NCERT Class 12 Mathematics — Matrices, Miscellaneous Exercise on Chapter 3. All 11 questions solved.

The Miscellaneous Exercise mixes the two halves of the chapter: proofs about symmetry that rely on the transpose rules, and computational questions where a matrix product is set equal to something and an unknown falls out.

Everything rests on three facts already established:

$$(\mathrm{AB})’ = \mathrm{B}’\mathrm{A}’, \qquad \mathrm{A} \text{ symmetric} \iff \mathrm{A}’ = \mathrm{A}, \qquad \mathrm{A} \text{ skew symmetric} \iff \mathrm{A}’ = -\mathrm{A}$$

Key insight. For any question of the form “show that this expression is symmetric or skew symmetric”, the method is always the same: transpose the whole expression and simplify until you get either the original back or its negative. Questions 1 and 2 are that method applied twice, and nothing else is needed for either.

Question 1

If A and B are symmetric matrices, prove that $\mathrm{AB} – \mathrm{BA}$ is a skew symmetric matrix.

Solution. Transpose the expression, using $(\mathrm{AB})’ = \mathrm{B}’\mathrm{A}’$ on each product.

$$(\mathrm{AB} – \mathrm{BA})’ = (\mathrm{AB})’ – (\mathrm{BA})’ = \mathrm{B}’\mathrm{A}’ – \mathrm{A}’\mathrm{B}’$$

Both matrices are symmetric, so $\mathrm{A}’ = \mathrm{A}$ and $\mathrm{B}’ = \mathrm{B}$:

$$= \mathrm{BA} – \mathrm{AB} = -(\mathrm{AB} – \mathrm{BA})$$

The transpose equals the negative of the original, which is precisely the definition of skew symmetric.

Question 2

Show that the matrix $\mathrm{B}’\mathrm{AB}$ is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Solution. Transpose the product. With three factors the rule $(\mathrm{XY})’ = \mathrm{Y}’\mathrm{X}’$ applies twice, reversing the whole order:

$$(\mathrm{B}’\mathrm{AB})’ = \mathrm{B}’\mathrm{A}'(\mathrm{B}’)’ = \mathrm{B}’\mathrm{A}’\mathrm{B}$$

using $(\mathrm{B}’)’ = \mathrm{B}$. Everything now depends on what $\mathrm{A}’$ is.

If A is symmetric, $\mathrm{A}’ = \mathrm{A}$, so

$$(\mathrm{B}’\mathrm{AB})’ = \mathrm{B}’\mathrm{AB}$$

and the product is symmetric.

If A is skew symmetric, $\mathrm{A}’ = -\mathrm{A}$, so

$$(\mathrm{B}’\mathrm{AB})’ = \mathrm{B}'(-\mathrm{A})\mathrm{B} = -\mathrm{B}’\mathrm{AB}$$

and the product is skew symmetric.

Notice that the outer factors $\mathrm{B}’$ and $\mathrm{B}$ come back in the same positions after transposing — that is the whole reason this construction preserves symmetry, and it would fail for $\mathrm{BAB}’$ in general.

Question 3

Find the values of $x$, $y$, $z$ if the matrix $\mathrm{A} = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}$ satisfies the equation $\mathrm{A}’\mathrm{A} = \mathrm{I}$.

Solution. Compute $\mathrm{A}’\mathrm{A}$. Its $(i,j)$ entry is the dot product of column $i$ with column $j$ of A, and the off-diagonal ones turn out to vanish automatically:

$$\mathrm{A}’\mathrm{A} = \begin{bmatrix} 2x^2 & 0 & 0 \\ 0 & 6y^2 & 0 \\ 0 & 0 & 3z^2 \end{bmatrix}$$

Setting this equal to I gives three independent equations:

$$2x^2 = 1, \qquad 6y^2 = 1, \qquad 3z^2 = 1$$

$$x = \pm\frac{1}{\sqrt{2}}, \qquad y = \pm\frac{1}{\sqrt{6}}, \qquad z = \pm\frac{1}{\sqrt{3}}$$

Each sign is free, so there are eight matrices satisfying the condition. The off-diagonal entries being zero is what makes this work: it says the columns of A are already mutually perpendicular, and only their lengths need fixing.

