Question

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

• 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 $AX=B$, where
$A=\left[\begin{array}{cc}2& -1\\ 3& 4\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right]$ and $B=\left[\begin{array}{c}r-2\\ 3\end{array}\right]$
Here, $|A|=\left|\begin{array}{cc}2& -1\\ 3& 4\end{array}\right|=2\times 4-(-3)=11\ne 0$
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 $A_{22}=2$.
$\mathrm{adj}(A)={\left[\begin{array}{cc}4& -3\\ 1& 2\end{array}\right]}^{\prime }=\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]$
$\therefore A^{-1}=\frac{1}{|A|}(\mathrm{adj}A)=\frac{1}{11}\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]$
Now, $X=A^{-1}B=\frac{1}{11}\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]\left[\begin{array}{c}-2\\ 3\end{array}\right]$
$=\frac{1}{11}\left[\begin{array}{c}-8+3\\ 6+6\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}-5\\ 12\end{array}\right]=\left[\begin{array}{c}-\frac{5}{11}\\ \frac{12}{11}\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-\frac{5}{11}\\ \frac{12}{11}\end{array}\right]$
Hence, $x=\frac{-5}{11}$ and $y=\frac{12}{11}$.