Question

If $A = \left( {\begin{array}{*{20}{c}}
  1&2&x \\
  0&1&0 \\
  0&0&1
\end{array}} \right)$ and $B = \left( {\begin{array}{*{20}{c}}
  1&{ - 2}&y \\
  0&1&0 \\
  0&0&1
\end{array}} \right)$ and $AB = I$ then $x + y$ equals to
a) $0$
b) $ - 1$
c) $2$
d) None of these

Answer 1

Hint: The general idea is to use the rule of multiplication of matrices and equality of matrices. If we know the definition of matrix multiplication and equality of two matrices then we can use it here and get the answer. $I$ is the identity matrix here.
$I = \left( {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right)$

Formula used:
The multiplication of two $3 \times 3$matrices $A = \left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right)$ and $B = \left( {\begin{array}{*{20}{c}}
  {{b_{11}}}&{{b_{12}}}&{{b_{13}}} \\
  {{b_{21}}}&{{b_{22}}}&{{b_{23}}} \\
  {{b_{31}}}&{{b_{32}}}&{{b_{33}}}
\end{array}} \right)$ is defined as:
$A \times B = \left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
  {{b_{11}}}&{{b_{12}}}&{{b_{13}}} \\
  {{b_{21}}}&{{b_{22}}}&{{b_{23}}} \\
  {{b_{31}}}&{{b_{32}}}&{{b_{33}}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {{c_{11}}}&{{c_{12}}}&{{c_{13}}} \\
  {{c_{21}}}&{{c_{22}}}&{{c_{23}}} \\
  {{c_{31}}}&{{c_{32}}}&{{c_{33}}}
\end{array}} \right)$ , where ${c_{ij}} = {a_{i1}}{b_{1j}} + {a_{i2}}{b_{2j}} + {a_{i3}}{b_{3j}},1 \leqslant i,j \leqslant 3$

Complete step-by-step answer:
The given matrices are:
$A = \left( {\begin{array}{*{20}{c}}
  1&2&x \\
  0&1&0 \\
  0&0&1
\end{array}} \right)$ and $B = \left( {\begin{array}{*{20}{c}}
  1&{ - 2}&y \\
  0&1&0 \\
  0&0&1
\end{array}} \right)$ and the identity matrix is: ${I_{3 \times 3}} = \left( {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right)$
According to the question $AB = I$
So,
$\left( {\begin{array}{*{20}{c}}
  1&2&x \\
  0&1&0 \\
  0&0&1
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
  1&{ - 2}&y \\
  0&1&0 \\
  0&0&1
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right)$
Lets first solve the LHS of the equation:
LHS
$ = \left( {\begin{array}{*{20}{c}}
  1&2&x \\
  0&1&0 \\
  0&0&1
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
  1&{ - 2}&y \\
  0&1&0 \\
  0&0&1
\end{array}} \right)$
$ = \left( {\begin{array}{*{20}{c}}
  {1 \times 1 + 2 \times 0 + x \times 0}&{1 \times - 2 + 2 \times 1 + x \times 0}&{1 \times y + 2 \times 0 + x \times 1} \\
  {0 \times 1 + 1 \times 0 + 0 \times 0}&{0 \times - 2 + 1 \times 1 + 0 \times 0}&{0 \times y + 1 \times 0 + 0 \times 1} \\
  {0 \times 1 + 0 \times 0 + 1 \times 0}&{0 \times - 2 + 0 \times 1 + 1 \times 0}&{0 \times y + 0 \times 0 + 1 \times 1}
\end{array}} \right)$
\[ = \left( {\begin{array}{*{20}{c}}
  1&0&{y + x} \\
  0&1&0 \\
  0&0&1
\end{array}} \right)\] --(1)
Equating LHS with RHS we get:
\[\left( {\begin{array}{*{20}{c}}
  1&0&{y + x} \\
  0&1&0 \\
  0&0&1
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right)\] --(2)
Two matrices are equal if each entry at a place of one matrix corresponds to the same entry at that place of another matric with the same element.
Or mathematically,
$[{a_{ij}}] = [{b_{ij}}]$
Clearly from (1) we have:
$y + x = 0$
So the value that we were asked to find is equal to $0$.
As $0$ corresponds to option ‘a’ so option ‘a’ is correct.
So, the correct answer is “Option A”.

Note: ARemember there are many other properties related to multiplication of matrices one example is if $AB = I$ the matrices $A$ and $B$ are inverses of each other and if you try to find the inverse of $A$ and equate with the value of $B$ then in our case you don’t get the required answer using this trick. Also if you try to do this method you may end up with miscalculations and eventually, you will get an unexpected answer.