Single Choice

The inverse of the matrix $$\begin{bmatrix} 1& 0 & 0\\ 3 & 3 & 0\\ 5 & 2 & -1\end{bmatrix}$$ is

A$$-\dfrac {1}{3}\begin{bmatrix} -3& 0 & 0\\ 3 & 1 & 0\\ 9 & 2 & -3\end{bmatrix}$$
B$$-\dfrac {1}{3}\begin{bmatrix} -3& 0 & 0\\ 3 & -1 & 0\\ -9 & -2 & 3\end{bmatrix}$$
Correct Answer
C$$-\dfrac {1}{3}\begin{bmatrix} 3& 0 & 0\\ 3 & -1 & 0\\ -9 & -2 & 3\end{bmatrix}$$
D$$-\dfrac {1}{3}\begin{bmatrix} -3& 0 & 0\\ -3 & -1 & 0\\ -9 & -2 & 3\end{bmatrix}$$

Solution

$$A = \begin{bmatrix} 1& 0 & 0\\ 3 & 3 & 0\\ 5 & 2 & -1\end{bmatrix} \quad \quad |A| = -3\neq 0$$
$$\therefore A^{-1}$$ exist
$$A^{-1} = \dfrac {1}{|A|}Adj\ A$$
$$Adj(A)=$$ Transpose of co-factor of $$A$$
$$M_{11}=(-1)^{1+1}(-3-0)=-3$$, $$M_{12}=(-1)^{1+2}(-3-0)=3$$, $$M_{13}=(-1)^{1+3}(6-15)=-9$$
$$M_{21}=(-1)^{2+1}(0-0)=0$$, $$M_{22}=(-1)^{2+2}(-1+0)=-1$$, $$M_{23}=(-1)^{2+3}(2-0)=-2$$
$$M_{31}=(-1)^{3+1}(0-0)=0$$, $$M_{32}=(-1)^{3+2}(0-0)=0$$, $$M_{33}=(-1)^{3+3}(3-0)=3$$
$$Adj(A)= \begin{bmatrix} -3& 0 & 0\\ 3 & -1 & 0\\ -9 & -2 & 3\end{bmatrix}$$
$$\therefore A^{-1}= -\dfrac {1}{3}\begin{bmatrix} -3& 0 & 0\\ 3 & -1 & 0\\ -9 & -2 & 3\end{bmatrix}$$

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