Question

Find\[AB\], if \[A = \left[ {\begin{array}{*{20}{c}}
  1&2&3 \\
  1&{ - 2}&3
\end{array}} \right]\]and \[B = \left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  1&2 \\
  1&{ - 2}
\end{array}} \right]\]. Examine whether AB has an inverse or not.

Answer 1

Hint: Here we have to solve the two matrices by using the dot multiplication and when we get them \[AB\] then we have to check whether the matrix has an inverse or not and so for this if the matrix of \[AB\] will not be equal to zero then we can say that the matrix is inverse.

Formula used:
The dot product of two matrices will be given by:
${x^T}y = \left[ {{x_1}{\text{ }}{{\text{x}}_2}{\text{ }}{{\text{x}}_3}.........{x_n}} \right]{\text{ }}\left[ {\begin{array}{*{20}{c}}
  {{y_1}} \\
  {{y_2}} \\
  . \\
  {{y_n}}
\end{array}} \right] = {x_1}{y_1} + {x_2}{y_2} + ....{x_n}{y_n} = XY$
Here,
$XY$, it will be the dot product of two matrices.
${x_1}...{x_n}$, will be the first matrix
${y_1}.....{y_n}$, will be the second matrix.

Complete step-by-step answer:
Since we have the matrix
\[A = \left[ {\begin{array}{*{20}{c}}
  1&2&3 \\
  1&{ - 2}&3
\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  1&2 \\
  1&{ - 2}
\end{array}} \right]\]
Now by applying the dot product, we get
\[ \Rightarrow AB = \left[ {\begin{array}{*{20}{c}}
  {1 \times 1{\text{ }} + {\text{ }}2 \times 1{\text{ }} + {\text{ }}3 \times 1}&{1 \times - 1{\text{ }} + {\text{ }}2 \times 2{\text{ }} + {\text{ }}3 \times (-)2} \\
  {1 \times 1{\text{ }} + {\text{ }} - 2 \times 1{\text{ }} + {\text{ }}3 \times 1}&{1 \times - 1{\text{ }} + {\text{ }} - 2 \times 2{\text{ }} + {\text{ }}3 \times - 2}
\end{array}} \right]\]
Now on solving the above matrix we get
$ \Rightarrow AB = \left[ {\begin{array}{*{20}{c}}
  6&{ - 3} \\
  2&{ - 11}
\end{array}} \right]$
On making the cross-multiplication, we get
$ \Rightarrow \left| {AB} \right| = - 66 + 6$
And therefore on doing the multiplication, we get
$ \Rightarrow \left| {AB} \right| = - 60$
Now we will check whether it has an inverse or not.
So for this, as we know if the matrix of \[AB\] will not be equal to zero then we can say that the matrix is inverse.
Therefore, we can see that $\left| {AB} \right| \ne 0$.
Hence, we can say that the inverse of it will exist.

Note: Matrix multiplication can be seen as the dot product between the rows of the first matrix and the columns of the second matrix. That is why the dimensions of the two matrices must be of the form $m \times n$ and $n \times p$ for matrix multiplication to be valid. Now and again, it is unimaginable to expect to ascertain a dot product between two networks of various measurements. Matrix multiplication does not support commutative law. It also does not depend on each element of the dot product.