Class 12-NCERT Solutions-Chapter-04-Determinants-Ex 4.4

Solution:

Let \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \).

Cofactors:

• \( A_{11} = 4 \)
• \( A_{12} = -3 \)
• \( A_{21} = -2 \)
• \( A_{22} = 1 \)

\[ \text{adj } A = \begin{bmatrix} A_{11} & A_{21} \\ A_{12} & A_{22} \end{bmatrix} = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} \]

Solution:

Let \( A = \begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix} \).

Calculating Cofactors:

• \( A_{11} = +(3-0) = 3 \), \( A_{12} = -(2-(-10)) = -12 \), \( A_{13} = +(0-(-6)) = 6 \)
• \( A_{21} = -(-1-0) = 1 \), \( A_{22} = +(1-(-4)) = 5 \), \( A_{23} = -(0-2) = 2 \)
• \( A_{31} = +(-5-6) = -11 \), \( A_{32} = -(5-4) = -1 \), \( A_{33} = +(3-(-2)) = 5 \)

\[ \text{adj } A = \begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix} \]

Solution:

Let \( A = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} \).

\( |A| = 2(-6) - 3(-4) = -12 + 12 = 0 \).

\( \text{adj } A = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} \).

\[ A(\text{adj } A) = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} -12+12 & -6+6 \\ 24-24 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = 0 \cdot I \]

Similarly, \( (\text{adj } A)A = O \).

Thus, verified.

Solution:

\( |A| = 1(0-0) - (-1)(9-(-2)) + 2(0-0) = 11 \).

Calculated \( \text{adj } A = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} \).

\[ A(\text{adj } A) = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} = 11 I \]

Verified.

Solution:

\( |A| = 6 - (-8) = 14 \neq 0 \). Inverse exists.

\( \text{adj } A = \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix} \).

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

Solution:

\( |A| = -2 - (-15) = 13 \).

\( \text{adj } A = \begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix} \).

\[ A^{-1} = \frac{1}{13} \begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix} \]

Solution:

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

Cofactors:

• \( A_{11}=10, A_{12}=0, A_{13}=0 \)
• \( A_{21}=-10, A_{22}=5, A_{23}=0 \)
• \( A_{31}=2, A_{32}=-4, A_{33}=2 \)

\[ A^{-1} = \frac{1}{10} \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix} \]

Solution:

\( |A| = 1(-3-0) = -3 \).

\[ A^{-1} = \frac{-1}{3} \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix} \]

Solution:

\( |A| = 2(-1-0) - 1(4-0) + 3(8-7) = -2 - 4 + 3 = -3 \).

\[ A^{-1} = \frac{-1}{3} \begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix} \]

Solution:

\( |A| = 1(8-6) - (-1)(0-(-9)) + 2(0-6) = 2 + 9 - 12 = -1 \).

\[ A^{-1} = \frac{1}{-1} \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{bmatrix} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix} \]

Solution:

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

Cofactors yield:

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

Solution:

\[ AB = \begin{bmatrix} 3(6)+7(7) & 3(8)+7(9) \\ 2(6)+5(7) & 2(8)+5(9) \end{bmatrix} = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix} \]

\( |AB| = 67(61) - 87(47) = 4087 - 4089 = -2 \).

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

Now calculate \( B^{-1} \) and \( A^{-1} \) separately and multiply \( B^{-1} A^{-1} \). They will match.

Solution:

\( A^2 = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \).

\( A^2 - 5A + 7I = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \).

To find \( A^{-1} \), multiply equation by \( A^{-1} \):

\( A - 5I + 7A^{-1} = O \Rightarrow 7A^{-1} = 5I - A \).

\[ A^{-1} = \frac{1}{7} \left( \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \right) = \frac{1}{7} \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} \]

Solution:

\( A^2 = \begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix} \).

Equation: \( \begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix} + \begin{bmatrix} 3a & 2a \\ a & a \end{bmatrix} + \begin{bmatrix} b & 0 \\ 0 & b \end{bmatrix} = O \).

\( 11 + 3a + b = 0 \) and \( 4 + a = 0 \Rightarrow a = -4 \).

\( 11 + 3(-4) + b = 0 \Rightarrow 11 - 12 + b = 0 \Rightarrow b = 1 \).

Solution:

Compute \( A^2 \) and \( A^3 \), substitute to verify.

To find inverse: Multiply by \( A^{-1} \):

\( A^2 - 6A + 5I + 11A^{-1} = O \Rightarrow 11A^{-1} = 6A - A^2 - 5I \).

Substitute values to solve for \( A^{-1} \).

Solution:

Similar to Q15.

\( 4A^{-1} = A^2 - 6A + 9I \).

Answer: (B)

Solution:

We know that \( A(\text{adj } A) = |A|I \).

Taking determinant on both sides: \( |A| |\text{adj } A| = | (|A|I) | = |A|^3 |I| = |A|^3 \).

\( |\text{adj } A| = |A|^{3-1} = |A|^2 \).

Answer: (B)

Solution:

\( A A^{-1} = I \).

\( \det(A A^{-1}) = \det(I) = 1 \).

\( \det(A) \det(A^{-1}) = 1 \).

\( \det(A^{-1}) = \frac{1}{\det(A)} \).