NCERT Exemplar Solutions Class 12 Maths Matrices — Short Answer and Long Answer Q25–52

NCERT Exemplar Problems, Class 12 Mathematics — Matrices, Short Answer questions 25 to 48 and Long Answer questions 49 to 52. All 28 questions solved.

Twenty-eight questions, mostly of five kinds. Questions 25 to 28, 32, 33 and 35 ask you to verify an identity on given matrices, which means computing both sides and comparing. Questions 29 to 31, 34, 36, 46 to 49 prove general rules about transposes and products. Questions 37 and 51 find inverses by row operations. Questions 38 to 45 and 50 find unknown entries by equating matrices, and Question 52 splits a matrix into symmetric and skew-symmetric parts.

A note on the syllabus. The rationalised NCERT Chapter 3 dropped elementary row and column operations, so Questions 37 and 51 (inverse by row operations) are no longer in the textbook; they are solved below by the book’s method, and each inverse is checked by $A\cdot A^{-1} = I$. Proof by mathematical induction is no longer in the rationalised Class 11 textbook or the JEE Main syllabus, which affects Questions 36 and 49. Everything else is inside the current textbook.

Key insight. Matrix multiplication is associative and distributes over addition, but it is not commutative. Every proof below uses only those rules plus $(AB)’ = B’A’$: reverse the order when you transpose a product, and never swap two factors unless the question tells you $AB = BA$.

Question 25

If $A = \begin{bmatrix} 2 & 1 \end{bmatrix}$, $B = \begin{bmatrix} 5 & 3 & 4 \\ 8 & 7 & 6 \end{bmatrix}$ and $C = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 0 & 2 \end{bmatrix}$, verify that $A(B + C) = AB + AC$.

Solution. Left side. Add first: $B + C = \begin{bmatrix} 4 & 5 & 5 \\ 9 & 7 & 8 \end{bmatrix}$. Then $$A(B+C) = \begin{bmatrix} 2 & 1 \end{bmatrix}\begin{bmatrix} 4 & 5 & 5 \\ 9 & 7 & 8 \end{bmatrix} = \begin{bmatrix} 8+9 & 10+7 & 10+8 \end{bmatrix} = \begin{bmatrix} 17 & 17 & 18 \end{bmatrix}.$$

Right side. $AB = \begin{bmatrix} 10+8 & 6+7 & 8+6 \end{bmatrix} = \begin{bmatrix} 18 & 13 & 14 \end{bmatrix}$ and $AC = \begin{bmatrix} -2+1 & 4+0 & 2+2 \end{bmatrix} = \begin{bmatrix} -1 & 4 & 4 \end{bmatrix}$, so $$AB + AC = \begin{bmatrix} 17 & 17 & 18 \end{bmatrix}.$$

The two sides are equal.

Question 26

If $A = \begin{bmatrix} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix}$, verify that $A^2 + A = A(A + I)$, where $I$ is the $3 \times 3$ unit matrix.

Solution. First $A^2$: $$A^2 = \begin{bmatrix} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4 \end{bmatrix}.$$

Left side: $A^2 + A = \begin{bmatrix} 2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5 \end{bmatrix}$.

Right side: $A + I = \begin{bmatrix} 2 & 0 & -1 \\ 2 & 2 & 3 \\ 0 & 1 & 2 \end{bmatrix}$, and $$A(A+I) = \begin{bmatrix} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix}\begin{bmatrix} 2 & 0 & -1 \\ 2 & 2 & 3 \\ 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5 \end{bmatrix}.$$

The two sides agree. In general $A(A + I) = A^2 + AI = A^2 + A$, because $AI = A$.

Question 27

If $A = \begin{bmatrix} 0 & -1 & 2 \\ 4 & 3 & -4 \end{bmatrix}$ and $B = \begin{bmatrix} 4 & 0 \\ 1 & 3 \\ 2 & 6 \end{bmatrix}$, verify that (i) $(A’)’ = A$, (ii) $(AB)’ = B’A’$, (iii) $(kA)’ = (kA’)$.

Solution. (i) Transposing swaps rows and columns, so doing it twice restores the matrix. $A’ = \begin{bmatrix} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{bmatrix}$ and $(A’)’ = \begin{bmatrix} 0 & -1 & 2 \\ 4 & 3 & -4 \end{bmatrix} = A$.

