1
COMEDK 2024 Morning Shift
+1
-0

If $$A=\left[\begin{array}{ccc}-1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$$ then the inverse of $$(A I)^t$$ (where $$\mathrm{I}$$ is an identity matrix) is

A
$$\left[\begin{array}{ccc} 1 & -8 & 5 \\ -1 & 7 & -4 \\ 1 & -5 & 3 \end{array}\right]$$
B
$$\left[\begin{array}{ccc} -1 & 1 & -1 \\ 8 & -7 & 5 \\ -5 & 4 & -3 \end{array}\right]$$
C
$$\left[\begin{array}{ccc} 1 & 8 & -5 \\ -1 & 7 & -4 \\ 0 & 5 & 3 \end{array}\right]$$
D
$$\left[\begin{array}{ccc} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{array}\right]$$
2
COMEDK 2024 Morning Shift
+1
-0

$$\text { If } x, y, z \text { are non zero real numbers, then inverse of matrix } A=\left[\begin{array}{lll} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right] \text { is }$$

A
$$\frac{1}{x y z}\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$
B
$$\left[\begin{array}{ccc} 1 / x & 0 & 0 \\ 0 & 1 / y & 0 \\ 0 & 0 & 1 / z \end{array}\right]$$
C
$$x y z\left[\begin{array}{ccc} 1 / x & 0 & 0 \\ 0 & 1 / y & 0 \\ 0 & 0 & 1 / z \end{array}\right]$$
D
$$\frac{1}{x y z}\left[\begin{array}{lll} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right]$$
3
COMEDK 2024 Morning Shift
+1
-0

$$\text { If the matrix } A=\left(\begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array}\right) \text { then } A^{n+1}=$$

A
$$2\left(\begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array}\right)$$
B
$$2 n\left(\begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array}\right)$$
C
$$2^{n+1}\left(\begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array}\right)$$
D
$$2^n\left(\begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array}\right)$$
4
COMEDK 2024 Morning Shift
+1
-0

If $$\left[\begin{array}{lll} 1 & x & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \\ x \end{array}\right]=[0]$$ then x is equal to

A
2, 14
B
$$-2,-14$$
C
7, 4
D
2, $$-14$$
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