Question

Given two matrices A and B
$A=\left[ \begin{matrix}
   1 & -2 & 3 \\
   1 & 4 & 1 \\
   1 & -3 & 2 \\
\end{matrix} \right]\ and\ B=\left[ \begin{matrix}
   11 & -5 & -14 \\
   -1 & -1 & 2 \\
   -7 & 1 & 6 \\
\end{matrix} \right]$
Find AB and use this result to solve the following system of equations:
$x-2y+3z=6,x+4y+z=12,x-3y+2z=1$.

Answer 1

Hint:The given problem is related to multiplication of matrices, and solution of simultaneous linear equations. Use the method of AX = B and the property of the identity matrix to solve the equations.

Complete step-by-step answer:
The given two matrices are $A=\left[ \begin{matrix}
   1 & -2 & 3 \\
   1 & 4 & 1 \\
   1 & -3 & 2 \\
\end{matrix} \right]\ $and $B=\left[ \begin{matrix}
   11 & -5 & -14 \\
   -1 & -1 & 2 \\
   -7 & 1 & 6 \\
\end{matrix} \right]$ . The product of the matrices is given as $AB=\left[ \begin{matrix}
   1 & -2 & 3 \\
   1 & 4 & 1 \\
   1 & -3 & 2 \\
\end{matrix} \right]\ \left[ \begin{matrix}
   11 & -5 & -14 \\
   -1 & -1 & 2 \\
   -7 & 1 & 6 \\
\end{matrix} \right]$ .
$\Rightarrow AB=\left[ \begin{matrix}
   \left( 1\times 11 \right)+\left( -2\times \left( -1 \right) \right)+\left( 3\times \left( -7 \right) \right) & \left( 1\times \left( -5 \right) \right)+\left( -2\times \left( -1 \right) \right)+\left( 3\times 1 \right) & \left( 1\times \left( -14 \right) \right)+\left( -2\times 2 \right)+\left( 3\times 6 \right) \\
   \left( 1\times 11 \right)+\left( 4\times \left( -1 \right) \right)+\left( 1\times \left( -7 \right) \right) & \left( 1\times \left( -5 \right) \right)+\left( 4\times \left( -1 \right) \right)+\left( 1\times 1 \right) & \left( 1\times \left( -14 \right) \right)+\left( 4\times 2 \right)+\left( 1\times 6 \right) \\
   \left( 1\times 11 \right)+\left( -3\times \left( -1 \right) \right)+\left( 2\times \left( -7 \right) \right) & \left( 1\times \left( -5 \right) \right)+\left( -3\times \left( -1 \right) \right)+\left( 2\times 1 \right) & \left( 1\times \left( -14 \right) \right)+\left( -3\times 2 \right)+\left( 2\times 6 \right) \\
\end{matrix} \right]$$\Rightarrow AB=\left[ \begin{matrix}
   -8 & 0 & 0 \\
   0 & -8 & 0 \\
   0 & 0 & -8 \\
\end{matrix} \right]$
$\Rightarrow AB=-8\left[ \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right]$
$\Rightarrow AB=-8I$, where I is the identity matrix.
 We observe that,
$AB=-8I$.
We write $I$ as $A.{{A}^{-1}}$.
\[\begin{align}
  & \Rightarrow AB=-8A{{A}^{-1}} \\
 & \Rightarrow -8A.{{A}^{-1}}=AB \\
 & \Rightarrow A{{A}^{-1}}=\dfrac{AB}{-8} \\
 & \Rightarrow {{A}^{-1}}=\dfrac{-B}{8} \\
 & \therefore {{A}^{-1}}=\dfrac{-B}{8} \\
\end{align}\]
Given, $x-2y+3z=6$
$\begin{align}
  & x+4y+z=12 \\
 & x-3y+2x=1 \\
\end{align}$
We write the above set of equations in matrix form.
\[\left[ \begin{matrix}
   1 & -2 & 3 \\
   1 & 4 & 1 \\
   1 & -3 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   6 \\
   12 \\
   1 \\
\end{matrix} \right]\]
We know, \[\left[ \begin{matrix}
   1 & -2 & 3 \\
   1 & 4 & 1 \\
   1 & -3 & 2 \\
\end{matrix} \right]=A\] and \[\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=X\] .
Then, \[AX=\left[ \begin{matrix}
   6 \\
   12 \\
   1 \\
\end{matrix} \right]\]
\[\begin{align}
  & \Rightarrow {{A}^{-1}}AX={{A}^{-1}}\left[ \begin{matrix}
   6 \\
   12 \\
   1 \\
\end{matrix} \right] \\
 & \Rightarrow X={{A}^{-1}}\left[ \begin{matrix}
   6 \\
   12 \\
   1 \\
\end{matrix} \right] \\
 & \Rightarrow X=\dfrac{-{{B}^{-1}}}{8}\left[ \begin{matrix}
   6 \\
   12 \\
   1 \\
\end{matrix} \right] \\
 & \Rightarrow X=\dfrac{-1}{8}\left[ \begin{matrix}
   11 & -5 & -14 \\
   -1 & -1 & 2 \\
   -7 & 1 & 6 \\
\end{matrix} \right]\left[ \begin{matrix}
   6 \\
   12 \\
   1 \\
\end{matrix} \right] \\
\end{align}\]
\[\Rightarrow X=\dfrac{-1}{8}\left[ \begin{matrix}
   \left( 11\times 6 \right)+\left( -5\times 12 \right)+\left( -14\times 1 \right) \\
   \left( -1\times 6 \right)+\left( -1\times 12 \right)+\left( 2\times 1 \right) \\
   \left( -7\times 6 \right)+\left( 1\times 12 \right)+\left( 6\times 1 \right) \\
\end{matrix} \right]\]
\[\Rightarrow X=\dfrac{-1}{8}\left[ \begin{matrix}
   66-60-14 \\
   -6-12+2 \\
   -42+12+6 \\
\end{matrix} \right]\]
\[\begin{align}
  & \Rightarrow X=\dfrac{-1}{8}\left[ \begin{matrix}
   -8 \\
   -16 \\
   -24 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 \\
   2 \\
   3 \\
\end{matrix} \right] \\
 & \Rightarrow \left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 \\
   2 \\
   3 \\
\end{matrix} \right] \\
\end{align}\]
Therefore, x = 1, y = 2, z = 3.

Note: While doing matrix multiplication, make sure the order in which the matrices are taken, because AB is not equal to BA.The number of columns of the 1st matrix must equal the number of rows of the 2nd matrix. And the result will have the same number of rows as the 1st matrix, and the same number of columns as the 2nd matrix.We can verify the answer by substituting values of x , y and z in given equation i.e $x-2y+3z=6$, we get $1-2(2)+3(3)=6$ ,L.H.S=R.H.S hence the answer is right.