(ii) $AB = \begin{bmatrix} 0-1+4 & 0-3+12 \\ 16+3-8 & 0+9-24 \end{bmatrix} = \begin{bmatrix} 3 & 9 \\ 11 & -15 \end{bmatrix}$, so $(AB)’ = \begin{bmatrix} 3 & 11 \\ 9 & -15 \end{bmatrix}$.

Now $B’ = \begin{bmatrix} 4 & 1 & 2 \\ 0 & 3 & 6 \end{bmatrix}$ and $$B’A’ = \begin{bmatrix} 4 & 1 & 2 \\ 0 & 3 & 6 \end{bmatrix}\begin{bmatrix} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{bmatrix} = \begin{bmatrix} -1+4 & 16+3-8 \\ -3+12 & 9-24 \end{bmatrix} = \begin{bmatrix} 3 & 11 \\ 9 & -15 \end{bmatrix}.$$ Equal to $(AB)’$.

(iii) Let $k$ be any scalar. The matrix $kA$ has entries $k\,a_{ij}$; transposing it moves the $(i,j)$ entry to position $(j,i)$, so $(kA)’$ has $(j,i)$ entry $k\,a_{ij}$. The matrix $kA’$ has $(j,i)$ entry $k$ times the $(j,i)$ entry of $A’$, which is $k\,a_{ij}$. The entries agree everywhere, so $(kA)’ = kA’$. For instance with $k = 2$, both equal $\begin{bmatrix} 0 & 8 \\ -2 & 6 \\ 4 & -8 \end{bmatrix}$.

Question 28

If $A = \begin{bmatrix} 1 & 2 \\ 4 & 1 \\ 5 & 6 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{bmatrix}$, verify that (i) $(2A + B)’ = 2A’ + B’$, (ii) $(A – B)’ = A’ – B’$.

Solution. (i) $2A + B = \begin{bmatrix} 2+1 & 4+2 \\ 8+6 & 2+4 \\ 10+7 & 12+3 \end{bmatrix} = \begin{bmatrix} 3 & 6 \\ 14 & 6 \\ 17 & 15 \end{bmatrix}$, so $(2A+B)’ = \begin{bmatrix} 3 & 14 & 17 \\ 6 & 6 & 15 \end{bmatrix}$.

Also $2A’ + B’ = \begin{bmatrix} 2 & 8 & 10 \\ 4 & 2 & 12 \end{bmatrix} + \begin{bmatrix} 1 & 6 & 7 \\ 2 & 4 & 3 \end{bmatrix} = \begin{bmatrix} 3 & 14 & 17 \\ 6 & 6 & 15 \end{bmatrix}$. Equal.

(ii) $A – B = \begin{bmatrix} 0 & 0 \\ -2 & -3 \\ -2 & 3 \end{bmatrix}$, so $(A-B)’ = \begin{bmatrix} 0 & -2 & -2 \\ 0 & -3 & 3 \end{bmatrix}$.

Also $A’ – B’ = \begin{bmatrix} 1 & 4 & 5 \\ 2 & 1 & 6 \end{bmatrix} – \begin{bmatrix} 1 & 6 & 7 \\ 2 & 4 & 3 \end{bmatrix} = \begin{bmatrix} 0 & -2 & -2 \\ 0 & -3 & 3 \end{bmatrix}$. Equal.

Question 29

Show that $A’A$ and $AA’$ are both symmetric matrices for any matrix $A$.

Solution. A matrix $S$ is symmetric when $S’ = S$. Use $(XY)’ = Y’X’$ and $(A’)’ = A$: $$(A’A)’ = A'(A’)’ = A’A, \qquad (AA’)’ = (A’)’A’ = AA’.$$ Each equals its own transpose. (If $A$ is $m \times n$, then $A’A$ is $n \times n$ and $AA’$ is $m \times m$, so both products are square and the question makes sense.)

Question 30

Let $A$ and $B$ be square matrices of the order $3 \times 3$. Is $(AB)^2 = A^2B^2$? Give reasons.

Solution. Write the squares out: $(AB)^2 = ABAB$, while $A^2B^2 = AABB$. To turn the first into the second, the middle $B$ and $A$ would have to swap places, and matrix multiplication does not allow that in general. So the equality is not true in general; it holds when $AB = BA$, because then $ABAB = A(BA)B = A(AB)B = A^2B^2$.

A counterexample: take $A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$. Then $AB$ has a single non-zero entry $2$ in the top-left corner, so $(AB)^2$ has $4$ there. But $A^2 = A$ and $B^2 = B$, so $A^2B^2 = AB$ has $2$ there. They differ.