Question 4

For what values of $x$: $\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = \mathrm{O}$?

Solution. Work left to right, since the first product is fully numerical.

$$\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 2 & 4 \end{bmatrix}$$

Then multiply by the column:

$$\begin{bmatrix} 6 & 2 & 4 \end{bmatrix}\begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = 0 + 4 + 4x = 4 + 4x$$

The result is a $1\times1$ matrix, and setting it to zero:

$$4 + 4x = 0 \quad\Rightarrow\quad x = -1$$

Question 5

If $\mathrm{A} = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, show that $\mathrm{A}^2 – 5\mathrm{A} + 7\mathrm{I} = 0$.

Solution.

$$\mathrm{A}^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$$

$$\mathrm{A}^2 – 5\mathrm{A} + 7\mathrm{I} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} – \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$

This same matrix reappears in Chapter 4, Exercise 4.4 question 13, where the identity is used to produce $\mathrm{A}^{-1}$ without computing an adjoint.

Question 6

Find $x$, if $\begin{bmatrix} x & -5 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}\begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = \mathrm{O}$.

Solution. Unlike question 4, here $x$ appears in both outer factors, so the result will be quadratic. Taking the first product:

$$\begin{bmatrix} x & -5 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} = \begin{bmatrix} x – 2 & -10 & 2x – 8 \end{bmatrix}$$

Then multiplying by the column:

$$(x-2)x + (-10)(4) + (2x-8)(1) = x^2 – 2x – 40 + 2x – 8 = x^2 – 48$$

$$x^2 – 48 = 0 \quad\Rightarrow\quad x = \pm 4\sqrt{3}$$

Both roots are valid — nothing in the question excludes the negative one.

Question 7

A manufacturer produces three products $x$, $y$, $z$ which he sells in two markets. Annual sales are: Market I — 10,000, 2,000, 18,000; Market II — 6,000, 20,000, 8,000. (a) If unit sale prices are ₹2.50, ₹1.50 and ₹1.00, find the total revenue in each market. (b) If the unit costs are ₹2.00, ₹1.00 and 50 paise, find the gross profit.

Solution. Write the sales as a $2\times3$ matrix, with one row per market, and the prices as a $3\times1$ column. The product is then revenue per market.

(a)

$$\begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix}\begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix} = \begin{bmatrix} 25000 + 3000 + 18000 \\ 15000 + 30000 + 8000 \end{bmatrix} = \begin{bmatrix} 46000 \\ 53000 \end{bmatrix}$$

(b) The same sales matrix against the cost column — note that 50 paise is ₹0.50:

$$\begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix}\begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix} = \begin{bmatrix} 31000 \\ 36000 \end{bmatrix}$$

Gross profit is revenue minus cost, market by market:

$$\begin{bmatrix} 46000 \\ 53000 \end{bmatrix} – \begin{bmatrix} 31000 \\ 36000 \end{bmatrix} = \begin{bmatrix} 15000 \\ 17000 \end{bmatrix}$$

Question 8

Find the matrix X so that $\mathrm{X}\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$.

Solution. First work out the order of X. It multiplies a $2\times3$ on the left to give a $2\times3$, so X must be $2\times2$. Write $\mathrm{X} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and multiply out:

$$\begin{bmatrix} a + 4b & 2a + 5b & 3a + 6b \\ c + 4d & 2c + 5d & 3c + 6d \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$$

From the first row, $a + 4b = -7$ and $2a + 5b = -8$. Substituting $a = -7 – 4b$ into the second gives $-14 – 8b + 5b = -8$, so $b = -2$ and $a = 1$. The third entry checks: $3(1) + 6(-2) = -9$ ✓

From the second row, $c + 4d = 2$ and $2c + 5d = 4$ give $d = 0$ and $c = 2$, and again the third entry checks: $3(2) + 6(0) = 6$ ✓

