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If the cofactors of the elements 3,7 and 6 of the matrix $\left[\begin{array}{ccc}1 & 2 & 3 \\ 4 & -1 & 7 \\ 2 & 4 & 6\end{array}\right]$ are a, b and c respectively, then
$\left[\begin{array}{lll}\mathrm{a} & \mathrm{b} & \mathrm{c}\end{array}\right]\left[\begin{array}{l}1 \\ 4 \\ 2\end{array}\right]+\left[\begin{array}{lll}\mathrm{a} & \mathrm{b} & \mathrm{c}\end{array}\right]\left[\begin{array}{l}3 \\ 7 \\ 6\end{array}\right]=$
Options:
$\left[\begin{array}{lll}\mathrm{a} & \mathrm{b} & \mathrm{c}\end{array}\right]\left[\begin{array}{l}1 \\ 4 \\ 2\end{array}\right]+\left[\begin{array}{lll}\mathrm{a} & \mathrm{b} & \mathrm{c}\end{array}\right]\left[\begin{array}{l}3 \\ 7 \\ 6\end{array}\right]=$
Solution:
1209 Upvotes
Verified Answer
The correct answer is:
$0$
In the given matrix
$\begin{aligned}
& C_{13}=(-1)^{1+3}(16+2) \Rightarrow a=18 \\
& C_{23}=(-1)^{2+3}(4-4) \Rightarrow b=0 \\
& C_{33}=(-1)^{3+3}(-1-8) \Rightarrow c=-9
\end{aligned}$
Now,
$\begin{aligned}
& {\left[\begin{array}{lll}
18 & 0 & -9
\end{array}\right]\left[\begin{array}{l}
1 \\
4 \\
2
\end{array}\right]=\left[\begin{array}{lll}
18 & 0 & 9
\end{array}\right]\left[\begin{array}{l}
3 \\
7 \\
6
\end{array}\right]} \\
& =\left[\begin{array}{lll}
18 & 0 & -9
\end{array}\right]\left[\begin{array}{c}
4 \\
11 \\
8
\end{array}\right]=[72-72]=[0]=0
\end{aligned}$
$\begin{aligned}
& C_{13}=(-1)^{1+3}(16+2) \Rightarrow a=18 \\
& C_{23}=(-1)^{2+3}(4-4) \Rightarrow b=0 \\
& C_{33}=(-1)^{3+3}(-1-8) \Rightarrow c=-9
\end{aligned}$
Now,
$\begin{aligned}
& {\left[\begin{array}{lll}
18 & 0 & -9
\end{array}\right]\left[\begin{array}{l}
1 \\
4 \\
2
\end{array}\right]=\left[\begin{array}{lll}
18 & 0 & 9
\end{array}\right]\left[\begin{array}{l}
3 \\
7 \\
6
\end{array}\right]} \\
& =\left[\begin{array}{lll}
18 & 0 & -9
\end{array}\right]\left[\begin{array}{c}
4 \\
11 \\
8
\end{array}\right]=[72-72]=[0]=0
\end{aligned}$
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