Question 31

Show that if $A$ and $B$ are square matrices such that $AB = BA$, then $(A + B)^2 = A^2 + 2AB + B^2$.

Solution. Expand without assuming anything about the order of factors: $$(A+B)^2 = (A+B)(A+B) = A^2 + AB + BA + B^2.$$ Now use $BA = AB$: $$(A+B)^2 = A^2 + AB + AB + B^2 = A^2 + 2AB + B^2.$$

Question 32

Let $A = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}$, $B = \begin{bmatrix} 4 & 0 \\ 1 & 5 \end{bmatrix}$, $C = \begin{bmatrix} 2 & 0 \\ 1 & -2 \end{bmatrix}$ and $a = 4$, $b = -2$. Show that

(a) $A + (B + C) = (A + B) + C$ (b) $A(BC) = (AB)C$ (c) $(a + b)B = aB + bB$ (d) $a(C – A) = aC – aA$ (e) $(A^T)^T = A$ (f) $(bA)^T = b\,A^T$ (g) $(AB)^T = B^TA^T$ (h) $(A – B)C = AC – BC$ (i) $(A – B)^T = A^T – B^T$

Solution. Each part is computed on both sides.

(a) $B + C = \begin{bmatrix} 6 & 0 \\ 2 & 3 \end{bmatrix}$, so $A + (B+C) = \begin{bmatrix} 7 & 2 \\ 1 & 6 \end{bmatrix}$. Also $A + B = \begin{bmatrix} 5 & 2 \\ 0 & 8 \end{bmatrix}$, so $(A+B) + C = \begin{bmatrix} 7 & 2 \\ 1 & 6 \end{bmatrix}$. Equal.

(b) $BC = \begin{bmatrix} 8 & 0 \\ 7 & -10 \end{bmatrix}$ and $A(BC) = \begin{bmatrix} 22 & -20 \\ 13 & -30 \end{bmatrix}$. Also $AB = \begin{bmatrix} 6 & 10 \\ -1 & 15 \end{bmatrix}$ and $(AB)C = \begin{bmatrix} 22 & -20 \\ 13 & -30 \end{bmatrix}$. Equal.

(c) $a + b = 2$, so $(a+b)B = 2B = \begin{bmatrix} 8 & 0 \\ 2 & 10 \end{bmatrix}$. And $aB + bB = \begin{bmatrix} 16 & 0 \\ 4 & 20 \end{bmatrix} + \begin{bmatrix} -8 & 0 \\ -2 & -10 \end{bmatrix} = \begin{bmatrix} 8 & 0 \\ 2 & 10 \end{bmatrix}$. Equal.

(d) $C – A = \begin{bmatrix} 1 & -2 \\ 2 & -5 \end{bmatrix}$, so $a(C-A) = \begin{bmatrix} 4 & -8 \\ 8 & -20 \end{bmatrix}$. And $aC – aA = \begin{bmatrix} 8 & 0 \\ 4 & -8 \end{bmatrix} – \begin{bmatrix} 4 & 8 \\ -4 & 12 \end{bmatrix} = \begin{bmatrix} 4 & -8 \\ 8 & -20 \end{bmatrix}$. Equal.

(e) Transposing twice restores $A$: $A^T = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$ and $(A^T)^T = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} = A$.

(f) $bA = -2A = \begin{bmatrix} -2 & -4 \\ 2 & -6 \end{bmatrix}$, so $(bA)^T = \begin{bmatrix} -2 & 2 \\ -4 & -6 \end{bmatrix}$. And $bA^T = -2\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} -2 & 2 \\ -4 & -6 \end{bmatrix}$. Equal.

(g) $(AB)^T = \begin{bmatrix} 6 & -1 \\ 10 & 15 \end{bmatrix}$. And $B^TA^T = \begin{bmatrix} 4 & 1 \\ 0 & 5 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 6 & -1 \\ 10 & 15 \end{bmatrix}$. Equal.

(h) $A – B = \begin{bmatrix} -3 & 2 \\ -2 & -2 \end{bmatrix}$, so $(A-B)C = \begin{bmatrix} -6+2 & -4 \\ -4-2 & 4 \end{bmatrix} = \begin{bmatrix} -4 & -4 \\ -6 & 4 \end{bmatrix}$. And $AC = \begin{bmatrix} 4 & -4 \\ 1 & -6 \end{bmatrix}$, $BC = \begin{bmatrix} 8 & 0 \\ 7 & -10 \end{bmatrix}$, so $AC – BC = \begin{bmatrix} -4 & -4 \\ -6 & 4 \end{bmatrix}$. Equal.

