Question

How do I use the inverse of a matrix to solve a system of equations?

Answer 1

Hint: The inverse of the matrix \[A\] is \[{{A}^{-1}}\]. Find \[A,B\] and \[C\] matrices in the matrix equation \[AB=C\], where you can get these matrices from the system of equations in the following form: \[{{a}_{1}}x+{{b}_{1}}y={{c}_{1}}\] and \[{{a}_{2}}x+{{b}_{2}}y={{c}_{2}}\] From this system the coefficient matrix \[A\]is: \[A=\left[ \begin{matrix}
   {{a}_{1}} & {{b}_{1}} \\
   {{a}_{2}} & {{b}_{2}} \\
\end{matrix} \right]\]. The variable matrix \[B\] is \[B=\left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]\] and the constant matrix is \[C=\left[ \begin{matrix}
   {{c}_{1}} \\
   {{c}_{2}} \\
\end{matrix} \right]\]. Which makes \[AB=C\] equal to \[\left[ \begin{matrix}
   {{a}_{1}} & {{b}_{1}} \\
   {{a}_{2}} & {{b}_{2}} \\
\end{matrix} \right]\text{ }\left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]=\left[ \begin{matrix}
   {{c}_{1}} \\
   {{c}_{2}} \\
\end{matrix} \right]\] and if the inverse of this equation exists then solve for B that is \[B={{A}^{-1}}C\]
And in matrix form \[\left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]\text{=}\dfrac{1}{{{a}_{1}}{{b}_{2}}-{{b}_{1}}{{a}_{2}}}\left[ \begin{matrix}
   {{b}_{2}} & -{{b}_{1}} \\
   -{{a}_{2}} & {{a}_{1}} \\
\end{matrix} \right]\text{ }\left[ \begin{matrix}
   {{c}_{1}} \\
   {{c}_{2}} \\
\end{matrix} \right]\].

Complete step by step solution:
Let us suppose following are the system of equations:
\[{{a}_{1}}x+{{b}_{1}}y={{c}_{1}}\] and
 \[{{a}_{2}}x+{{b}_{2}}y={{c}_{2}}\]
Now we can convert this system of equations to the matrix equation that is
 \[AB=C...\left( 1 \right)\]
Where the coefficient matrix \[A\]is:
\[A=\left[ \begin{matrix}
   {{a}_{1}} & {{b}_{1}} \\
   {{a}_{2}} & {{b}_{2}} \\
\end{matrix} \right]\].
The variable matrix \[B\] is
 \[B=\left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]\]
and the constant matrix is \[C=\left[ \begin{matrix}
   {{c}_{1}} \\
   {{c}_{2}} \\
\end{matrix} \right]\].
Now substituting the values of \[A,B\] and \[C\] in equation \[\left( 1 \right)\] we get:
\[\left[ \begin{matrix}
   {{a}_{1}} & {{b}_{1}} \\
   {{a}_{2}} & {{b}_{2}} \\
\end{matrix} \right]\text{ }\left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]=\left[ \begin{matrix}
   {{c}_{1}} \\
   {{c}_{2}} \\
\end{matrix} \right]\]
To solve this matrix equation, we check if the inverse of the matrix \[A\]exists or not by checking the determinant of the matrix. If the determinant of the matrix is equal to zero it implies that the inverse of the matrix doesn’t exist. And if the coefficient matrix is non invertible then the system could be inconsistent and have no solution, or be dependent and have infinitely many solutions.
If the inverse the coefficient matrix is invertible, we solve the matrix equation as follows:
\[\Rightarrow AB=C\]
Multiply both sides by the inverse of \[A\] to obtain the solution.
 \[\begin{align}
  & \Rightarrow \left( {{A}^{-1}} \right)AB=\left( {{A}^{-1}} \right)C \\
 & \Rightarrow \left[ \left( {{A}^{-1}} \right)A \right]B=\left( {{A}^{-1}} \right)C \\
 & \Rightarrow \left[ I \right]B=\left( {{A}^{-1}} \right)C \\
 & \Rightarrow B=\left( {{A}^{-1}} \right)C...\left( 2 \right) \\
\end{align}\]
Now substituting the values of \[A,B\] and \[C\] in equation \[\left( 2 \right)\] we get:
\[\Rightarrow \left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]={{\left[ \begin{matrix}
   {{a}_{1}} & {{b}_{1}} \\
   {{a}_{2}} & {{b}_{2}} \\
\end{matrix} \right]}^{-1}}\text{ }\left[ \begin{matrix}
   {{c}_{1}} \\
   {{c}_{2}} \\
\end{matrix} \right]\]
Where \[{{A}^{-1}}={{\left[ \begin{matrix}
   {{a}_{1}} & {{b}_{1}} \\
   {{a}_{2}} & {{b}_{2}} \\
\end{matrix} \right]}^{-1}}\text{=}\dfrac{1}{{{a}_{1}}{{b}_{2}}-{{b}_{1}}{{a}_{2}}}\left[ \begin{matrix}
   {{b}_{2}} & -{{b}_{1}} \\
   -{{a}_{2}} & {{a}_{1}} \\
\end{matrix} \right]\]
Therefore, with the help of inverse of the matrix we can find the values of variable matrix that is: \[\left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]\text{=}\dfrac{1}{{{a}_{1}}{{b}_{2}}-{{b}_{1}}{{a}_{2}}}\left[ \begin{matrix}
   {{b}_{2}} & -{{b}_{1}} \\
   -{{a}_{2}} & {{a}_{1}} \\
\end{matrix} \right]\text{ }\left[ \begin{matrix}
   {{c}_{1}} \\
   {{c}_{2}} \\
\end{matrix} \right]\].

Note:
Students can go wrong by forgetting to check whether the coefficient matrix is invertible or not which leads to the wrong answer. So, it’s important to remember to check the invertibility of the coefficient matrix by looking if the determinant is zero or not.