1
MHT CET 2026 19th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
The inverse of the matrix $A = \begin{bmatrix} 2 & -1 & 4 \\ 4 & -3 & 1 \\ 1 & 2 & 1 \end{bmatrix}$ is $B = \dfrac{1}{37}\begin{bmatrix} -5 & 9 & 11 \\ -3 & -2 & 14 \\ 11 & -5 & k \end{bmatrix}$, then the value of $k$ is...
A
$1$
B
$-1$
C
$2$
D
$-2$
2
MHT CET 2026 19th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
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 $\text{adj}(A) = \ldots$ ..
A
$\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$
B
$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \end{bmatrix}$
C
$\begin{bmatrix} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$
D
$\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$
3
MHT CET 2026 19th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
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
$\begin{bmatrix} 1 & p & p^2 \\ q & 1+pq & p+p^2q \\ q^2 & q+pq^2 & 1+pq+p^2q^2 \end{bmatrix}$
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
$\begin{bmatrix} 1 & -p & p^2 \\ -q & 1+pq & p+p^2q \\ q^2 & q+pq^2 & 1+pq+p^2q^2 \end{bmatrix}$
4
MHT CET 2026 18th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
If matrix A and its inverse $A^{-1}$ are given by $A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} \dfrac{1}{2} & -\dfrac{1}{2} & \dfrac{1}{2} \\ -4 & 3 & y \\ \dfrac{5}{2} & -\dfrac{3}{2} & \dfrac{1}{2} \end{bmatrix}$, then the polar co-ordinates of the points whose Cartesian co-ordinates are $(x, y)$ are $\ldots$
A
$\left(2, \dfrac{7\pi}{4}\right)$
B
$\left(\sqrt{2}, \dfrac{\pi}{4}\right)$
C
$\left(\sqrt{2}, \dfrac{7\pi}{4}\right)$
D
$\left(2, \dfrac{\pi}{4}\right)$

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