(i) $(A-B)^T = \begin{bmatrix} -3 & -2 \\ 2 & -2 \end{bmatrix}$ and $A^T – B^T = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} – \begin{bmatrix} 4 & 1 \\ 0 & 5 \end{bmatrix} = \begin{bmatrix} -3 & -2 \\ 2 & -2 \end{bmatrix}$. Equal.

Question 33

If $A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$, show that $A^2 = \begin{bmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{bmatrix}$.

Solution. Multiply $A$ by itself: $$A^2 = \begin{bmatrix} \cos^2\theta – \sin^2\theta & \cos\theta\sin\theta + \sin\theta\cos\theta \\ -\sin\theta\cos\theta – \cos\theta\sin\theta & -\sin^2\theta + \cos^2\theta \end{bmatrix}.$$ Using $\cos^2\theta – \sin^2\theta = \cos 2\theta$ and $2\sin\theta\cos\theta = \sin 2\theta$: $$A^2 = \begin{bmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{bmatrix}.$$

Question 34

If $A = \begin{bmatrix} 0 & -x \\ x & 0 \end{bmatrix}$, $B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ and $x^2 = -1$, show that $(A + B)^2 = A^2 + B^2$.

Solution. Expanding, $(A+B)^2 = A^2 + AB + BA + B^2$, so the claim is equivalent to $AB + BA = O$. Compute: $$AB = \begin{bmatrix} -x & 0 \\ 0 & x \end{bmatrix}, \qquad BA = \begin{bmatrix} x & 0 \\ 0 & -x \end{bmatrix}, \qquad AB + BA = O.$$ So $(A+B)^2 = A^2 + B^2$. This cancellation happens for every $x$; the condition $x^2 = -1$ is not needed for the identity itself. With it, $A^2 = \begin{bmatrix} -x^2 & 0 \\ 0 & -x^2 \end{bmatrix} = I$ and $B^2 = I$, so both sides equal $2I$. A direct check agrees: $A + B = \begin{bmatrix} 0 & 1-x \\ 1+x & 0 \end{bmatrix}$, whose square is $(1-x^2)I = 2I$.

Question 35

Verify that $A^2 = I$ when $A = \begin{bmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{bmatrix}$.

Solution. Multiply row by column: $$A^2 = \begin{bmatrix} 0+4-3 & 0-3+3 & 0+4-4 \\ 0-12+12 & 4+9-12 & -4-12+16 \\ 0-12+12 & 3+9-12 & -3-12+16 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I.$$

Question 36

Prove by mathematical induction that $(A’)^n = (A^n)’$, where $n \in \mathbf{N}$, for any square matrix $A$.

Solution. Let $P(n)$ be the statement $(A’)^n = (A^n)’$.

Base case. For $n = 1$ both sides are $A’$, so $P(1)$ holds.

Inductive step. Assume $P(k)$: $(A’)^k = (A^k)’$. Then $$(A’)^{k+1} = (A’)^k A’ = (A^k)’A’ = (A\,A^k)’ = (A^{k+1})’.$$ The third equality is the reversal rule $(XY)’ = Y’X’$ read backwards with $X = A$ and $Y = A^k$. So $P(k)$ implies $P(k+1)$, and by induction $P(n)$ holds for all $n \in \mathbf{N}$.

Question 37

Find the inverse, by elementary row operations (if possible), of the following matrices: (i) $\begin{bmatrix} 1 & 3 \\ -5 & 7 \end{bmatrix}$, (ii) $\begin{bmatrix} 1 & -3 \\ -2 & 6 \end{bmatrix}$.

Solution. Place $A$ beside $I$ and apply the same row operations to both until the left side becomes $I$; the right side is then $A^{-1}$.

(i) Start with $\left[\begin{array}{cc|cc} 1 & 3 & 1 & 0 \\ -5 & 7 & 0 & 1 \end{array}\right]$.

$R_2 \to R_2 + 5R_1$: $\left[\begin{array}{cc|cc} 1 & 3 & 1 & 0 \\ 0 & 22 & 5 & 1 \end{array}\right]$.

$R_2 \to \tfrac{1}{22}R_2$: $\left[\begin{array}{cc|cc} 1 & 3 & 1 & 0 \\ 0 & 1 & \tfrac{5}{22} & \tfrac{1}{22} \end{array}\right]$.

