Question
It is given that $X\left[\begin{array}{cc}3 & 2 \\ 1 & -1\end{array}\right]=\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]$. Then matrix $X$ is :
✓
Answer
Let $X=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
We have, $x\left[\begin{array}{cc}3 & 2 \\ 1 & -1\end{array}\right]=\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]$
$ \Rightarrow\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\left[\begin{array}{cc}
3 & 2 \\ 1 & -1 \end{array}\right]=\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \end{array}\right] $
$ \Rightarrow\left[\begin{array}{ll} 3 a+b & 2 a-b \\ 3 c+d & 2 c-d \end{array}\right]=\left[\begin{array}{ll}
4 & 1 \\ 2 & 3 \end{array}\right]$
On comparing the element of matrices, we get
$ \begin{array}{l} 3 a+b=4 \\ 2 a-b=1 \\ 3 c+d=2 \\ 2 c-d=3 \end{array} $
Adding $(i)$ and $(ii),$ we get $5 a=5 \Rightarrow a=1$
Putting $a=1$ in $(i),$ we get $3(1)+b=4 \Rightarrow b=1$
Adding $(iii)$ and $(iv),$ we get $5 c=5 \Rightarrow c=1$
Putting $c=1$ in $(iii),$ we get $3+d=2 \Rightarrow d=-1$
$ \therefore X=\left[\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right]$
We have, $x\left[\begin{array}{cc}3 & 2 \\ 1 & -1\end{array}\right]=\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]$
$ \Rightarrow\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\left[\begin{array}{cc}
3 & 2 \\ 1 & -1 \end{array}\right]=\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \end{array}\right] $
$ \Rightarrow\left[\begin{array}{ll} 3 a+b & 2 a-b \\ 3 c+d & 2 c-d \end{array}\right]=\left[\begin{array}{ll}
4 & 1 \\ 2 & 3 \end{array}\right]$
On comparing the element of matrices, we get
$ \begin{array}{l} 3 a+b=4 \\ 2 a-b=1 \\ 3 c+d=2 \\ 2 c-d=3 \end{array} $
Adding $(i)$ and $(ii),$ we get $5 a=5 \Rightarrow a=1$
Putting $a=1$ in $(i),$ we get $3(1)+b=4 \Rightarrow b=1$
Adding $(iii)$ and $(iv),$ we get $5 c=5 \Rightarrow c=1$
Putting $c=1$ in $(iii),$ we get $3+d=2 \Rightarrow d=-1$
$ \therefore X=\left[\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right]$
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