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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?

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

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

Now continue your answer and see you will get the right answer.

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