$R_1 \to R_1 – 3R_2$: $\left[\begin{array}{cc|cc} 1 & 0 & \tfrac{7}{22} & -\tfrac{3}{22} \\ 0 & 1 & \tfrac{5}{22} & \tfrac{1}{22} \end{array}\right]$.

So $A^{-1} = \dfrac{1}{22}\begin{bmatrix} 7 & -3 \\ 5 & 1 \end{bmatrix}$. Check: $A A^{-1} = \tfrac{1}{22}\begin{bmatrix} 7+15 & -3+3 \\ -35+35 & 15+7 \end{bmatrix} = I$.

(ii) Start with $\left[\begin{array}{cc|cc} 1 & -3 & 1 & 0 \\ -2 & 6 & 0 & 1 \end{array}\right]$. Apply $R_2 \to R_2 + 2R_1$: $$\left[\begin{array}{cc|cc} 1 & -3 & 1 & 0 \\ 0 & 0 & 2 & 1 \end{array}\right].$$ The left side now has a row of zeros, so it can never be reduced to $I$. The inverse does not exist. This agrees with $|A| = 6 – 6 = 0$.

Question 38

If $\begin{bmatrix} xy & 4 \\ z+6 & x+y \end{bmatrix} = \begin{bmatrix} 8 & w \\ 0 & 6 \end{bmatrix}$, then find the values of $x, y, z$ and $w$.

Solution. Equal matrices have equal corresponding entries, so $xy = 8$, $w = 4$, $z + 6 = 0$ and $x + y = 6$. Hence $z = -6$, $w = 4$.

For $x$ and $y$ we know the sum $6$ and product $8$, so they are the roots of $t^2 – 6t + 8 = 0$, i.e. $(t-2)(t-4) = 0$. Therefore $\{x, y\} = \{2, 4\}$ in either order.

Question 39

If $A = \begin{bmatrix} 1 & 5 \\ 7 & 12 \end{bmatrix}$ and $B = \begin{bmatrix} 9 & 1 \\ 7 & 8 \end{bmatrix}$, find a matrix $C$ such that $3A + 5B + 2C$ is a null matrix.

Solution. Setting $3A + 5B + 2C = O$ gives $C = -\tfrac{1}{2}(3A + 5B)$. Now $$3A + 5B = \begin{bmatrix} 3 & 15 \\ 21 & 36 \end{bmatrix} + \begin{bmatrix} 45 & 5 \\ 35 & 40 \end{bmatrix} = \begin{bmatrix} 48 & 20 \\ 56 & 76 \end{bmatrix},$$ so $C = -\tfrac{1}{2}\begin{bmatrix} 48 & 20 \\ 56 & 76 \end{bmatrix}$.

Question 40

If $A = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix}$, then find $A^2 – 5A – 14I$. Hence, obtain $A^3$.

Solution. Compute $A^2 = \begin{bmatrix} 9+20 & -15-10 \\ -12-8 & 20+4 \end{bmatrix} = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix}$. Then $$A^2 – 5A – 14I = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} – \begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix} – \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix} = O.$$

So $A^2 = 5A + 14I$. Multiply by $A$ to get $A^3 = 5A^2 + 14A$, and substitute $A^2$ again: $A^3 = 5(5A + 14I) + 14A = 39A + 70I$. Therefore $$A^3 = \begin{bmatrix} 117 & -195 \\ -156 & 78 \end{bmatrix} + \begin{bmatrix} 70 & 0 \\ 0 & 70 \end{bmatrix} = \begin{bmatrix} 187 & -195 \\ -156 & 148 \end{bmatrix}.$$

Question 41

Find the values of $a, b, c$ and $d$, if $3\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a & 6 \\ -1 & 2d \end{bmatrix} + \begin{bmatrix} 4 & a+b \\ c+d & 3 \end{bmatrix}$.

Solution. Add the matrices on the right and equate entries with $\begin{bmatrix} 3a & 3b \\ 3c & 3d \end{bmatrix}$:

• $3a = a + 4$, so $a = 2$.
• $3b = 6 + a + b$, so $2b = 6 + a = 8$ and $b = 4$.
• $3d = 2d + 3$, so $d = 3$.
• $3c = -1 + c + d$, so $2c = -1 + 3 = 2$ and $c = 1$.

Question 42

Find the matrix $A$ such that $\begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix} A = \begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}$.

Solution. The left factor is $3 \times 2$ and the product is $3 \times 3$, so $A$ must be $2 \times 3$. Let $A = \begin{bmatrix} p & q & r \\ s & t & u \end{bmatrix}$.

