1
MHT CET 2022 11th August Evening Shift
+2
-0

Given $$A=\left[\begin{array}{ccc}x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & z\end{array}\right]$$, if $$x y z=60$$ and $$8 x+4 y+3 z=20$$, then $$A$$.(adjA)

A
$$\left[\begin{array}{ccc}60 & 0 & 0 \\ 0 & 60 & 0 \\ 0 & 0 & 60\end{array}\right]$$
B
$$\left[\begin{array}{ccc}108 & 0 & 0 \\ 0 & 108 & 0 \\ 0 & 0 & 108\end{array}\right]$$
C
$$\left[\begin{array}{ccc}20 & 0 & 0 \\ 0 & 20 & 0 \\ 0 & 0 & 20\end{array}\right]$$
D
$$\left[\begin{array}{ccc}68 & 0 & 0 \\ 0 & 68 & 0 \\ 0 & 0 & 68\end{array}\right]$$
2
MHT CET 2021 24th September Evening Shift
+2
-0

If $$\mathrm{A}=\left[\begin{array}{cc}\lambda & \mathrm{i} \\ \mathrm{i} & -\lambda\end{array}\right]$$ and $$\mathrm{A}^{-1}$$ does not exist, then $$\lambda=$$ (where $$\mathrm{i}=\sqrt{-1}$$)

A
$$\pm 2$$
B
$$\pm 1$$
C
0
D
$$\pm 3$$
3
MHT CET 2021 24th September Evening Shift
+2
-0

If $$A=\left[\begin{array}{ccc}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]$$, and $$A(\operatorname{adj} A)=k I$$, then the value of $$(k+1)^4$$ is

A
256
B
81
C
16
D
625
4
MHT CET 2021 24th September Evening Shift
+2
-0

IF $$A X=B$$, where $$A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right], B=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$$, then $$2 x+y-z=$$

A
2
B
1
C
4
D
$$-$$2
EXAM MAP
Medical
NEET