NCERT Exemplar Solutions Class 12 Maths Matrices — Objective Type and Fill in the Blanks Q53–81

NCERT Exemplar Problems, Class 12 Mathematics — Matrices, Objective Type questions 53 to 67 and Fill in the Blanks questions 68 to 81. All 29 questions solved.

Twenty-nine short questions that test whether the definitions are secure. Questions 53 to 61 ask you to name a matrix (square, diagonal, symmetric, skew symmetric) or to count and compare orders. Questions 62 to 65 and 74 to 80 are about transposes: the two rules that do almost all the work are $(AB)^{\prime} = B^{\prime}A^{\prime}$ and $(A \pm B)^{\prime} = A^{\prime} \pm B^{\prime}$. Questions 66, 67 and 81 use elementary operations, and Questions 68 to 73 are vocabulary.

A note on the syllabus. The rationalised NCERT Chapter 3 dropped elementary row and column operations. Questions 66, 67 and 81 depend on them, so they are outside the current textbook, though the idea is still used in JEE-style problems. They are kept here and solved by the method the question names. Everything else on this page is inside the current textbook.

Key insight. To classify a matrix, take its transpose and compare. If $A^{\prime} = A$ it is symmetric; if $A^{\prime} = -A$ it is skew symmetric. For an expression such as $AB^{\prime} – BA^{\prime}$, transpose the whole expression using $(XY)^{\prime} = Y^{\prime}X^{\prime}$ and see whether it comes back unchanged or with a minus sign.

Question 53

The matrix $P = \begin{bmatrix} 0 & 0 & 4 \\ 0 & 4 & 0 \\ 4 & 0 & 0 \end{bmatrix}$ is a

Solution. $P$ has 3 rows and 3 columns, so it is square. It is not diagonal, because the entries $4$ in the corners (positions $(1,3)$ and $(3,1)$) lie off the main diagonal. It is not a unit matrix either, since the diagonal entries are not all $1$ and the off-diagonal entries are not all $0$. So (A) is the only true description.

Question 54

Total number of possible matrices of order $3 \times 3$ with each entry $2$ or $0$ is

Solution. A $3 \times 3$ matrix has $9$ entries, and each can be filled in $2$ independent ways ($2$ or $0$). The number of matrices is therefore $2^9 = 512$.

Question 55

If $\begin{bmatrix} 2x+y & 4x \\ 5x-7 & 4x \end{bmatrix} = \begin{bmatrix} 7 & 7y-13 \\ y & x+6 \end{bmatrix}$, then the value of $x + y$ is

Solution. Equal matrices have equal corresponding entries. The $(2,2)$ entries give $4x = x + 6$, so $x = 2$. The $(1,1)$ entries give $2x + y = 7$, so $y = 3$.

Check with the other two entries: $5x – 7 = 3 = y$ and $4x = 8 = 7y – 13 = 21 – 13$. All four agree. So $x = 2$, $y = 3$, and $x + y = 5$.

Question 56

If $A = \dfrac{1}{\pi}\begin{bmatrix} \sin^{-1}(x\pi) & \tan^{-1}\!\left(\dfrac{x}{\pi}\right) \\[2mm] \sin^{-1}\!\left(\dfrac{x}{\pi}\right) & \cot^{-1}(\pi x) \end{bmatrix}$, $B = \dfrac{1}{\pi}\begin{bmatrix} -\cos^{-1}(x\pi) & \tan^{-1}\!\left(\dfrac{x}{\pi}\right) \\[2mm] \sin^{-1}\!\left(\dfrac{x}{\pi}\right) & -\tan^{-1}(\pi x) \end{bmatrix}$, then $A – B$ is equal to

Solution. Subtract entry by entry. The off-diagonal entries are identical in $A$ and $B$, so they cancel to $0$. The diagonal entries become

$$\frac{1}{\pi}\left[\sin^{-1}(x\pi) + \cos^{-1}(x\pi)\right] = \frac{1}{\pi}\cdot\frac{\pi}{2} = \frac12, \qquad \frac{1}{\pi}\left[\cot^{-1}(\pi x) + \tan^{-1}(\pi x)\right] = \frac{1}{\pi}\cdot\frac{\pi}{2} = \frac12,$$

using $\sin^{-1}t + \cos^{-1}t = \frac{\pi}{2}$ and $\tan^{-1}t + \cot^{-1}t = \frac{\pi}{2}$. So $A – B = \begin{bmatrix} \frac12 & 0 \\ 0 & \frac12 \end{bmatrix} = \frac12 I$.