The second row of the left factor is $\begin{bmatrix} 1 & 0 \end{bmatrix}$, so the second row of the product is just the first row of $A$: $p = 1$, $q = -2$, $r = -5$.

The third row gives $-3p + 4s = 9$, so $s = 3$; $-3q + 4t = 22$, so $t = 4$; and $-3r + 4u = 15$, so $u = 0$.

Check with the first row: $2p – s = -1$, $2q – t = -8$, $2r – u = -10$. All three hold.

Question 43

If $A = \begin{bmatrix} 1 & 2 \\ 4 & 1 \end{bmatrix}$, find $A^2 + 2A + 7I$.

Solution. $A^2 = \begin{bmatrix} 1+8 & 2+2 \\ 4+4 & 8+1 \end{bmatrix} = \begin{bmatrix} 9 & 4 \\ 8 & 9 \end{bmatrix}$. Then $$A^2 + 2A + 7I = \begin{bmatrix} 9 & 4 \\ 8 & 9 \end{bmatrix} + \begin{bmatrix} 2 & 4 \\ 8 & 2 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}.$$

Question 44

If $A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$ and $A^{-1} = A’$, find the value of $\alpha$.

Solution. The condition $A^{-1} = A’$ is the same as $AA’ = I$. Now $$AA’ = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} = \begin{bmatrix} \cos^2\alpha + \sin^2\alpha & 0 \\ 0 & \sin^2\alpha + \cos^2\alpha \end{bmatrix} = I$$ whatever $\alpha$ is. So the condition puts no restriction on $\alpha$: it holds for every real $\alpha$ (here $|A| = 1 \neq 0$, so $A^{-1}$ always exists).

Question 45

If the matrix $\begin{bmatrix} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{bmatrix}$ is a skew-symmetric matrix, find the values of $a$, $b$ and $c$.

Solution. A skew-symmetric matrix satisfies $a_{ji} = -a_{ij}$, so every diagonal entry is zero and mirror entries are negatives of each other.

• Diagonal: the entry $b$ must be $0$.
• $a_{12} = a$ and $a_{21} = 2$, so $a = -2$.
• $a_{13} = 3$ and $a_{31} = c$, so $c = -3$.
• Check the remaining pair: $a_{23} = -1$ and $a_{32} = 1$, which are negatives, as required.

Question 46

If $P(x) = \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix}$, then show that $P(x)\cdot P(y) = P(x+y) = P(y)\cdot P(x)$.

Solution. Multiply: $$P(x)P(y) = \begin{bmatrix} \cos x\cos y – \sin x\sin y & \cos x\sin y + \sin x\cos y \\ -\sin x\cos y – \cos x\sin y & -\sin x\sin y + \cos x\cos y \end{bmatrix} = \begin{bmatrix} \cos(x+y) & \sin(x+y) \\ -\sin(x+y) & \cos(x+y) \end{bmatrix} = P(x+y).$$ Interchanging $x$ and $y$ in the same working gives $P(y)P(x) = P(y+x) = P(x+y)$, because $x + y = y + x$. So all three are equal, and rotation matrices of this kind commute.

Question 47

If $A$ is a square matrix such that $A^2 = A$, show that $(I + A)^3 = 7A + I$.

Solution. $I$ commutes with every matrix, so the binomial expansion is valid: $$(I+A)^3 = I + 3A + 3A^2 + A^3.$$ From $A^2 = A$ we get $A^3 = A^2A = AA = A^2 = A$. Substituting, $$(I+A)^3 = I + 3A + 3A + A = 7A + I.$$

Question 48

If $A$, $B$ are square matrices of the same order and $B$ is a skew-symmetric matrix, show that $A’BA$ is skew-symmetric.

Solution. We must show $(A’BA)’ = -A’BA$. Reverse the order of the three factors and transpose each: $$(A’BA)’ = A’\,B’\,(A’)’ = A’B’A.$$ Since $B$ is skew-symmetric, $B’ = -B$, so $$(A’BA)’ = A'(-B)A = -A’BA.$$

Question 49

If $AB = BA$ for any two square matrices, prove by mathematical induction that $(AB)^n = A^nB^n$.

Solution. Preliminary. If $AB = BA$ then $AB^k = B^kA$ for every $k \ge 1$. This is true for $k = 1$; if $AB^k = B^kA$, then $AB^{k+1} = (AB^k)B = B^kAB = B^k(BA) = B^{k+1}A$.

