NCERT Solutions Class 12 Maths Chapter 4 Ex 4.4 Full | LearnCBSEHub

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NCERT EXERCISE 4.4 • Q1 — Q18 FULL SOLUTIONS

💡 Essential Logic

$\text{adj } A$: The transpose of the matrix of cofactors.
$A^{-1}$: Calculated as $\frac{1}{|A|} \text{adj } A$.
Square Property: $A(\text{adj } A) = |A|I$.

Question 01 & 02 • Adjoint

Find adjoint of the given matrices.

Q1 Solution (2×2 Matrix)

We swap the elements of the primary diagonal and change the signs of the secondary diagonal:

$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \implies \text{adj } A = \mathbf{\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}}$

Q2 Solution (3×3 Matrix)

Calculate cofactors $A_{ij}$ for all elements and then take the transpose:

$\text{adj } A = \begin{bmatrix} 3 & 1 & 1 \\ -12 & 5 & -1 \\ -11 & 2 & 5 \end{bmatrix}$

Question 03 & 04 • Verification

Verify $A (\text{adj } A) = (\text{adj } A) A = |A| I$.

Q3 Verification

For $A = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}$, $|A| = -12 – (-12) = 0$.

$\text{adj } A = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} \implies A(\text{adj } A) = \begin{bmatrix} 2(-6)+3(4) & 2(-3)+3(2) \\ -4(-6)-6(4) & -4(-3)-6(2) \end{bmatrix} = \mathbf{\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}}$

✅ LHS = RHS. Verified.

Question 05 — 11 • Inverse

Find the inverse of each matrix (if it exists).

Q5 Solution

$|A| = 6 – (-8) = 14$. Since $|A| \neq 0$, $A^{-1}$ exists.

$A^{-1} = \frac{1}{14} \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}$

Q7 Solution (3×3)

$|A| = 1(10-0) = 10$.

$\text{adj } A = \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix} \implies A^{-1} = \mathbf{\frac{1}{10} \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix}}$

Q11 Solution (Trigonometry)

$|A| = 1(-\cos^2\alpha – \sin^2\alpha) = -1$.

$A^{-1} = \frac{1}{-1} \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha \end{bmatrix} = \mathbf{\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}}$

Question 12

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

Step 1: Compute AB

$AB = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix} \implies |AB| = -2$.

$(AB)^{-1} = -\frac{1}{2} \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix}$

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

$B^{-1} = -\frac{1}{2} \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}, A^{-1} = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$.

Multiplying these yields the same result. Verified.

Question 13

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

Verification follows by substituting $A^2, A, \text{ and } I$. To find $A^{-1}$:

$A^2 – 5A + 7I = O \implies 7I = 5A – A^2$

Multiply by $A^{-1}$: $7A^{-1} = 5I – A$.

$7A^{-1} = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} – \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$
$A^{-1} = \mathbf{\frac{1}{7} \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}}$

Question 15 & 16

Verify Cubic Matrix Equations and find $A^{-1}$.

For Q15, multiply $A^3 – 6A^2 + 5A + 11I = O$ by $A^{-1}$:

$A^2 – 6A + 5I + 11A^{-1} = O \implies 11A^{-1} = 6A – A^2 – 5I$.

$A^{-1} = \frac{1}{11} \begin{bmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{bmatrix}$

Question 17 & 18 • MCQs

Theoretical properties.

17. For order $n$, $|\text{adj } A| = |A|^{n-1}$. For $n=3$, $|\text{adj } A| = |A|^2$.
Correct Answer: (B)

18. Since $AA^{-1} = I$, then $|A||A^{-1}| = 1 \implies |A^{-1}| = 1/|A|$.
Correct Answer: (B)