Question

Solve the equations using matrices: $2x+y=3$ and $x-y=1$

• A. $(\frac{2}{3},\frac{1}{3})$
• B. $(\frac{4}{3},\frac{1}{3})$
• C. $(\frac{1}{3},\frac{1}{3})$
• D. $(\frac{4}{3},\frac{2}{3})$

Answer 1

Answer: (B)

$(\frac{4}{3},\frac{1}{3})$

$2x+y=3$ and $x-y=1$

Write the equation in the form of matrices,

$\left[\begin{array}{cc}2& 1\\ 1& -1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}3\\ 1\end{array}\right]$

Finding the determinant of the matrix, $D=\left[\begin{array}{cc}2& 1\\ 1& -1\end{array}\right]$

$= -1 \times 2 - 1 \times 1

$=-2-1=-3$

First, find the inverse of the coefficient matrix, the inverse of $\left[\begin{array}{cc}2& 1\\ 1& -1\end{array}\right]$ is $\frac{1}{D}\times \left[\begin{array}{cc}-1& -1\\ -1& 2\end{array}\right]=\frac{1}{-3}\times \left[\begin{array}{cc}-1& -1\\ -1& 2\end{array}\right]$

$=\left[\begin{array}{cc}\frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{-2}{3}\end{array}\right]$

$\therefore \left[\begin{array}{cc}\frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{-2}{3}\end{array}\right]\left[\begin{array}{cc}2& 1\\ 1& -1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{cc}\frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{-2}{3}\end{array}\right]\left[\begin{array}{c}3\\ 1\end{array}\right]$

$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}\frac{4}{3}\\ \frac{1}{3}\end{array}\right]$

The identity matrix on the left verifies that the inverse matrix was calculated correctly.

$\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}\frac{4}{3}\\ \frac{1}{3}\end{array}\right]$

The solution is $(\frac{4}{3},\frac{1}{3})$