NCERT Solutions Class 12 Maths Chapter 4 Miscellaneous | LearnCBSEHub

Determinants Finale

MISCELLANEOUS EXERCISE • CHAPTER 4 • ALL QUESTIONS

💡 Final Chapter Insights

This exercise combines all properties of determinants and matrices. Key things to remember:

Independence: If a final simplified expression contains no $\theta$, it is independent of $\theta$.
$(AB)^{-1} = B^{-1}A^{-1}$: Multiplication of inverses follows the reversal law.
Determinant Property: Expansion should be done along rows/columns with maximum zeros.

Question 01 • Proof

Prove that the determinant $\begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}$ is independent of $\theta$.

Expanding along $R_1$:

$$\Delta = x(-x^2 – 1) – \sin\theta(-x\sin\theta – \cos\theta) + \cos\theta(-\sin\theta + x\cos\theta)$$ $$\Delta = -x^3 – x + x\sin^2\theta + \sin\theta\cos\theta – \sin\theta\cos\theta + x\cos^2\theta$$ $$\Delta = -x^3 – x + x(\sin^2\theta + \cos^2\theta)$$ $$\Delta = -x^3 – x + x(1) = -x^3$$

Result: $-x^3$ (Independent of $\theta$)

Question 03

If $A^{-1}$ and $B$ are given, find $(AB)^{-1}$.

Property: $(AB)^{-1} = B^{-1}A^{-1}$.

Step 1: Find $B^{-1}$

$|B| = 1(3-0) – 2(1-0) – 2(2-0) = 3 – 2 – 4 = -3$.

$$\text{adj } B = \begin{bmatrix} 3 & -2 & 6 \\ -1 & 1 & -2 \\ -2 & 2 & -5 \end{bmatrix}^T = \begin{bmatrix} 3 & -1 & -2 \\ -2 & 1 & 2 \\ 6 & -2 & -5 \end{bmatrix}$$ $$B^{-1} = -\frac{1}{3} \begin{bmatrix} 3 & -1 & -2 \\ -2 & 1 & 2 \\ 6 & -2 & -5 \end{bmatrix}$$

Step 2: Multiply $B^{-1}A^{-1}$

Perform matrix multiplication to get final $(AB)^{-1}$.

Question 07 • System of Equations

Solve: $\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4$, etc.

Let $u = 1/x, v = 1/y, w = 1/z$. The system becomes linear in $u, v, w$.

$A = \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix}, B = \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix}$ $$|A| = 1200 \neq 0$$ Solving $X = A^{-1}B$ gives $u = 1/2, v = 1/3, w = 1/5$.

Solution: $x = 2, y = 3, z = 5$

Question 08 • MCQ

Inverse of diagonal matrix $A = \text{diag}(x, y, z)$.

For a diagonal matrix, the inverse is simply the reciprocal of the diagonal elements.

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

Correct Option: (A)

Question 09 • Range MCQ

Find $\text{det}(A)$ for $A = \begin{bmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{bmatrix}$.

$$\text{det}(A) = 1(1+\sin^2\theta) – \sin\theta(-\sin\theta+\sin\theta) + 1(\sin^2\theta+1)$$ $$\text{det}(A) = 2(1+\sin^2\theta)$$

Since $0 \leq \sin^2\theta \leq 1$, then $2(1+0) \leq \text{det}(A) \leq 2(1+1)$.

Correct Option: (D) $[2, 4]$