Solve the system by solving the corresponding matrix equation with an inverse.

Question

Solve the system by solving the corresponding matrix equation with an inverse. ​  ​
​

Quizplus · Accepted Answer

We can write the system of equations in matrix form:
\begin{bmatrix}
1 & 3 \ 
2 & 5
\end{bmatrix}
\begin{bmatrix}
x \ 
y
\end{bmatrix}
=
\begin{bmatrix}
5 \
11
\end{bmatrix}
To solve this, we can find the inverse of the coefficient matrix:
\begin{bmatrix}
1 & 3 \ 
2 & 5
\end{bmatrix}^{-1}
=
\frac{1}{1\cdot 5 - 3\cdot 2}
\begin{bmatrix}
5 & -3 \ 
-2 & 1
\end{bmatrix}
=
\begin{bmatrix}
\frac{5}{1\cdot 5-3\cdot 2} & \frac{-3}{1\cdot 5-3\cdot 2}\
\frac{-2}{1\cdot 5-3\cdot 2} & \frac{1}{1\cdot 5-3\cdot 2}
\end{bmatrix}
=
\begin{bmatrix}
\frac{5}{-1} & \frac{-3}{-1} \
\frac{-2}{-1} & \frac{1}{-1}
\end{bmatrix}
=
\begin{bmatrix}
-5 & 3 \
2 & -1
\end{bmatrix}
Then, we can left-multiply both sides of the equation by this inverse:
\begin{bmatrix}
-5 & 3 \
2 & -1
\end{bmatrix}
\begin{bmatrix}
1 & 3 \ 
2 & 5
\end{bmatrix}
\begin{bmatrix}
x \ 
y
\end{bmatrix}
=
\begin{bmatrix}
-5 & 3 \
2 & -1
\end{bmatrix}
\begin{bmatrix}
5 \
11
\end{bmatrix}
Simplifying:
\begin{bmatrix}
1 & 0 \
0 & 1
\end{bmatrix}
\begin{bmatrix}
x \ 
y
\end{bmatrix}
=
\begin{bmatrix}
-32 \
13
\end{bmatrix}
Therefore, the solution is (x,y) = (-32,13).

Solve the System by Solving the Corresponding Matrix Equation with an Inverse