NCERT Class 12 Maths Chapter 3 Miscellaneous Solutions | LearnCBSEHub

Miscellaneous Universe

CHAPTER 3 • MATRICES • FINAL EXERCISE

💡 Mastery Summary

Symmetry: $A$ is symmetric if $A’ = A$; skew-symmetric if $A’ = -A$.
Transpose Rule: Remember the reversal law $(AB)’ = B’A’$.
Zero Matrix: If a matrix equation equals $O$, it represents a matrix where every element is $0$.

Question 01 • Proof

If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.

Proof: Given that $A$ and $B$ are symmetric matrices, we have $A’ = A$ and $B’ = B$.

To check if $(AB – BA)$ is skew-symmetric, we take its transpose:

$$(AB – BA)’ = (AB)’ – (BA)’$$ $$\text{Using the reversal law: } = B’A’ – A’B’$$ $$\text{Substitute } A’=A \text{ and } B’=B: = BA – AB$$ $$\text{Factor out -1: } = -(AB – BA)$$

✅ Since $X’ = -X$, the matrix is Skew-Symmetric.

Question 02 • Proof

Show that $B’AB$ is symmetric or skew symmetric according as $A$ is symmetric or skew symmetric.

Proof: Let $X = B’AB$. Then,

$$X’ = (B'(AB))’ = (AB)'(B’)’$$ $$X’ = (B’A’)B = B’A’B$$

Case 1: If $A$ is symmetric, then $A’ = A$. Hence, $X’ = B’AB = X$. (Symmetric)

Case 2: If $A$ is skew-symmetric, then $A’ = -A$. Hence, $X’ = B'(-A)B = -B’AB = -X$. (Skew-Symmetric)

Question 03 • Find Unknowns

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

$$A’A = \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix} \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} = \begin{bmatrix} 2x^2 & 0 & 0 \\ 0 & 6y^2 & 0 \\ 0 & 0 & 3z^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

Equating corresponding elements:

$2x^2 = 1 \implies \mathbf{x = \pm \frac{1}{\sqrt{2}}}$
$6y^2 = 1 \implies \mathbf{y = \pm \frac{1}{\sqrt{6}}}$
$3z^2 = 1 \implies \mathbf{z = \pm \frac{1}{\sqrt{3}}}$

Question 04

Find $x$ if $\begin{bmatrix} 1 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = O$.

Step 1: Multiply first two matrices

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

Step 2: Multiply with the column matrix $$\begin{bmatrix} 6 & 2 & 4 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = 0 + 4 + 4x = 0$$

Answer: $4x = -4 \implies x = -1$

Question 05

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

$$A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$$ $$-5A = \begin{bmatrix} -15 & -5 \\ 5 & -10 \end{bmatrix}, \quad 7I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$$ $$\text{Result} = \begin{bmatrix} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$$

✅ Hence Proved.

Question 06

Find $x$ for the 3-matrix product equaling zero.

$$\begin{bmatrix} x & -5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} = \begin{bmatrix} x-2 & -10 & 2x-3 \end{bmatrix}$$ $$\begin{bmatrix} x-2 & -10 & 2x-3 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = x^2 – 2x – 40 + 2x – 3 = 0$$

Answer: $x^2 – 43 = 0 \implies x = \pm \sqrt{43}$

Question 07 • Revenue

Find total revenue and profit using matrix algebra.

(a) Revenue: [Quantity Matrix] $\times$ [Price Matrix]

$$\begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix} = \mathbf{\begin{bmatrix} 46000 \\ 53000 \end{bmatrix}}$$

(b) Profit: (Revenue – Cost). Total profit is Market I: ₹15,000 and Market II: ₹17,000.

Question 08

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

Let $X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Multiplying and comparing:

$a + 4b = -7, \ 2a + 5b = -8 \implies \mathbf{a = 1, b = -2}$
$c + 4d = 2, \ 2c + 5d = 4 \implies \mathbf{c = 2, d = 0}$

Result: $X = \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}$

MCQs (Q9 — Q11)

Theoretical concepts of Chapter 3.

Q9: $A^2 = I \implies \begin{bmatrix} \alpha^2+\beta\gamma & 0 \\ 0 & \beta\gamma+\alpha^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \implies 1 – \alpha^2 – \beta\gamma = 0$.
Correct: (C)

Q10: A matrix both symmetric ($A’=A$) and skew-symmetric ($A’=-A$) means $A = -A \implies 2A = O \implies A = O$.
Correct: (B) Zero Matrix

Q11: Given $A^2 = A$. Expand $(I+A)^3 – 7A$.
$(I + 3A + 3A^2 + A^3) – 7A = (I + 3A + 3A + A) – 7A = I + 7A – 7A = I$.
Correct: (C) I