Question

Assertion (A) The system of equations
$2 x-y=-2 ; 3 x+4 y=3$
has unique solution and $x=-\frac{5}{11}$ and $y=\frac{12}{11}$.
Reason (R) The system of equation $A X=B$ has a unique solution, if $|A| \neq 0$

• A. Both $A$ and $R$ are correct; $R$ is the correct explanation of $A$
• B. Both A and R are correct; $R$ is not the correct explanation of $A$
• C. $A$ is correct; $R$ is incorrect
• D. $R$ is correct; $A$ is incorrect

Answer 1

Answer: (A)

The given system can be written as $A X=B$, where
$A=\begin{bmatrix}2 & -1 \\ 3 & 4\end{bmatrix}, X=\begin{bmatrix} x \\ y\end{bmatrix}$ and $B=\begin{bmatrix}{r}-2 \\ 3\end{bmatrix}$
Here, $|A|=\begin{vmatrix}2 & -1 \\ 3 & 4\end{vmatrix}=2 \times 4-(-3)=11 \neq 0$
Thus, $A$ is non-singular. Therefore, its inverse exists.
Therefore, the given system is consistent and has a unique solution given by $X-A^{-1} B$.
Cofactors of $A$ are $A_{11}=4, A_{12}=-3, A_{21}=1$ and $A_{22}=2$.
$\operatorname{adj}(A) =\begin{bmatrix}4 & -3 \\1 & 2\end{bmatrix}^{\prime}=\begin{bmatrix}4 & 1 \\-3 & 2\end{bmatrix}$
$\therefore A^{-1} =\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{11}\begin{bmatrix}4 & 1 \\-3 & 2\end{bmatrix}$
Now, $X =A^{-1} B=\frac{1}{11}\begin{bmatrix}4 & 1 \\-3 & 2\end{bmatrix}\begin{bmatrix}-2 \\3\end{bmatrix} $
$=\frac{1}{11}\begin{bmatrix}-8+3 \\6+6\end{bmatrix}=\frac{1}{11}\begin{bmatrix}-5 \\12\end{bmatrix}=\begin{bmatrix}-\frac{5}{11} \\\frac{12}{11}\end{bmatrix}$
$\Rightarrow \begin{bmatrix}x \\y\end{bmatrix}=\begin{bmatrix}-\frac{5}{11} \\\frac{12}{11}\end{bmatrix}$
Hence, $x=\frac{-5}{11}$ and $y=\frac{12}{11}$.