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$\operatorname{adj}\left[\begin{array}{ccc}1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1\end{array}\right]=\left[\begin{array}{ccc}5 & a & -2 \\ 1 & 1 & 0 \\ -2 & -2 & b\end{array}\right], \quad$ then $\left[\begin{array}{ll}a & b\end{array}\right]$ is equal to
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1659 Upvotes
Verified Answer
The correct answer is:
[4 1]
Given that
$\begin{aligned}
& \text { Cofactor of }\left[\begin{array}{ccc}
1 & 0 & 2 \\
-1 & 1 & -2 \\
0 & 2 & 1
\end{array}\right] \text { are } \\
& C_{11}=5, C_{12}=1, C_{13}=-2 \\
& C_{21}=4, C_{22}=1, C_{23}=-2 \\
& C_{31}=-2, C_{32}=0, C_{33}=1 \\
& \Rightarrow\left[\begin{array}{ccc}
5 & 4 & -2 \\
1 & 1 & 0 \\
-2 & -2 & 1
\end{array}\right]=\left[\begin{array}{ccc}
5 & a & -2 \\
1 & 1 & 0 \\
-2 & -2 & b
\end{array}\right]
\end{aligned}$
On comparing the corresponding elements, we get
$\begin{array}{rlrl}
& a & =4, b=1 \\
\therefore & {\left[\begin{array}{ll}
a & b
\end{array}\right]} & =\left[\begin{array}{ll}
4 & 1
\end{array}\right]
\end{array}$
$\begin{aligned}
& \text { Cofactor of }\left[\begin{array}{ccc}
1 & 0 & 2 \\
-1 & 1 & -2 \\
0 & 2 & 1
\end{array}\right] \text { are } \\
& C_{11}=5, C_{12}=1, C_{13}=-2 \\
& C_{21}=4, C_{22}=1, C_{23}=-2 \\
& C_{31}=-2, C_{32}=0, C_{33}=1 \\
& \Rightarrow\left[\begin{array}{ccc}
5 & 4 & -2 \\
1 & 1 & 0 \\
-2 & -2 & 1
\end{array}\right]=\left[\begin{array}{ccc}
5 & a & -2 \\
1 & 1 & 0 \\
-2 & -2 & b
\end{array}\right]
\end{aligned}$
On comparing the corresponding elements, we get
$\begin{array}{rlrl}
& a & =4, b=1 \\
\therefore & {\left[\begin{array}{ll}
a & b
\end{array}\right]} & =\left[\begin{array}{ll}
4 & 1
\end{array}\right]
\end{array}$
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