Let $P(n)$ be the statement $(AB)^n = A^nB^n$.

Base case. $P(1)$ reads $AB = AB$, which is true.

Inductive step. Assume $P(k)$. Then $$(AB)^{k+1} = (AB)^k(AB) = A^kB^k\,AB = A^k(B^kA)B = A^k(AB^k)B = A^{k+1}B^{k+1}.$$ So $P(k)$ implies $P(k+1)$, and $P(n)$ holds for all $n \in \mathbf{N}$.

Question 50

Find $x, y, z$ if $A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}$ satisfies $A’ = A^{-1}$.

Solution. $A’ = A^{-1}$ means $A’A = I$. Here $A’ = \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix}$, and $$A’A = \begin{bmatrix} 2x^2 & 0 & 0 \\ 0 & 6y^2 & 0 \\ 0 & 0 & 3z^2 \end{bmatrix}.$$ The off-diagonal entries vanish on their own (for instance the $(1,2)$ entry is $2xy – xy – xy = 0$), so we only need the diagonal to be $1$: $$2x^2 = 1,\quad 6y^2 = 1,\quad 3z^2 = 1.$$

Question 51

If possible, using elementary row transformations, find the inverse of the following matrices: (i) $\begin{bmatrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{bmatrix}$, (ii) $\begin{bmatrix} 2 & 3 & -3 \\ -1 & -2 & 2 \\ 1 & 1 & -1 \end{bmatrix}$, (iii) $\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}$.

Solution. In each part write $[A \mid I]$ and reduce the left block to $I$ using row operations. The first goal is to get a row like $[0\ 0\ 1]$ quickly by combining rows.

(i) Start with $\left[\begin{array}{ccc|ccc} 2 & -1 & 3 & 1 & 0 & 0 \\ -5 & 3 & 1 & 0 & 1 & 0 \\ -3 & 2 & 3 & 0 & 0 & 1 \end{array}\right]$.

$R_1 \to R_1 + R_2$ gives $[-3\ \ 2\ \ 4 \mid 1\ \ 1\ \ 0]$; then $R_1 \to R_1 – R_3$ gives $[0\ \ 0\ \ 1 \mid 1\ \ 1\ \ {-1}]$.

$R_2 \to R_2 – R_1$ and $R_3 \to R_3 – 3R_1$ clear the third column below that row: $R_2 = [-5\ \ 3\ \ 0 \mid -1\ \ 0\ \ 1]$, $R_3 = [-3\ \ 2\ \ 0 \mid -3\ \ -3\ \ 4]$.

$R_2 \to R_2 – R_3$: $R_2 = [-2\ \ 1\ \ 0 \mid 2\ \ 3\ \ {-3}]$.

$R_3 \to R_3 – 2R_2$: $R_3 = [1\ \ 0\ \ 0 \mid -7\ \ -9\ \ 10]$.

$R_2 \to R_2 + 2R_3$: $R_2 = [0\ \ 1\ \ 0 \mid -12\ \ -15\ \ 17]$.

Swap $R_1 \leftrightarrow R_3$ so the identity appears in order: $$\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & -7 & -9 & 10 \\ 0 & 1 & 0 & -12 & -15 & 17 \\ 0 & 0 & 1 & 1 & 1 & -1 \end{array}\right].$$

So $A^{-1} = \begin{bmatrix} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{bmatrix}$. Check, first row of $A$ times the columns: $2(-7) + (-1)(-12) + 3(1) = 1$, $2(-9) + 15 + 3 = 0$, $2(10) – 17 – 3 = 0$; the other rows agree, so $AA^{-1} = I$.

(ii) Start with $\left[\begin{array}{ccc|ccc} 2 & 3 & -3 & 1 & 0 & 0 \\ -1 & -2 & 2 & 0 & 1 & 0 \\ 1 & 1 & -1 & 0 & 0 & 1 \end{array}\right]$.

$R_1 \to R_1 + R_2$ gives $[1\ \ 1\ \ {-1} \mid 1\ \ 1\ \ 0]$, which is the same as $R_3$ on the left. Then $R_3 \to R_3 – R_1$ gives $[0\ \ 0\ \ 0 \mid -1\ \ -1\ \ 1]$.

The right block of that row is $[0\ 0\ 1] – [1\ 1\ 0] = [-1\ -1\ 1]$, but the left block has a zero row, so reduction to $I$ is impossible. The inverse does not exist. (Indeed $R_1 + R_2 = R_3$, so the determinant is $0$.)