Question 57

If $A$ and $B$ are two matrices of the order $3 \times m$ and $3 \times n$, respectively, and $m = n$, then the order of matrix $(5A – 2B)$ is

Solution. Since $m = n$, $A$ and $B$ have the same order $3 \times n$, so they can be subtracted. Multiplying by the scalars $5$ and $2$ does not change the order, and the difference has the order of its parts: $3 \times n$.

Question 58

If $A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, then $A^2$ is equal to

Solution.

$$A^2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0\cdot0+1\cdot1 & 0\cdot1+1\cdot0 \\ 1\cdot0+0\cdot1 & 1\cdot1+0\cdot0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$$

Question 59

If matrix $A = [a_{ij}]_{2 \times 2}$, where $a_{ij} = 1$ if $i \neq j$ and $a_{ij} = 0$ if $i = j$, then $A^2$ is equal to

Solution. The rule puts $0$ on the diagonal and $1$ off it, so $A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, the matrix of Question 58. Hence $A^2 = I$.

Question 60

The matrix $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix}$ is a

Solution. The matrix is diagonal, but “diagonal” is not offered. It is not an identity matrix, because the diagonal entries are $1, 2, 4$ rather than all $1$. Its transpose equals itself (every off-diagonal entry is $0$ on both sides), so it is symmetric. It is not skew symmetric, since that would need $a_{ii} = -a_{ii}$, that is, a zero diagonal.

Question 61

The matrix $\begin{bmatrix} 0 & -5 & 8 \\ 5 & 0 & 12 \\ -8 & -12 & 0 \end{bmatrix}$ is a

Solution. The diagonal is all zeros, and each entry is the negative of its mirror image: $a_{12} = -5 = -a_{21}$, $a_{13} = 8 = -a_{31}$, $a_{23} = 12 = -a_{32}$. So $A^{\prime} = -A$, and the matrix is skew symmetric. It has non-zero off-diagonal entries, so it is neither diagonal nor scalar, and $a_{12} \neq a_{21}$ rules out symmetric.

Question 62

If $A$ is a matrix of order $m \times n$ and $B$ is a matrix such that $AB^{\prime}$ and $B^{\prime}A$ are both defined, then the order of matrix $B$ is

Solution. Let $B$ be $p \times q$, so $B^{\prime}$ is $q \times p$. For $AB^{\prime}$ ($m \times n$ times $q \times p$) the inner orders must match: $n = q$. For $B^{\prime}A$ ($q \times p$ times $m \times n$) we need $p = m$. So $B$ is $p \times q = m \times n$.

Question 63

If $A$ and $B$ are matrices of the same order, then $(AB^{\prime} – BA^{\prime})$ is a

Solution. Transpose the expression, reversing the order of each product:

$$(AB^{\prime} – BA^{\prime})^{\prime} = (AB^{\prime})^{\prime} – (BA^{\prime})^{\prime} = BA^{\prime} – AB^{\prime} = -(AB^{\prime} – BA^{\prime}).$$

The transpose is the negative of the matrix, so it is skew symmetric. It need not be null, since $AB^{\prime}$ and $BA^{\prime}$ are different matrices in general.

Question 64

If $A$ is a square matrix such that $A^2 = I$, then $(A – I)^3 + (A + I)^3 – 7A$ is equal to

Solution. $A$ commutes with $I$, so the usual binomial expansions apply:

$$(A-I)^3 + (A+I)^3 = 2A^3 + 6A.$$

(The terms in $A^2$ and $I$ cancel in pairs.) Since $A^2 = I$, $A^3 = A\cdot A^2 = A$, so the sum is $2A + 6A = 8A$. Then $8A – 7A = A$.

Question 65

For any two matrices $A$ and $B$, we have

Solution. Matrix multiplication is not commutative in general, so (A) is not always true. But $AB \neq BA$ is not always true either: $A = I$ gives $AB = BA$. And $AB = O$ certainly does not hold for every pair. None of (A), (B), (C) holds for all $A$ and $B$.

