1. ## Help with matrices

HI I'm really close to the answer, but do not know where I went wrong..

Consider the two equations 4x+8y=1 and 2x-ay=11 ( One is on top of the other) so in matrix form it is
$\displaystyle \begin{bmatrix} 4 & 8 \\2 & -a \end{bmatrix}$$\displaystyle \begin{bmatrix} x \\y \end{bmatrix} = \displaystyle \begin{bmatrix} 1 \\11 \end{bmatrix} For what values of a does the system have a unique solution? Arranging in augmented matrix and using row operations, I reduced the matrix to \displaystyle \begin{bmatrix} 4 & 8 \\0 & 2a+8 \end{bmatrix}$$\displaystyle \begin{bmatrix} x \\y \end{bmatrix}$ = $\displaystyle \begin{bmatrix} 1 \\-22 \end{bmatrix}$

I had to invert the matrix premultiplied to the matrix containing x and y, so I had to get the det, I got 8a+32

$\displaystyle \begin{bmatrix} x \\y \end{bmatrix}$ = $\displaystyle \frac{1}{8a+32}\begin{bmatrix} 2a+8 & -8 \\0 & 4 \end{bmatrix}\begin{bmatrix} 1\\-22\end{bmatrix}$
Then I multiplied the determinant into the inversed matrix, simplified, was left with
$\displaystyle \begin{bmatrix}\frac{a+4}{4a+16} &\frac{-1}{a+4} \\0 &\frac{1}{2a+8} \end{bmatrix}\begin{bmatrix} 1 \\-22 \end{bmatrix}$
multiplying them, I get
$\displaystyle \begin{bmatrix}\frac{a+4}{4a+16}+\frac{22}{a+4}\\0 +\frac{-22}{2a+8}\end{bmatrix}$
So my answer is x = $\displaystyle \frac{a+92}{4a+16}$.. and y = $\displaystyle \frac{-22}{2a+8}$
but it's wrong.. can someone show me where I went wrong?

Will really appreciate it, thank you so much!

2. ## Re: Help with matrices

Hello, Tutu!

First of all, you didn't answer the question.

$\displaystyle \text{Given: }\:\begin{bmatrix} 4 & 8 \\2 & \text{-}a \end{bmatrix}$$\displaystyle \begin{bmatrix} x \\y \end{bmatrix}$ = $\displaystyle \begin{bmatrix} 1 \\11 \end{bmatrix}$

$\displaystyle \text{For what values of }{\color{red}a}\text{ does the system have a unique solution?}$

The system does not have a unique solution if its determinant equals zero.

. . $\displaystyle D \:=\:\begin{vmatrix}4&8 \\ 2&\text{-}a \end{vmatrix} \;=\;(4)(\text{-}a) - (8)(2) \;=\;\text{-}4a - 16$

. . If $\displaystyle D = 0\!:\;\text{-}4a - 16 \:=\:0 \quad\Rightarrow\quad a \:=\:\text{-}4$

$\displaystyle \text{The system has a unique solution for all }a \ne \text{-}4.$
. . . . $\displaystyle a \,\in\,(\text{-}\infty,\text{-}4) \cup (\text{-}4,\infty)$