Q4(i):

Let $A = \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}$. Verify that (i) $[adj A]^{-1} = adj (A^{-1})$

Step 1: Computing the Determinant of Matrix $A$

Given the matrix $A = \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}$.

The determinant $|A|$ is calculated by expanding along the first row [Laplace Expansion Theorem]:

$|A| = 1(3(5) - 1(1)) - (-2)((-2)(5) - 1(1)) + 1((-2)(1) - 3(1))$

$|A| = 1(15 - 1) + 2(-10 - 1) + 1(-2 - 3)$

$|A| = 1(14) + 2(-11) + 1(-5) = 14 - 22 - 5 = -13$

Since $|A| \neq 0$, the inverse $A^{-1}$ exists.

Step 2: Computing the Adjoint of Matrix $A$

The adjoint, $adj(A)$, is the transpose of the cofactor matrix $C = [C_{ij}]$.

• $C_{11} = +(15-1) = 14$; $C_{12} = -(-10-1) = 11$; $C_{13} = +(-2-3) = -5$
• $C_{21} = -(-10-1) = 11$; $C_{22} = +(5-1) = 4$; $C_{23} = -(1+2) = -3$
• $C_{31} = +(-2-3) = -5$; $C_{32} = -(1+2) = -3$; $C_{33} = +(3-4) = -1$

$adj(A) = \begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix}^T = \begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix}$

Step 3: Calculating $[adj A]^{-1}$

Let $B = adj(A)$. To find $B^{-1} = \frac{1}{|B|} adj(B)$, we first find $|B|$:

$|B| = 14(-4 - 9) - 11(-11 - 15) - 5(-33 + 20) = 14(-13) - 11(-26) - 5(-13)$

$|B| = -182 + 286 + 65 = 169$

Note that $|adj(A)| = |A|^{n-1} = (-13)^{3-1} = (-13)^2 = 169$ [Property: $|adj(A)| = |A|^{n-1}$].

Computing the adjugate of $B$ (cofactors of $adj(A)$):

$[adj(adj(A))] = \begin{bmatrix} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{bmatrix} \implies [adj A]^{-1} = \frac{1}{169} \begin{bmatrix} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{bmatrix} = \begin{bmatrix} -1/13 & 2/13 & -1/13 \\ 2/13 & -3/13 & -1/13 \\ -1/13 & -1/13 & -5/13 \end{bmatrix}$

Step 4: Calculating $adj(A^{-1})$

Using the property $A^{-1} = \frac{1}{|A|} adj(A) = -\frac{1}{13} \begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix}$:

$A^{-1} = \begin{bmatrix} -14/13 & -11/13 & 5/13 \\ -11/13 & -4/13 & 3/13 \\ 5/13 & 3/13 & 1/13 \end{bmatrix}$

The inverse of a matrix $X$ is $\frac{1}{|X|}adj(X)$. For a $3 \times 3$ matrix, $adj(A^{-1}) = |A^{-1}| (A^{-1})^{-1}$.

Since $(A^{-1})^{-1} = A$ and $|A^{-1}| = \frac{1}{|A|} = -\frac{1}{13}$:

$adj(A^{-1}) = -\frac{1}{13} A = -\frac{1}{13} \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix} = \begin{bmatrix} -1/13 & 2/13 & -1/13 \\ 2/13 & -3/13 & -1/13 \\ -1/13 & -1/13 & -5/13 \end{bmatrix}$

Comparison Table

Final Solution: Since the matrices are identical, the identity $[adj A]^{-1} = adj (A^{-1})$ is verified.