(iii) Start with $\left[\begin{array}{ccc|ccc} 2 & 0 & -1 & 1 & 0 & 0 \\ 5 & 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 3 & 0 & 0 & 1 \end{array}\right]$.

$R_2 \to R_2 – 2R_1$: $[1\ \ 1\ \ 2 \mid -2\ \ 1\ \ 0]$.

$R_1 \to R_1 – 2R_2$: $[0\ \ {-2}\ \ {-5} \mid 5\ \ {-2}\ \ 0]$.

$R_1 \to R_1 + 2R_3$: $[0\ \ 0\ \ 1 \mid 5\ \ {-2}\ \ 2]$.

$R_3 \to R_3 – 3R_1$: $[0\ \ 1\ \ 0 \mid -15\ \ 6\ \ {-5}]$.

$R_2 \to R_2 – R_3 – 2R_1$: $[1\ \ 0\ \ 0 \mid 3\ \ {-1}\ \ 1]$.

The rows now hold the unit rows in the order $e_3, e_1, e_2$ (top to bottom, left block). Reordering them as $R_2, R_3, R_1$ gives $$\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 3 & -1 & 1 \\ 0 & 1 & 0 & -15 & 6 & -5 \\ 0 & 0 & 1 & 5 & -2 & 2 \end{array}\right],$$ so $A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}$. Check: first row of $A$ times the columns: $2(3) + 0 – 5 = 1$, $2(-1) + 0 + 2 = 0$, $2(1) + 0 – 2 = 0$; second row $[5\ 1\ 0]$ gives $15 – 15 = 0$, $-5 + 6 = 1$, $5 – 5 = 0$; third row $[0\ 1\ 3]$ gives $-15 + 15 = 0$, $6 – 6 = 0$, $-5 + 6 = 1$. So $AA^{-1} = I$.

Question 52

Express the matrix $\begin{bmatrix} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.

Solution. For any square matrix $A$, $P = \tfrac{1}{2}(A + A’)$ is symmetric and $Q = \tfrac{1}{2}(A – A’)$ is skew-symmetric, and $P + Q = A$. Here $A’ = \begin{bmatrix} 2 & 1 & 4 \\ 3 & -1 & 1 \\ 1 & 2 & 2 \end{bmatrix}$, so $$P = \tfrac{1}{2}\begin{bmatrix} 4 & 4 & 5 \\ 4 & -2 & 3 \\ 5 & 3 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 2 & \tfrac52 \\ 2 & -1 & \tfrac32 \\ \tfrac52 & \tfrac32 & 2 \end{bmatrix}, \qquad Q = \tfrac{1}{2}\begin{bmatrix} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -\tfrac32 \\ -1 & 0 & \tfrac12 \\ \tfrac32 & -\tfrac12 & 0 \end{bmatrix}.$$ $P$ equals its transpose and $Q$ equals the negative of its transpose, as needed.

Common mistakes

• Not reversing the order in a transposed product. $(AB)’ = B’A’$, not $A’B’$. In Question 27, $AB$ is $2 \times 2$ but $A’B’$ is $3 \times 3$, so it cannot equal $(AB)’$.
• Treating matrices as commuting numbers. $(AB)^2 \neq A^2B^2$ and $(A+B)^2 \neq A^2 + 2AB + B^2$ unless $AB = BA$ (Questions 30, 31, 49).
• Forgetting that a zero row ends the row reduction. In Question 37(ii) and 51(ii) a row of zeros in the left block means the inverse does not exist; carrying on gives nonsense.
• Searching for $\alpha$ in Question 44. The condition $AA’ = I$ turns into $\sin^2\alpha + \cos^2\alpha = 1$, which is always true, so no particular value of $\alpha$ is singled out.
• Forgetting the final row swap. In Question 51 the unit rows can appear in a shuffled order; the right block must be reordered the same way, or the inverse has its rows in the wrong places.

Practise next

• Exemplar Short Answer Q1–24, the first half of the Short Answer set: order, equality, and matrix products.
• Exemplar Objective Type and Fill in the Blanks Q53–81, which tests the same transposes and inverses in one or two lines each.
• Exemplar True or False Q82–101, where most statements fail because matrices do not commute.
• NCERT textbook Exercise 3.3 for transpose and symmetric or skew-symmetric matrices, as in Questions 27 to 29, 48 and 52.
• NCERT textbook Exercise 3.4 for invertible matrices, as in Questions 44 and 50.
• NCERT textbook Miscellaneous Exercise for further proofs with matrix powers.