IfA=⎡⎢⎣201110211⎤⎥⎦ and adjA=⎡⎢⎣12−1xyz−1−22�

1.	IfA=⎡⎢⎣201110211⎤⎥⎦ and adjA=⎡⎢⎣12−1xyz−1−22⎤⎥⎦ then (x,y,z)
Answer» If $A = ⎡ ⎢ ⎣ \begin{matrix} 2 & 0 & 1 1 & 1 & 0 2 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦$ and $a d j A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & - 1 x & y & z - 1 & - 2 & 2 \end{matrix} ⎤ ⎥ ⎦$ then $(x, y, z)$

IfA=⎡⎢⎣201110211⎤⎥⎦ and adjA=⎡⎢⎣12−1xyz−1−22⎤⎥⎦ then (x,y,z)

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If $A = ⎡ ⎢ ⎣ \begin{matrix} 2 & 0 & 1 1 & 1 & 0 2 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦$ and $a d j A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & - 1 x & y & z - 1 & - 2 & 2 \end{matrix} ⎤ ⎥ ⎦$ then $(x, y, z)$

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IfA=⎡⎢⎣201110211⎤⎥⎦ and adjA=⎡⎢⎣12−1xyz−1−22⎤⎥⎦ then (x,y,z)

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