COMEDK 2025 Morning Shift | Matrices and Determinants Question 9 | Mathematics

If $A=\left[\begin{array}{cc}1 & -2 \\ 4 & 5\end{array}\right] ; f(t)=t^2-3 t+7$ then $f(A)+\left[\begin{array}{cc}3 & 6 \\ -12 & -9\end{array}\right]=$

$\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$

$\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]$

$\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

COMEDK 2025 Morning Shift

MCQ (Single Correct Answer)

-0

For two matrices $A$ and $B$, given that $A^{-1}=\frac{1}{8} B$ then inverse of $(8 A)$ is

$\frac{1}{8} B$

8 B

$\frac{1}{64} B$

COMEDK 2025 Morning Shift

MCQ (Single Correct Answer)

-0

If $A=\left[\begin{array}{ccc}0 & 1 & -2 \\ -1 & 0 & 3 \\ 2 & -3 & 0\end{array}\right]$ then $A^{-1}$

equal to $-\frac{1}{12}(\operatorname{adj} A)$

equal to $-$12

equal to $\frac{1}{12}(\operatorname{adj} A)$

doesn't exit

COMEDK 2024 Evening Shift

MCQ (Single Correct Answer)

-0

$$ \text { If } 3 A+4 B^t=\left(\begin{array}{ccc} 7 & -10 & 17 \\ 0 & 6 & 31 \end{array}\right) \text { and } 2 B-3 A^t=\left(\begin{array}{cc} -1 & 18 \\ 4 & -6 \\ -5 & -7 \end{array}\right) \text { then }(5 B)^t= $$

$$ \left(\begin{array}{ccc} 5 & 5 & 10 \\ 15 & 0 & 20 \end{array}\right) $$

$$ \left(\begin{array}{ccc} -5 & 5 & 10 \\ -15 & 0 & 20 \end{array}\right) $$

$$ \left(\begin{array}{ccc} 5 & -5 & -10 \\ 15 & 0 & -20 \end{array}\right) $$

$$ \left(\begin{array}{ccc} 5 & -5 & 10 \\ 15 & 0 & 20 \end{array}\right) $$

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