# Where did I go wrong when calculating the x and y values using these matrices?

A problem is stated as such:

Solve the system of equations.

$$\begin{bmatrix}x+2y=16\\3x-y=6\end{bmatrix}$$

So, I started by partitioning the matrix up into its constant and coefficient parts.

$$\begin{bmatrix}1 & 2 \\ 3 & 1\end{bmatrix}$$ $$\begin{bmatrix}x\\y\end{bmatrix}$$ $$\begin{bmatrix}16\\6\end{bmatrix}$$

Then I calculated the determinant of the square matrix.

$$\begin{vmatrix}1&2\\3&1\end{vmatrix} = -5$$

Then I calculated the coefficient matrix of the square matrix.

$$(-1)^{1+1}\times1=1 \\ (-1)^{1+2}\times3=-3\\(-1)^{2+1}\times2=-2\\(-1)^{2+2}\times1=1\\ \begin{bmatrix}1&-3\\-2&1\end{bmatrix}$$

Transposing this coefficient matrix, I got the adjoint matrix.

$$\begin{bmatrix}1&-2\\-3&1\end{bmatrix}$$

Multiplying the adjoint matrix with the determinant, I got the inverse matrix of the original square matrix.

$$\begin{bmatrix}1&-2\\-3&1\end{bmatrix} \div -5 = \begin{bmatrix}-0.2&0.4\\0.6&-0.2\end{bmatrix}$$

Multiplying this last matrix, the inverse of the square matrix, with the matrix containing the constants $$(16,6)$$, I will get the $$(x,y)$$ matrix, only it will contain numerical values, thus revealing the values of the two unknowns.

$$\begin{bmatrix}-0.2&0.4\\0.6&-0.2\end{bmatrix} \times \begin{bmatrix}16\\6\end{bmatrix} = \begin{bmatrix}2.4\\8.4\end{bmatrix}$$

These values don't work within the original equations though. Where did I go wrong?

• The first matrix should be $\begin{bmatrix}1&2\\3&-1\end{bmatrix}$, right? May 22 at 16:38
• @player3236 yes it should, that was my mistake. May 23 at 10:43

Your mistake is in the first equation. You have taken coefficient of $$y$$ in $$3x-y=6$$ as $$1$$ but should be $$-1$$. Thus the first matrix should be
$$\begin{bmatrix}1 & 2 \\ 3 & -1\end{bmatrix}$$