MHT CET 2025 23rd April Evening Shift | Matrices and Determinants Question 3 | Mathematics

If $A=\left[\begin{array}{rrr}1 & -1 & 1 \\ 0 & 2 & -3 \\ 2 & 1 & 0\end{array}\right], B=\operatorname{adj} A$ and $C=5 A$, then $\frac{|\operatorname{adjB}|}{|\mathrm{C}|}=$

MHT CET 2025 23rd April Morning Shift

MCQ (Single Correct Answer)

-0

If A and B are non-singular matrices of order 2 such that $\quad(A B)^{-1}=\frac{1}{6}\left[\begin{array}{cc}-1 & -3 \\ 2 & 3\end{array}\right] \quad$ and $A^{-1}=\frac{1}{3}\left[\begin{array}{cc}4 & 3 \\ -1 & 0\end{array}\right]$ then $B^{-1}=$

$\frac{1}{2}\left[\begin{array}{cc}2 & 3 \\ 1 & -1\end{array}\right]$

$\frac{1}{2}\left[\begin{array}{ll}3 & 1 \\ 2 & 4\end{array}\right]$

$\frac{1}{2}\left[\begin{array}{cc}-1 & 3 \\ 1 & 2\end{array}\right]$

$\frac{1}{6}\left[\begin{array}{ll}1 & 1 \\ 2 & 3\end{array}\right]$

MHT CET 2025 22nd April Evening Shift

MCQ (Single Correct Answer)

-0

If matrix $\quad A=\frac{1}{11}\left[\begin{array}{rrr}-1 & 7 & -24 \\ 2 & a & 4 \\ 2 & -3 & 15\end{array}\right] \quad$ and $A^{-1}=\left[\begin{array}{rrr}3 & 3 & 4 \\ 2 & -3 & 4 \\ b & -1 & c\end{array}\right]$, then the values of $a, b, c$ respectively are ……

$3,1,0$

$\frac{-6}{11}, 0, \frac{1}{11}$

$-3,0,1$

$\frac{-3}{11}, 0, \frac{1}{11}$

MHT CET 2025 22nd April Morning Shift

MCQ (Single Correct Answer)

-0

If $A=\left[\begin{array}{cc}1 & \cot \frac{\theta}{2} \\ -\cot \frac{\theta}{2} & 1\end{array}\right]$ then $A^{-1}=$

$\operatorname{cosec}^2 \frac{\theta}{2} \mathrm{~A}^{\mathrm{T}}$

$\frac{-\sin ^2 \theta}{2} A^T$

$\left(\frac{1+\cos \theta}{2}\right) A^T$

$\left(\frac{1-\cos \theta}{2}\right) A^T$

PREVIOUS NEXT

MHT CET Subjects