MCQ
Suppose $X =\left[x_{i j}\right]$ a matrix, where
$
X=\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 4 & -5 \\
2 & -1 & 3
\end{array}\right]
$
Then matrix $Y =\left[m_{i j}\right]$, where $m_{i j}=$ minor of $x_{i j}$ :
  • A
    $\left[\begin{array}{ccc}7 & -5 & -3 \\ 19 & 1 & -11 \\ -11 & 1 & 7\end{array}\right]$
  • B
    $\left[\begin{array}{ccc}7 & -19 & -11 \\ 5 & -1 & -1 \\ 3 & 11 & 7\end{array}\right]$
  • C
    $\left[\begin{array}{ccc}7 & 19 & -11 \\ -3 & 11 & 7 \\ -5 & -1 & -1\end{array}\right]$
  • $\left[\begin{array}{ccc}7 & 19 & -11 \\ -1 & -1 & 1 \\ -3 & -11 & 7\end{array}\right]$

Answer

Correct option: D.
$\left[\begin{array}{ccc}7 & 19 & -11 \\ -1 & -1 & 1 \\ -3 & -11 & 7\end{array}\right]$
(D)
Here $X=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3\end{array}\right]$$
\begin{aligned}
\therefore \quad m_{11} & =\left|\begin{array}{cc}
4 & -5 \\
-1 & 3
\end{array}\right|=12-5=7 \\
m_{12} & =\left|\begin{array}{cc}
3 & -5 \\
2 & 3
\end{array}\right|=9+10=19 \\
m_{13} & =\left|\begin{array}{cc}
3 & 4 \\
2 & -1
\end{array}\right|=-3-8=-11
\end{aligned}
$
Thus $\quad m_{21}=-1, m_{22}=-1, m_{23}=1$$
\begin{aligned}
m_{31} & =-3, m_{32}=-11, m_{33}=7 \\
Y=\left[m_{i j}\right] & =\left[\begin{array}{lll}
m_{11} & m_{12} & m_{13} \\
m_{21} & m_{22} & m_{23} \\
m_{31} & m_{32} & m_{33}
\end{array}\right]=\left[\begin{array}{ccc}
7 & 19 & -11 \\
-1 & -1 & 1 \\
-3 & -11 & 7
\end{array}\right]
\end{aligned}
$

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