Question 66

On using elementary column operations $C_2 \to C_2 – 2C_1$ in the following matrix equation $\begin{bmatrix} 1 & -3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix}$, we have:

Solution. For an equation $A = BC$, a column operation is applied to $A$ and to the last factor $C$ only; the first factor stays as it is. (Doing the operation to $A$ and to $C$ keeps the equation true because the column operation on $BC$ is the same as the operation on $C$ followed by multiplication by $B$.)

Apply $C_2 \to C_2 – 2C_1$ to the left side: $-3 – 2(1) = -5$ and $4 – 2(2) = 0$, giving $\begin{bmatrix} 1 & -5 \\ 2 & 0 \end{bmatrix}$.

Apply it to the last factor: $1 – 2(3) = -5$ and $4 – 2(2) = 0$, giving $\begin{bmatrix} 3 & -5 \\ 2 & 0 \end{bmatrix}$, with $\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}$ unchanged. That is option (D).

Check: $\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 3 & -5 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 3-2 & -5 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 1 & -5 \\ 2 & 0 \end{bmatrix}$.

Question 67

On using elementary row operation $R_1 \to R_1 – 3R_2$ in the following matrix equation $\begin{bmatrix} 4 & 2 \\ 3 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}\begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$, we have:

Solution. For $A = BC$, a row operation is applied to $A$ and to the first factor $B$ only; the last factor stays as it is.

On the left, $R_1 \to R_1 – 3R_2$ gives $4 – 9 = -5$ and $2 – 9 = -7$, so $\begin{bmatrix} -5 & -7 \\ 3 & 3 \end{bmatrix}$. On the first factor it gives $1 – 3(0) = 1$ and $2 – 3(3) = -7$, so $\begin{bmatrix} 1 & -7 \\ 0 & 3 \end{bmatrix}$, with $\begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$ unchanged. That is option (A).

Check: $\begin{bmatrix} 1 & -7 \\ 0 & 3 \end{bmatrix}\begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2-7 & -7 \\ 3 & 3 \end{bmatrix} = \begin{bmatrix} -5 & -7 \\ 3 & 3 \end{bmatrix}$.

Question 68

Fill in the blank: __ matrix is both symmetric and skew symmetric matrix.

Solution. If $A^{\prime} = A$ and $A^{\prime} = -A$, then $A = -A$, so $2A = O$ and every entry is $0$. Only the null matrix satisfies both conditions (and it must be square for the idea of symmetry to apply).

Question 69

Fill in the blank: Sum of two skew symmetric matrices is always __ matrix.

Solution. If $A^{\prime} = -A$ and $B^{\prime} = -B$, then $(A + B)^{\prime} = A^{\prime} + B^{\prime} = -A – B = -(A + B)$.

Question 70

Fill in the blank: The negative of a matrix is obtained by multiplying it by __.

Solution. The negative of $A$ is the matrix whose entries are $-a_{ij}$. That is $(-1)A$.

Question 71

Fill in the blank: The product of any matrix by the scalar __ is the null matrix.

Solution. Multiplying by a scalar multiplies every entry, and $0 \cdot a_{ij} = 0$ for every entry.

Question 72

Fill in the blank: A matrix which is not a square matrix is called a __ matrix.

Solution. A matrix whose number of rows differs from its number of columns is called rectangular.

Question 73

Fill in the blank: Matrix multiplication is __ over addition.

Solution. For matrices of suitable orders, $A(B + C) = AB + AC$ and $(A + B)C = AC + BC$. This is the distributive law.

Question 74

Fill in the blank: If $A$ is a symmetric matrix, then $A^3$ is a __ matrix.

Solution. $(A^3)^{\prime} = (A^{\prime})^3 = A^3$, since $A^{\prime} = A$. So $A^3$ equals its own transpose.

Question 75

Fill in the blank: If $A$ is a skew symmetric matrix, then $A^2$ is a __.

Solution. $(A^2)^{\prime} = (A^{\prime})^2 = (-A)^2 = A^2$. The minus signs cancel in a square, so $A^2$ is symmetric.

Question 76

Fill in the blanks: If $A$ and $B$ are square matrices of the same order, then (i) $(AB)^{\prime} = $ _, (ii) $(kA)^{\prime} = $ _ ($k$ is any scalar), (iii) $[k(A – B)]^{\prime} = $ __.

