Q64 / 150 · MHT CET 2026 (19 April Shift 1) · Mathematics

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

  • A
    $\begin{bmatrix} 1+pq & p & 0 \\ q & 1+pq & p \\ 0 & q & 1 \end{bmatrix}$
  • B
    $\begin{bmatrix} 1 & p & p^2 \\ q & 1+pq & p+p^2q \\ q^2 & q+pq^2 & 1+pq+p^2q^2 \end{bmatrix}$
  • Correct answer
    C
    $\begin{bmatrix} 1 & -p & p^2 \\ -q & 1+pq & -(p+p^2q) \\ q^2 & -(q+pq^2) & 1+pq+p^2q^2 \end{bmatrix}$
  • D
    $\begin{bmatrix} 1 & -p & p^2 \\ -q & 1+pq & p+p^2q \\ q^2 & q+pq^2 & 1+pq+p^2q^2 \end{bmatrix}$

Solution

Let the given matrix be $A = \begin{bmatrix} 1+pq & p & 0 \\ q & 1+pq & p \\ 0 & q & 1 \end{bmatrix}$.

The inverse of matrix $A$ is given by $A^{-1} = \dfrac{1}{|A|}\text{adj}(A)$.

First, evaluate the determinant of $A$:

$|A| = (1+pq)[(1+pq)(1) - (p)(q)] - p[q(1) - (p)(0)] + 0$

$|A| = (1+pq)[1 + pq - pq] - p[q]$

$|A| = (1+pq)(1) - pq = 1 + pq - pq = 1$

Since $|A| = 1$, the inverse matrix is $A^{-1} = \text{adj}(A)$.

To find the adjoint of $A$, we calculate the cofactors of each element:

$C_{11} = (1+pq)(1) - pq = 1$

$C_{12} = -(q(1) - 0) = -q$

$C_{13} = q(q) - 0 = q^2$

$C_{21} = -(p(1) - 0) = -p$

$C_{22} = (1+pq)(1) - 0 = 1+pq$

$C_{23} = -((1+pq)q - 0) = -(q + pq^2)$

$C_{31} = p(p) - 0 = p^2$

$C_{32} = -((1+pq)p - 0) = -(p + p^2q)$

$C_{33} = (1+pq)^2 - pq = 1 + 2pq + p^2q^2 - pq = 1 + pq + p^2q^2$

The adjoint matrix is the transpose of the cofactor matrix:

$\text{adj}(A) = \begin{bmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{bmatrix} = \begin{bmatrix} 1 & -p & p^2 \\ -q & 1+pq & -(p+p^2q) \\ q^2 & -(q+pq^2) & 1+pq+p^2q^2 \end{bmatrix}$

Therefore, $A^{-1} = \begin{bmatrix} 1 & -p & p^2 \\ -q & 1+pq & -(p+p^2q) \\ q^2 & -(q+pq^2) & 1+pq+p^2q^2 \end{bmatrix}$.

Hence, the correct answer is option: 'C'.

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