MHT CET 2026 19th April Morning Shift | Matrices and Determinants Question 7 | Mathematics

If $A = [a_{ij}]_{3\times3}$, where $a_{ij} = \begin{cases} 1, & \text{if } i+j \text{ is even} \\ 0, & \text{if } i+j \text{ is odd} \end{cases}$, then $\text{adj}(A) = \ldots$ ..

$\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$

$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \end{bmatrix}$

$\begin{bmatrix} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$

$\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$

MHT CET 2026 19th April Morning Shift

MCQ (Single Correct Answer)

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

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

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

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

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

MHT CET 2025 5th May Evening Shift

MCQ (Single Correct Answer)

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If $A=\left[\begin{array}{rrr}1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$ then $A(I+\operatorname{adj} A)=$

$\left[\begin{array}{ccc}9 & -2 & 2 \\ 0 & 10 & -3 \\ 3 & -2 & 11\end{array}\right]$

$\left[\begin{array}{ccc}8 & -2 & 2 \\ 0 & 9 & -3 \\ 3 & -2 & 10\end{array}\right]$

$\left[\begin{array}{rrr}9 & -2 & 2 \\ 0 & 10 & -3 \\ 3 & -2 & 12\end{array}\right]$

$\left[\begin{array}{ccc}3 & 2 & -2 \\ 0 & 10 & 3 \\ -3 & 2 & 12\end{array}\right]$

MHT CET 2025 5th May Evening Shift

MCQ (Single Correct Answer)

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The vectors $\bar{p}=\hat{i}+a \hat{j}+a^2 \hat{k}, \bar{q}=\hat{i}+b \hat{j}+b^2 \hat{k}$ and $\overline{\mathrm{r}}=\hat{\mathrm{i}}+\mathrm{c} \hat{\mathrm{j}}+\mathrm{c}^2 \hat{\mathrm{k}}$ are non-coplanar and $\left|\begin{array}{lll}a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \\ c & c^2 & 1+c^3\end{array}\right|=0$ then the value of $(a b c)$ is

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