Solution. (i) The transpose of a product reverses the order: $(AB)^{\prime} = B^{\prime}A^{\prime}$.

(ii) A scalar passes through the transpose: $(kA)^{\prime} = kA^{\prime}$.

(iii) Combining the rules, $[k(A – B)]^{\prime} = k(A – B)^{\prime} = k(A^{\prime} – B^{\prime})$.

Question 77

Fill in the blank: If $A$ is skew symmetric, then $kA$ is a __ ($k$ is any scalar).

Solution. $(kA)^{\prime} = kA^{\prime} = k(-A) = -(kA)$.

Question 78

Fill in the blanks: If $A$ and $B$ are symmetric matrices, then (i) $AB – BA$ is a _, (ii) $BA – 2AB$ is a _.

Solution. Use $A^{\prime} = A$, $B^{\prime} = B$ and $(XY)^{\prime} = Y^{\prime}X^{\prime}$, so that $(AB)^{\prime} = BA$ and $(BA)^{\prime} = AB$.

(i) $(AB – BA)^{\prime} = BA – AB = -(AB – BA)$. It is skew symmetric.

(ii) $(BA – 2AB)^{\prime} = AB – 2BA$. This is neither $BA – 2AB$ nor its negative $2AB – BA$ in general, so $BA – 2AB$ is neither symmetric nor skew symmetric. (Equating $AB – 2BA$ with $BA – 2AB$ gives $AB = BA$, so it is symmetric only in that special case.)

Question 79

Fill in the blank: If $A$ is a symmetric matrix, then $B^{\prime}AB$ is __.

Solution. $(B^{\prime}AB)^{\prime} = B^{\prime}A^{\prime}(B^{\prime})^{\prime} = B^{\prime}AB$, using $(XYZ)^{\prime} = Z^{\prime}Y^{\prime}X^{\prime}$ and $A^{\prime} = A$. The matrix equals its transpose.

Question 80

Fill in the blank: If $A$ and $B$ are symmetric matrices of the same order, then $AB$ is symmetric if and only if __.

Solution. $(AB)^{\prime} = B^{\prime}A^{\prime} = BA$. So $AB$ is symmetric exactly when $(AB)^{\prime} = AB$, that is, when $BA = AB$.

Question 81

Fill in the blank: In applying one or more row operations while finding $A^{-1}$ by elementary row operations, we obtain all zeros in one or more rows of the matrix $A$ on the left, then $A^{-1}$ __.

Solution. In the row-operation method we write $A = IA$ and reduce the left side to $I$. If a row of zeros appears on the left, the rows of $A$ were linearly dependent, which means $|A| = 0$. A matrix with zero determinant is singular, so no inverse exists, and the reduction cannot reach $I$.

Common mistakes

• Choosing “diagonal” or “identity” for a diagonal matrix with unequal entries. In Question 60 the matrix is diagonal but “diagonal” is not offered; check each option in turn. It is symmetric, not an identity matrix.
• Applying row and column operations to the wrong factor. In $A = BC$, a row operation goes on $A$ and the first factor $B$ (Question 67); a column operation goes on $A$ and the last factor $C$ (Question 66). Applying it to the other factor, as in options (B) and (C) of Question 67, breaks the equation.
• Forgetting to reverse the order in a transpose. $(AB)^{\prime} = B^{\prime}A^{\prime}$, not $A^{\prime}B^{\prime}$. Question 63 and Question 78 turn on exactly this.
• Assuming $AB = BA$. Question 65 is a reminder that it is not true in general, but also that $AB \neq BA$ is not true in general.
• Reading the options in Question 55 as the answer to $x + y$. The options are pairs $(x, y)$; solve for $x$ and $y$ first, then pick the matching pair.

Practise next

• Exemplar Short Answer Q1–24 for order, equality and multiplication.
• Exemplar Short Answer Q25–52 and Long Answer for transposes, symmetric and skew symmetric parts, and inverses.
• Exemplar True or False Q82–101, the last part of the unit.
• NCERT textbook Exercise 3.3 for transpose and symmetric and skew symmetric matrices, as in Questions 60 to 63 and 74 to 80.
• NCERT textbook Exercise 3.4 for invertible matrices.