If $\left[\begin{array}{ccc}5 & a & -7 \\ b & -7 & c \\ -7 & d & -1\end{array}\right]$ is the adjoint of the…
If $\left[\begin{array}{ccc}5 & a & -7 \\ b & -7 & c \\ -7 & d & -1\end{array}\right]$ is the adjoint of the matrix $\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2\end{array}\right]$, then $a+b+c+d=$
$8$
$10$
$0$
$2$
Solution
$\begin{aligned} & C_{11}=(-1)^{1+1}(3 \times 2-1 \times 1)=5 \\ & C_{12}=(-1)^{1+2}(2 \times 2-3 \times 1)=-1 \\ & C_{13}=(-1)^{1+3}(2 \times 1-3 \times 3)=-7 \\ & C_{21}=(-1)^{2+1}(4-3)=-1 \\ & C_{22}=(-1)^{2+2}(2-9)=-7 \\ & C_{23}=(-1)^{2+3}(1-6)=5 \\ & C_{31}=(-1)^{3+1}(2-9)=-7 \\ & C_{32}=(-1)^{3+2}(1-6)=5 \\ & C_{33}=(-1)^{3+3}(3-4)=-1\end{aligned}$
Cofactor of given matrix is given by,
$\left[\begin{array}{ccc}5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1\end{array}\right]=B($ say $)$
Adjoint of given matrix
$=B^T=\left[\begin{array}{ccc}5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1\end{array}\right]=\left[\begin{array}{ccc}5 & a & -7 \\ b & -7 & c \\ -7 & d & -1\end{array}\right]$
After comparing, we get $a=b=-1$ and $c=d=5$
Thus, $a+b+c+d=-1-1+5+5=8$