Question

How do you solve the system $x+2y-z=6,-3x-2y+5z=-12,x-2z=3$?

Answer 1

Hint: We first form the multiplication form of matrices to find the coefficient matrix $AX=B$. Then we use the inverse matrix to find the variable matrix with the form of $X={{A}^{-1}}B$. We find the inverse after finding the matrix being singular or not.

Complete step-by-step answer:
We have three unknowns and three equations to solve. We use the matrix multiplication form and its inverse form to solve the variables.
We take 3 matrices A, X, B where they denote the coefficient matrix, variable matrix and solution matrix respectively.
Therefore, $A=\left[ \begin{matrix}
   1 & 2 & -1 \\
   -3 & -2 & 5 \\
   1 & 0 & -2 \\
\end{matrix} \right];X=\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right];B=\left[ \begin{matrix}
   6 \\
   -12 \\
   3 \\
\end{matrix} \right]$.
Matrix multiplication form gives $AX=B$.
We multiply the equation $AX=B$ with ${{A}^{-1}}$ to get
$\begin{align}
  & {{A}^{-1}}.AX={{A}^{-1}}.B \\
 & \Rightarrow IX=X={{A}^{-1}}B \\
\end{align}$
The variables matrix will be in the form of $X={{A}^{-1}}B$. We have to multiply the inverse matrix of A with the solution matrix.
The inverse of any matrix $A=\left[ {{a}_{ij}} \right]$ will be ${{A}^{-1}}=\dfrac{adj\left( A \right)}{\left| A \right|}$. Here ${{a}_{ij}}$ denotes the element of ${{i}^{th}}$ row and ${{j}^{th}}$ column.
$\left| A \right|$ is defined as the determinant value of the matrix A.
The $adj\left( A \right)$ is defined by the $adj\left( A \right)={{\left[ {{A}_{ij}} \right]}^{T}}=\left[ {{A}_{ji}} \right]$. Here, ${{A}_{ij}}$ denotes the co-factor of the element ${{a}_{ij}}$. The term ‘T’ denotes the transpose of the matrix.
We find the cofactors of the matrix A and get $\left[ {{A}_{ij}} \right]=\left[ \begin{matrix}
   4 & -1 & 2 \\
   4 & -1 & 2 \\
   8 & -2 & 4 \\
\end{matrix} \right]$ which gives
$adj\left( A \right)={{\left[ {{A}_{ij}} \right]}^{T}}=\left[ {{A}_{ji}} \right]=\left[ \begin{matrix}
   4 & 4 & 8 \\
   -1 & -1 & -2 \\
   2 & 2 & 4 \\
\end{matrix} \right]$.
Now we find the determinant which is $\left| A \right|=1\times 4+2\times \left( -1 \right)+\left( -1 \right)\times 2=4-2-2=0$.
As the matrix A is a singular matrix the inverse of the matrix is not possible.
Therefore, the system of equations $x+2y-z=6,-3x-2y+5z=-12,x-2z=3$ have no solutions.

Note: The addition of first two equations $x+2y-z=6,-3x-2y+5z=-12$ give
$\begin{align}
  & x+2y-z-3x-2y+5z=6-12 \\
 & \Rightarrow -2x+4z=-6 \\
 & \Rightarrow x-2z=3 \\
\end{align}$
This is the same as the third equations. Therefore, there is no exact solution for the system of equations.