Question 9

If $\mathrm{A} = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$ is such that $\mathrm{A}^2 = \mathrm{I}$, then

Solution. Square the matrix. The off-diagonal terms cancel because of the $-\alpha$ in the corner:

$$\mathrm{A}^2 = \begin{bmatrix} \alpha^2 + \beta\gamma & \alpha\beta – \beta\alpha \\ \gamma\alpha – \alpha\gamma & \gamma\beta + \alpha^2 \end{bmatrix} = \begin{bmatrix} \alpha^2 + \beta\gamma & 0 \\ 0 & \alpha^2 + \beta\gamma \end{bmatrix}$$

Setting $\mathrm{A}^2 = \mathrm{I}$ requires the diagonal entries to be 1:

$$\alpha^2 + \beta\gamma = 1 \quad\Rightarrow\quad 1 – \alpha^2 – \beta\gamma = 0$$

Question 10

If the matrix A is both symmetric and skew symmetric, then

Solution. Both conditions hold at once, so write them down together:

$$\mathrm{A}’ = \mathrm{A} \quad\text{and}\quad \mathrm{A}’ = -\mathrm{A}$$

Therefore $\mathrm{A} = -\mathrm{A}$, so $2\mathrm{A} = \mathrm{O}$ and $\mathrm{A} = \mathrm{O}$.

Option (C) is true but not an answer to the question — A must be square for the transpose conditions to make sense at all, so it is a precondition rather than a conclusion. The zero matrix is the only matrix that is both.

Question 11

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

Solution. Expand the cube. This is legitimate in the usual binomial form because I commutes with everything:

$$(\mathrm{I}+\mathrm{A})^3 = \mathrm{I} + 3\mathrm{A} + 3\mathrm{A}^2 + \mathrm{A}^3$$

Now use $\mathrm{A}^2 = \mathrm{A}$. It follows that $\mathrm{A}^3 = \mathrm{A}^2\mathrm{A} = \mathrm{A}\mathrm{A} = \mathrm{A}^2 = \mathrm{A}$, so every power of A collapses to A:

$$= \mathrm{I} + 3\mathrm{A} + 3\mathrm{A} + \mathrm{A} = \mathrm{I} + 7\mathrm{A}$$

$$(\mathrm{I}+\mathrm{A})^3 – 7\mathrm{A} = \mathrm{I} + 7\mathrm{A} – 7\mathrm{A} = \mathrm{I}$$

The $7$ in the question is chosen precisely to cancel this coefficient — a hint that the expansion is the intended route.

Common mistakes

• Failing to reverse the order when transposing a product. Question 2 needs $(\mathrm{B}’\mathrm{AB})’ = \mathrm{B}’\mathrm{A}’\mathrm{B}$; taking it as $\mathrm{B}\mathrm{A}’\mathrm{B}’$ breaks the argument entirely.
• Multiplying a triple product in the wrong order. In questions 4 and 6 the orders only fit one way. Work left to right and the intermediate is a row matrix each time.
• Giving only the positive root in question 6. $x^2 = 48$ has two solutions, and nothing in the question rules either out — unlike question 4, which is linear and has one.
• Forgetting the eight sign combinations in question 3. Each of $x$, $y$, $z$ is determined only up to sign.
• Leaving 50 paise as 50 in question 7. The other costs are in rupees, so it must be ₹0.50.
• Expanding $(\mathrm{I}+\mathrm{A})^3$ carelessly. The binomial expansion is valid here only because I commutes with A; for two general matrices $(\mathrm{X}+\mathrm{Y})^2 \neq \mathrm{X}^2 + 2\mathrm{XY} + \mathrm{Y}^2$.

Practise next

• Chapter 4, Determinants — where the inverse defined in Exercise 3.4 is actually computed, and matrices are used to solve systems of equations.
• Exercise 3.3 — worth a second pass if the transpose rules in questions 1 and 2 felt mechanical rather than obvious.