Question

Find matrix $A$ such that $\left[ {\begin{array}{*{20}{c}}
  2&{ - 1} \\
  1&0 \\
  { - 3}&4
\end{array}} \right]A = \left[ {\begin{array}{*{20}{c}}
  { - 1}&{ - 8} \\
  1&{ - 2} \\
  9&{22}
\end{array}} \right]$

Answer 1

Hint: In this question, the multiplication of two matrices are given and its product as well. Here one thing that is much more important about a matrix is “order of a matrix”. So, multiplication of two matrices will be only possible when $\left( {{M_{a \times b }} \times {R_{ b \times c}}} \right)$ column of first matrix will be equal to row of second matrix. To solve it we have to use this property.

Step-by-step Solution:
Given:$\left[ {\begin{array}{*{20}{c}}
  2&{ - 1} \\
  1&0 \\
  { - 3}&4
\end{array}} \right]A = \left[ {\begin{array}{*{20}{c}}
  { - 1}&{ - 8} \\
  1&{ - 2} \\
  9&{22}
\end{array}} \right]$ so find $A$
As the multiplication of two matrices is given in which the first matrix is the order of $3 \times 2$ . Second matrix $A$ will be the order of $2 \times n$ where $n$ be a number. When we look at the multiplication of two matrices and its result will reach at the conclusion:
$
  {\left[ {\begin{array}{*{20}{c}}
  2&{ - 1} \\
  1&0 \\
  { - 3}&4
\end{array}} \right]_{3 \times 2}}{\left[ {\begin{array}{*{20}{c}}
  {}&{} \\
  {}&{} \\
  {}&{}
\end{array}} \right]_{2 \times n}} = {\left[ {\begin{array}{*{20}{c}}
  { - 1}&{ - 8} \\
  1&{ - 2} \\
  9&{22}
\end{array}} \right]_{3 \times 2}} \\
  \because 3 \times 2,2 \times n = 3 \times 2 \\
 $
$\therefore $ Value of $n = 2$
So, matrix $A$ will be order of $2 \times 2,A = \left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}}
\end{array}} \right)$
$\because \left[ {\begin{array}{*{20}{c}}
  2&{ - 1} \\
  1&0 \\
  { - 3}&4
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  { - 1}&{ - 8} \\
  1&{ - 2} \\
  9&{22}
\end{array}} \right]$
Performing multiplication operation,
$\left[ {\begin{array}{*{20}{c}}
  {2{a_{11}} - {a_{21}}}&{2{a_{12}} - {a_{22}}} \\
  {{a_{11}} + 0}&{{a_{12}} + 0} \\
  { - 3{a_{11}} + 4{a_{21}}}&{ - 3{a_{12}} + 4{a_{22}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  { - 1}&{ - 8} \\
  1&{ - 2} \\
  9&{22}
\end{array}} \right]$
Now, according to position of elements in the matrix, we equate them correspondingly,
$
\Rightarrow 2{a_{11}} - {a_{21}} = - 1 \to \left( 1 \right) \\
\Rightarrow 2{a_{12}} - {a_{22}} = - 8 \to \left( 2 \right) \\
\Rightarrow {a_{11}} = 1 \to \left( 3 \right) \\
\Rightarrow {a_{12}} = - 2 \to \left( 4 \right) \\
\Rightarrow - 3{a_{11}} + 4{a_{21}} = 9 \to \left( 5 \right) \\
\Rightarrow - 3{a_{12}} + 4{a_{22}} = 22 \to \left( 6 \right) \\
 $
Now, putting the value of equation $\left( 3 \right)$ in equation$\left( 1 \right)$, we have
$
  \because 2{a_{11}} - {a_{21}} = - 1 \\
   \Rightarrow 2 \times 1 - {a_{21}} = - 1 \\
   \Rightarrow - {a_{21}} = - 1 - 2 \\
   \Rightarrow - {a_{21}} = - 3 \\
  \therefore {a_{21}} = 3 \\
 $
Now, putting the value of equation $\left( 4 \right)$ in equation$\left( 2 \right)$, we have
$
  \because 2{a_{12}} - {a_{22}} = - 8 \\
   \Rightarrow 2 \times \left( { - 2} \right) - {a_{22}} = - 8 \\
   \Rightarrow - {a_{22}} = - 8 + 4 \\
   \Rightarrow - {a_{22}} = - 4 \\
  \therefore {a_{22}} = 4 \\
 $
So, matrix $A = \left[ {\begin{array}{*{20}{c}}
  1&{ - 2} \\
  3&4
\end{array}} \right]$

Note:
This is a matrix question so for a student matrix multiplication should be known. So, order is important. After that it can be solved very easily.