The inverse of the matrix A = begin{bmatrix} | vedclass

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(C) Let $A = \begin{bmatrix} 1 + pq & p & 0 \ q & 1 + pq & p \ 0 & q & 1 \end{bmatrix}$.First, calculate the determinant $|A| = (1+pq)((1+pq)(1) - p

Vedclass · Accepted Answer

$\begin{bmatrix} 1 & -p & p^2 \ -q & 1 + pq & -(p + p^2q) \ q^2 & -(q + pq^2) & 1 + pq + p^2q^2 \end{bmatrix}$

Vedclass · Answer

(C) Let $A = \begin{bmatrix} 1 + pq & p & 0 \ q & 1 + pq & p \ 0 & q & 1 \end{bmatrix}$.First, calculate the determinant $|A| = (1+pq)((1+pq)(1) - pq) - p(q(1) - 0) + 0 = (1+pq)(1) - pq = 1 + pq - pq = 1$.Since $|A| = 1$, the inverse $A^{-1}$ is the adjugate matrix $adj(A)$.The cofactors $C_{ij}$ are:$C_{11} = (1+pq)(1) - pq = 1$, $C_{12} = -(q - 0) = -q$, $C_{13} = q^2 - 0 = q^2$.$C_{21} = -(p - 0) = -p$, $C_{22} = (1+pq) - 0 = 1+pq$, $C_{23} = -(q(1+pq) - 0) = -(q+pq^2)$.$C_{31} = p^2 - 0 = p^2$, $C_{32} = -(p(1+pq) - 0) = -(p+p^2q)$, $C_{33} = (1+pq)^2 - pq = 1 + 2pq + p^2q^2 - pq = 1 + pq + p^2q^2$.Thus, $A^{-1} = \begin{bmatrix} C_{11} & C_{21} & C_{31} \ C_{12} & C_{22} & C_{32} \ C_{13} & C_{23} & C_{33} \end{bmatrix} = \begin{bmatrix} 1 & -p & p^2 \ -q & 1 + pq & -(p + p^2q) \ q^2 & -(q + pq^2) & 1 + pq + p^2q^2 \end{bmatrix}$.

The inverse of the matrix $A = \begin{bmatrix} 1 + pq & p & 0 \\ q & 1 + pq & p \\ 0 & q & 1 \end{bmatrix}$ is

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