1
MHT CET 2025 5th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

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

A

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

B

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

C

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

D

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

2
MHT CET 2025 5th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

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

A

0

B

-1

C

1

D

2

3
MHT CET 2025 26th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $A$ is a matrix of order 2 and $I$ is the identity matrix of order 2 such that $A^2-4 A+3 I=0$ then $(A+3 I)^{-1}=$

A

$\frac{\mathrm{A}}{24}-\frac{7}{24} \mathrm{I}$

B

$\frac{\mathrm{A}}{21}-\frac{7}{21} \mathrm{I}$

C

$\frac{7 \mathrm{I}}{24}-\frac{1}{24} \mathrm{~A}$

D

A-3I

4
MHT CET 2025 26th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

Matrix A is non-singular matrix and $(A-3 I)(A-5 I)=0$, then $\frac{15}{8} A^{-1}=\ldots \ldots$

A

$\mathrm{I}-8 \mathrm{~A}$

B

$2 \mathrm{I}-\frac{1}{15} \mathrm{~A}$

C

$\mathrm{I}-\frac{1}{8} \mathrm{~A}$

D

$8 \mathrm{I}-15 \mathrm{~A}$

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