If
$A = \left( {\begin{array}{*{20}{c}}
  1&2&1 \\
  3&4&2 \\
  {}&{}&{}
\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}
  3&{ - 2}&4 \\
  1&5&0 \\
  {}&{}&{}
\end{array}} \right)$ ,
then find the matrix $X$ from $X + A + B = 0$ .

Hint: In the given question, we have been asked to find the value of ‘X’ and it is given that $X + A + B = 0$. To solve this question, we need to get ‘X’ on one side of the “equals” sign, and all the other numbers on the other side. To solve this equation for a given variable ‘X’, we have to undo the matrix operations such as addition and subtraction , that has been done to the variables.

Complete step by step solution:
It is given that ,
$X + A + B = 0$
Substituting given values , we will get ,
$ \Rightarrow X + \left( {\begin{array}{*{20}{c}}
  1&2&1 \\
  3&4&2 \\
  {}&{}&{}
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  3&{ - 2}&4 \\
  1&5&0 \\
  {}&{}&{}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  0&0&0 \\
  0&0&0 \\
  {}&{}&{}
\end{array}} \right)$
Add the like terms , two matrices must have an equal number of rows and columns to be added. within which case, the sum of two matrices ‘A’ and ‘B’ are going to be a matrix which has the identical number of rows and columns as ‘A’ and ‘B’ . The sum of ‘A’ and ‘B’ , denoted ‘A+B’ , is computed by adding corresponding elements of ‘A’ and ‘B’ ,
$ \Rightarrow X + \left( {\begin{array}{*{20}{c}}
  4&0&5 \\
  4&9&2 \\
  {}&{}&{}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  0&0&0 \\
  0&0&0 \\
  {}&{}&{}
\end{array}} \right)$
Now, subtract the sum ‘A+B’ from both the side of the equation,
$ \Rightarrow X = \left( {\begin{array}{*{20}{c}}
  0&0&0 \\
  0&0&0 \\
  {}&{}&{}
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
  4&0&5 \\
  4&9&2 \\
  {}&{}&{}
\end{array}} \right)$
Two matrices must have an equal number of rows and columns to be subtracted. Within which case, the difference of two matrices is going to be a matrix which has the identical number of rows and columns . The difference is computed by subtracting corresponding elements ,
$\therefore X = \left( {\begin{array}{*{20}{c}}
  { - 4}&0&{ - 5} \\
  { - 4}&{ - 9}&{ - 2} \\
  {}&{}&{}
\end{array}} \right)$

Note: The important thing to recollect about any equation is that the ‘equals’ sign represents a balance. What the sign says is that what’s on the left-hand side is strictly an equal to what’s on the right-hand side. It is the type of question where only mathematical operations such as addition, subtraction, multiplication and division is used.