MCQ to identify your strengths and weak areas.

AP Inter 2nd Year Maths Matrices MCQ

I. Select the correct option from the given choices.

Question 1.
A = [aij]m×n is a square matrix, if
1) m < n
2) m > n
3) m = n
4) m = n2
Solution:
3) m = n
In a square matrix, the number of rows is equal to the number of columns.
∴ A = [aij]m×n is a square matrix, if m = n.

Question 2.
Which of the given values of x and y make the following pair of matrices equal \(\left[\begin{array}{cc}
3 x+2 & 5 \\
y+1 & 2-3 x
\end{array}\right]=\left[\begin{array}{cc}
0 & y-2 \\
8 & 4
\end{array}\right]\)
1) x = \(\frac{-1}{3}\), y = 7
2) y = 7, x = \(\frac{2}{3}\)
3) y = 7, x = \(\frac{-2}{3}\)
4) x = \(\frac{-1}{3}\), y = \(\frac{-2}{3}\)
Solution:
3) y = 7, x = \(\frac{-2}{3}\)
Equating the corresponding elements, in the given matrices, we get
3x + 2 = 0 ⇒ 3x = -2 ⇒ x = \(\frac{-2}{3}\); 2 – 3x = 4 ⇒ x = \(\frac{-2}{3}\)
y – 2 = 5 ⇒ y = 7; y + 1 = 8 ⇒ y = 7

Matrices MCQ AP Inter 2nd Year Maths Chapter 3

Question 3.
The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:
1) 27
2) 18
3) 81
4) 512
Solution:
4) 512
Matrix of the order 3 × 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.
Hence, by the multiplication principle, the required number of possible matrices is 29 = 512.

Question 4.
Assume Y, W, P are matrices of order 3 × k, n ×3, p × k respectively.
The restriction on n, k and p so that PY + WY will be defined are:
1) k = 3, p = n
2) k is arbitrary, p = 2
3) p is arbitrary, k = 3
4) k = 2, p = 3
Solution:
1) k = 3, p = n
\(\underset{p \times k}{P} \underset{3 \times k}{Y}+\underset{n \times 3}{W} \underset{3 \times k}{Y}\) has to be deflned
i) If k = 3 then PY is defined (PYp×k)
ii) If n = p then WY is defined (WYp×k )
⇒ PY + WY (addition of matrices of same order) will be defined when k = 3, p = n.

Matrices MCQ AP Inter 2nd Year Maths Chapter 3

Question 5.
Assume X, Z are matrices of order 2 × n, 2 × p respectively.
If n = p, then the order of the matrix 7X – 5Z is
1) p ×2
2) 2 × n
3) n × 3
4) p × n
Solution:
2) 2 × n
Given X2×n ⇒ order of 7X is 2 × n
Given Z2×p ⇒ order of 5Z is 2 × p.
Also p = n ⇒ order of 7X – 5Z is 2 × p (or) 2 × n

Question 6.
If A, B are symmetric matrices of same order, then AB – BA is a
1) Skew symmetric matrix
2) Symmetric matrix
3) Zero matrix
4) Identity matrix
Solution:
1) Skew symmetric matrix
If A, B are symmetric matrices of same order, then A’ =A and B’ = B ………(1)
Now consider(AB – BA)’ = (AB)’ – (BA)’ [∵ (A – B)’ = A’ – B’ ]
=B’A’ – A’B’ [∵ (AB)’ = B’A’]
= BA – AB [from(1)] = -(AB – BA) ∴ (AB – BA)’ = -(AB – BA)
Thus, AB – BA is a skew symmetric matrix.

Matrices MCQ AP Inter 2nd Year Maths Chapter 3

Question 7.
If A = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) and A + A’ = I, then the value of α is
1) \(\frac{\pi}{6}\)
2) \(\frac{\pi}{3}\)
3) π
4) \(\frac{3\pi}{2}\)
Solution:
2) \(\frac{\pi}{3}\)
Given that A = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right] \Rightarrow A^{\prime}=\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\)
Now, A + A’ = I
∴ \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]+\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \Rightarrow\left[\begin{array}{cc}
2 \cos \alpha & 0 \\
0 & 2 \cos \alpha
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Equating the corresponding elements of the two matrices, we get
2cosα = 1 ⇒ cos α = \(\frac{1}{2}\) ⇒ α = cos-1\(\left(\frac{1}{2}\right)\) ⇒ α = \(\frac{\pi}{3}\)

Question 8.
Matrices A and B will be inverse of each other only if
1) AB = BA
2) AB = BA = O
3) AB = O, BA = I
4) AB = BA = I
Solution:
4) AB = BA = I
From the definition of Inverse of a matrix, two matrices A and B are inverses of each other only when AB = BA = I.

Matrices MCQ AP Inter 2nd Year Maths Chapter 3

Question 9.
If A = \(\left[\begin{array}{cc}
\alpha & \beta \\
-\gamma & \alpha
\end{array}\right]\) is such that A2 = I, then
1) 1 + α2 + βγ = 0
2) 1 – α2 + βγ = 0
3) 1 – α2 – βγ = 0
4) 1 + α2 – βγ = 0
Solution:
3) 1 – α2 – βγ = 0
Given that A = \(\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]\)
A2 = A.A = \(\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]=\left[\begin{array}{cc}
\alpha^2+\beta \gamma & \alpha \beta-\alpha \beta \\
\alpha \gamma-\alpha \gamma & \beta \gamma+\alpha^2
\end{array}\right]=\left[\begin{array}{cc}
\alpha^2+\beta \gamma & 0 \\
0 & \beta \gamma+\alpha^2
\end{array}\right]\)
Now A2 = I. Hence \(\left[\begin{array}{cc}
\alpha^2+\beta \gamma & 0 \\
0 & \beta \gamma+\alpha^2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Equating the corresponding elements, we get
α2 + βγ = 1 ⇒ α2 + βγ – 1 = 0 ⇒ 1 – α2 – βγ = 0

Question 10.
If the matrix A is both symmetric and skew symmetric, then
1) A is a diagonal matrix
2) A is a zero matrix
3) A is a square matrix
4) None of these
Solution:
2) A is a zero matrix
Given that the matrix A is both symmetric and skew symmetric.
So A’ = A and A’ = -A
∴ A + A = O ⇒ 2A = O ⇒ A = O ∴ A is a zero matrix

Matrices MCQ AP Inter 2nd Year Maths Chapter 3

Question 11.
If A is square matrix such that A2 = A, then (I + A)3 – 7 A is equal to
1) A
2) I – A
3) I
4) 3A
Solution:
3) 1
Given that the matrix A is a square matrix such that A2 = A
Now, (I + A)3 – 7A = I3 + A3 + 3I2A + 3A2I – 7A
= I + A2.A + 3A + 3A2 – 7A
= I + A.A + 3A + 3A – 7A [A2 = A]
= I + A2 – A = I + A – A = I [∵ A2 = A]
Hence, (I + A)3 – 7A = I