Question

Find matrix $A$ such that$\left[ \begin{matrix} 2 & -1 \\\ 1 & 0 \\\ -3 & 4 \\\ \end{matrix} \right]A=\left[ \begin{matrix} -1 & -8 \\\ 1 & -2 \\\ 9 & 22 \\\ \end{matrix} \right]$.

Answer 1

In the question, the given matrix consists of three rows and 2 columns. Which we call the $3\times 2$matrix. Consider any $2\times 2$ matrix as$A$ and multiply $\left[ \begin{matrix} 2 & -1 \\\ 1 & 0 \\\ -3 & 4 \\\ \end{matrix} \right]$ with matrix $A$ and equate the corresponding terms with $\left[ \begin{matrix} -1 & -8 \\\ 1 & -2 \\\ 9 & 22 \\\ \end{matrix} \right]$. We have to know the multiplication of matrices.

Complete step by step answer:
From the question it is clear that we have to find the Find matrix $A$ such that$\left[ \begin{matrix} 2 & -1 \\\ 1 & 0 \\\ -3 & 4 \\\ \end{matrix} \right]A=\left[ \begin{matrix} -1 & -8 \\\ 1 & -2 \\\ 9 & 22 \\\ \end{matrix} \right]$.

2 & -1 \\\ 1 & 0 \\\ -3 & 4 \\\ \end{matrix} \right]A=\left[ \begin{matrix} -1 & -8 \\\ 1 & -2 \\\ 9 & 22 \\\ \end{matrix} \right]$$……………………(1) Now let us try to solve the matrix $$A$$ by taking the$$2\times 2$$ matrix as $$A$$. Let us consider matrix $$A={{\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]}_{2\times 2}}$$ Now put matrix $$A={{\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]}_{2\times 2}}$$in equation(1) So equation(1) can be written as $$\Rightarrow \left[ \begin{matrix} 2 & -1 \\\ 1 & 0 \\\ -3 & 4 \\\ \end{matrix} \right]\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]=\left[ \begin{matrix} -1 & -8 \\\ 1 & -2 \\\ 9 & 22 \\\ \end{matrix} \right]$$…………………(2) Now applying multiplication of matrices on $$\left[ \begin{matrix} 2 & -1 \\\ 1 & 0 \\\ -3 & 4 \\\ \end{matrix} \right]$$and $$\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]$$ we get $$\Rightarrow \left[ \begin{matrix} 2 & -1 \\\ 1 & 0 \\\ -3 & 4 \\\ \end{matrix} \right]\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]=\left[ \begin{matrix} 2a-c & 2b-d \\\ a & b \\\ -3a+4c & -3b+4d \\\ \end{matrix} \right]$$ Now put $$\left[ \begin{matrix} 2 & -1 \\\ 1 & 0 \\\ -3 & 4 \\\ \end{matrix} \right]\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]=\left[ \begin{matrix} -1 & -8 \\\ 1 & -2 \\\ 9 & 22 \\\ \end{matrix} \right]$$ in equation (2) $$\Rightarrow \left[ \begin{matrix} 2a-c & 2b-d \\\ a & b \\\ -3a+4c & -3b+4d \\\ \end{matrix} \right]=\left[ \begin{matrix} -1 & -8 \\\ 1 & -2 \\\ 9 & 22 \\\ \end{matrix} \right]$$ Now we got $$3\times 2$$matrices on both sides. So, now we can equate the corresponding terms of LHS to RHS. Now we will get $$\begin{aligned} & \Rightarrow 2a-c=-1 \\\ & \Rightarrow a=1 \\\ & \Rightarrow -3a+4c=9 \\\ & \Rightarrow 2b-d=-8 \\\ & \Rightarrow b=-2 \\\ & \Rightarrow -3b+4d=22 \\\ \end{aligned}$$ Now put $$a=1$$in $$2a-c=-1$$ to get value of $$c$$ $$\Rightarrow 2a-c=-1$$ After simplification we get, $$\begin{aligned} & \Rightarrow c=2+1 \\\ & \Rightarrow c=3 \\\ \end{aligned}$$ So, we got $$c=3$$. Now put $$b=-2$$ in $$2b-d=-8$$. Now We will get $$\Rightarrow 2b-d=-8$$ On simplification we get, $$\begin{aligned} & \Rightarrow 2\left( -2 \right)-d=-8 \\\ & \Rightarrow d=-4+8 \\\ & \Rightarrow d=4 \\\ \end{aligned}$$ So, we got $$d=4$$ Now, we know all the values of $$a,b,c,d$$ Now put $$a=1$$, $$b=-2$$, $$c=3$$, $$d=4$$ in $$A=\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]$$ So now we will get $$\Rightarrow A=\left[ \begin{matrix} 1 & -2 \\\ 3 & 4 \\\ \end{matrix} \right]$$ From there we were asked to find the matrix $$A$$. As per the question we got$$A=\left[ \begin{matrix} 1 & -2 \\\ 3 & 4 \\\ \end{matrix} \right]$$. **Note:** Students should be careful while doing matrix multiplication. During multiplication we have to multiply rows of the first matrix to the columns of the second matrix. Students should be accurate in calculations. Even small calculation mistakes can make large errors in the answer while finding the matrix $$A$$.