The inverse of matrix ≤ft[{array}{ccc}1+p q & p & 0 \\ q & 1+p q & p \\ 0 & q & 1{array}] is ...

Correct Answer is : [1−pp2−q1+pq−(p+p2q)q2−(q+pq2)1+pq+p2q2]≤ft[{array}{ccc}1 & -p & p² \\ -q & 1+p q & -≤ft(p+p² q) \\ q² & -≤ft(q+p q²) & 1+p q+p² q²{array}]1−qq2−p1+pq−(q+pq2)p2−(p+p2q)1+pq+p2q2

≤ft[{array}{ccc}1+p q & p & 0 \\ q & 1+p q & p \\ 0 & q & 1{array}]

≤ft[{array}{ccc}1 & p & p² \\ q & 1+p q & p+p² q \\ q² & q+p q² & 1+p q+p² q²{array}]

≤ft[{array}{ccc}1 & -p & p² \\ -q & 1+p q & -≤ft(p+p² q) \\ q² & -≤ft(q+p q²) & 1+p q+p² q²{array}]

≤ft[{array}{ccc}1 & -p & p² \\ -q & 1+p q & p+p² q \\ q² & q+p q² & 1+p q+p² q²{array}]

The inverse of matrix ≤ft[{array}{ccc}1+p q & p & 0 \\ q & 1+p q & p \\ 0 & q & 1{array}] is ...

Correct Answer is : [1−pp2−q1+pq−(p+p2q)q2−(q+pq2)1+pq+p2q2]≤ft[{array}{ccc}1 & -p & p² \\ -q & 1+p q & -≤ft(p+p² q) \\ q² & -≤ft(q+p q²) & 1+p q+p² q²{array}]1−qq2−p1+pq−(q+pq2)p2−(p+p2q)1+pq+p2q2

Test Index

MHTCET 19 Apr 2026 Shift 1 Paper

Question : 117 of 150

Marks: +1, -0

The inverse of matrix

\left[\begin{array}{ccc}1+p q & p & 0 \\ q & 1+p q & p \\ 0 & q & 1\end{array}\right]

is ...

$\left[\begin{array}{ccc}1+p q & p & 0 \\ q & 1+p q & p \\ 0 & q & 1\end{array}\right]$
$\left[\begin{array}{ccc}1 & p & p^{2} \\ q & 1+p q & p+p^{2} q \\ q^{2} & q+p q^{2} & 1+p q+p^{2} q^{2}\end{array}\right]$
$\left[\begin{array}{ccc}1 & -p & p^{2} \\ -q & 1+p q & -\left(p+p^{2} q\right) \\ q^{2} & -\left(q+p q^{2}\right) & 1+p q+p^{2} q^{2}\end{array}\right]$
$\left[\begin{array}{ccc}1 & -p & p^{2} \\ -q & 1+p q & p+p^{2} q \\ q^{2} & q+p q^{2} & 1+p q+p^{2} q^{2}\end{array}\right]$

Validate

Solution:

Concept:

For a square matrix with non-zero determinant,

A^{-1} = \frac{1}{|A|}\operatorname{adj} A

Explanation:

Let

A = \begin{bmatrix} 1+pq & p & 0 \\ q & 1+pq & p \\ 0 & q & 1 \end{bmatrix}

First compute the determinant by expanding along the first row:

|A| = (1+pq)\begin{vmatrix} 1+pq & p \\ q & 1 \end{vmatrix} - p\begin{vmatrix} q & p \\ 0 & 1 \end{vmatrix}

This simplifies to

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

Since

|A| = 1 \neq 0

, the inverse exists.

The adjoint of

A

is the transpose of the cofactor matrix, which is:

\operatorname{adj} A = \begin{bmatrix} 1 & -p & p^{2} \\ -q & 1+pq & -(p+p^{2}q) \\ q^{2} & -(q+pq^{2}) & 1+pq+p^{2}q^{2} \end{bmatrix}

Therefore,

A^{-1} = \frac{1}{1}\operatorname{adj} A = \operatorname{adj} A

Answer:

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

Hence, the correct option is C.

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