Question

If $\left[\begin{array}{cc}3& 1\\ 4& 1\end{array}\right]X=\left[\begin{array}{cc}5& -1\\ 2& 3\end{array}\right]$ then X is equal to:

• A. $\left[\begin{array}{cc}-3& 4\\ 14& -13\end{array}\right]$
• B. $\left[\begin{array}{cc}3& -4\\ -14& 13\end{array}\right]$
• C. $\left[\begin{array}{cc}3& 4\\ 14& 13\end{array}\right]$
• D. $\left[\begin{array}{cc}-3& 4\\ -14& 13\end{array}\right]$

Answer 1

Answer: (A)

Given : $\left[\begin{array}{cc}3& 1\\ 4& 1\end{array}\right]X=\left[\begin{array}{cc}5& -1\\ 2& 3\end{array}\right]$
Thus $X={\left[\begin{array}{cc}3& 1\\ 4& 1\end{array}\right]}^{-1}\left[\begin{array}{cc}5& -1\\ 2& 3\end{array}\right]$
Let $A=\left[\begin{array}{cc}3& 1\\ 4& 1\end{array}\right]$
$a_{11}$ = co-factor of $a_{11}=1$
$a_{12}$ = co-factor of $a_{12}=(-1{)}^{1+2}.4=-4$
$a_{21}$ = co-factor of $a_{21}=(-1{)}^{2+1}.1=-1$
$a_{22}$ = co-factor of $a_{22}=3$
$|A|=3-4=-1$
So $A^{-1}=\frac{\left[\begin{array}{cc}1& -4\\ -1& 3\end{array}\right]}{-1}=\left[\begin{array}{cc}-1& 1\\ 4& -3\end{array}\right]$
Thus $X=\left[\begin{array}{cc}-1& 1\\ 4& -13\end{array}\right]\left[\begin{array}{cc}5& -1\\ 2& 3\end{array}\right]$
$X=\left[\begin{array}{cc}-3& 4\\ 14& -13\end{array}\right]$