Question

Using matrices, solve the following system of linear equation:
$\begin{align}
  & x-y+2z=7 \\
 & 3x+4y-5z=-5 \\
 & 2x-y+3z=12 \\
\end{align}$

Answer 1

Hint: In this question we have been given a system of linear equations, which we need to solve using matrices. So, for that first we will represent these equations in the form of a matrix as $AX=B$. After this we will find out the determinant of the matrix A to find out if ${{A}^{-1}}$exist or not. If we exist we will find out the ${{A}^{-1}}$from which we will be getting the values of x, y and z.

Complete step by step answer:
We have been provided with a system of linear equations which are:
$\begin{align}
  & x-y+2z=7 \\
 & 3x+4y-5z=-5 \\
 & 2x-y+3z=12 \\
\end{align}$
In the above equation A= $\left( \begin{matrix}
   1 & -1 & 2 \\
   3 & 4 & -5 \\
   2 & -1 & 3 \\
\end{matrix} \right)$ , X=$\left( \begin{align}
  & x \\
 & y \\
 & z \\
\end{align} \right)$ and B=$\left( \begin{align}
  & 7 \\
 & -5 \\
 & 12 \\
\end{align} \right)$
Now we will be representing these equations in matrix form that is $AX=B$,
$\left( \begin{matrix}
   1 & -1 & 2 \\
   3 & 4 & -5 \\
   2 & -1 & 3 \\
\end{matrix} \right)$ $\left( \begin{align}
  & x \\
 & y \\
 & z \\
\end{align} \right)$ =$\left( \begin{align}
  & 7 \\
 & -5 \\
 & 12 \\
\end{align} \right)$
Now we will be finding the determinant of A matrix to find out whether ${{A}^{-1}}$ exist or not,
$\left| A \right|=\left( \begin{matrix}
   1 & -1 & 2 \\
   3 & 4 & -5 \\
   2 & -1 & 3 \\
\end{matrix} \right)=1(12-5)+1(9+10)+2(-3-8)$
Which is equal to, $\begin{align}
  & 7+19-22=4 \\
 & 4\ne 0 \\
\end{align}$
And as the determinant of A is not equal to 0, ${{A}^{-1}}$would exist and system would have unique solutions,
Now we will be finding the value of ${{A}^{-1}}$by using the matrix form:$AX=B$,
So, we will multiply the whole equation by ${{A}^{-1}}$,
So, we will get,
$A{{A}^{-1}}X=B{{A}^{-1}}$
And we know the product of $A{{A}^{-1}}$ is equal to identity matrix that is I.
So, now the equation would become:$IX=B{{A}^{-1}}$
Now we will be finding ${{A}^{-1}}$, for that we will find the cofactors
$\begin{align}
  & {{C}_{11}}={{(-1)}^{1+1}}\left( \begin{matrix}
   4 & -5 \\
   -1 & 3 \\
\end{matrix} \right) \\
 & \Rightarrow 12-5=7 \\
\end{align}$
$\begin{align}
  & {{C}_{12}}={{(-1)}^{1+2}}\left( \begin{matrix}
   3 & -5 \\
   2 & 3 \\
\end{matrix} \right) \\
 & \Rightarrow -(9+10)=-19 \\
 & {{C}_{13}}={{(-1)}^{1+3}}\left( \begin{matrix}
   3 & 4 \\
   2 & -1 \\
\end{matrix} \right) \\
 & \Rightarrow (-3-8)=-11 \\
\end{align}$
$\begin{align}
  & {{C}_{21}}={{(-1)}^{2+1}}\left( \begin{matrix}
   -1 & 2 \\
   -1 & 3 \\
\end{matrix} \right) \\
 & \Rightarrow -(-3+2)=1 \\
 & {{C}_{22}}={{(-1)}^{2+2}}\left( \begin{matrix}
   1 & 2 \\
   2 & 3 \\
\end{matrix} \right) \\
 & \Rightarrow 3-4=-1 \\
 & {{C}_{23}}={{(-1)}^{2+3}}\left( \begin{matrix}
   1 & -1 \\
   2 & -1 \\
\end{matrix} \right) \\
 & \Rightarrow -(-1+2)=-1 \\
 & \\
\end{align}$
$\begin{align}
  & {{C}_{31}}={{(-1)}^{3+1}}\left( \begin{matrix}
   -1 & 2 \\
   4 & -5 \\
\end{matrix} \right) \\
 & \Rightarrow 5-8=-3 \\
 & {{C}_{32}}={{(-1)}^{3+2}}\left( \begin{matrix}
   1 & 2 \\
   3 & -5 \\
\end{matrix} \right) \\
 & \Rightarrow 5+6=11 \\
 & {{C}_{33}}={{(-1)}^{3+3}}\left( \begin{matrix}
   1 & -1 \\
   3 & 4 \\
\end{matrix} \right) \\
 & \Rightarrow 4+3=7 \\
\end{align}$

Now we have found all the cofactors so these will form the adjoint A
So, for finding adjoint A we need to do the transpose of all the cofactors which we have found,
$AdjA={{\left( \begin{matrix}
   7 & -19 & -11 \\
   1 & -1 & -1 \\
   -3 & 11 & 7 \\
\end{matrix} \right)}^{T}}$
For which the value of adjoint A comes out to be:
$AdjA=\left( \begin{matrix}
   7 & 1 & -3 \\
   -19 & -1 & 11 \\
   -11 & -1 & 7 \\
\end{matrix} \right)$
Now we will be using the formula:${{A}^{-1}}=\dfrac{1}{\left| A \right|}AdjA$,
So, the value comes out to be: ${{A}^{-1}}=\dfrac{1}{4}\left( \begin{matrix}
   7 & 1 & -3 \\
   -19 & -1 & 11 \\
   -11 & -1 & 7 \\
\end{matrix} \right)$
Now we will be using the relation:$X=B{{A}^{-1}}$
$\left( \begin{align}
  & x \\
 & y \\
 & z \\
\end{align} \right)$=$\dfrac{1}{4}\left( \begin{matrix}
   7 & 1 & -3 \\
   -19 & -1 & 11 \\
   -11 & -1 & 7 \\
\end{matrix} \right)\left( \begin{align}
  & 7 \\
 & -5 \\
 & 12 \\
\end{align} \right)$

Therefore, simplifying these values we will get:$\dfrac{1}{4}\left( \begin{align}
  & 49-5-36 \\
 & -133+5+132 \\
 & -77+5+84 \\
\end{align} \right)$
Now solving the above equations, we will get:$\dfrac{1}{4}\left( \begin{align}
  & 8 \\
 & 4 \\
 & 12 \\
\end{align} \right)$
So, we will get the values of x, y and z:
$\left( \begin{align}
  & x \\
 & y \\
 & z \\
\end{align} \right)$= $\left( \begin{align}
  & 2 \\
 & 1 \\
 & 3 \\
\end{align} \right)$

So, the corresponding values of x=2, y=1 and z=3.
So, this is the solution for this question.

Note:
In this question we must be a little bit cautious while finding the cofactors as the chances of mistakes are more over there, we must remember the relation between adjoint A and ${{A}^{-1}}$. Also don’t forget to transpose the matrix before finding the values.