# Help with matrices

• Jan 19th 2013, 06:57 AM
Tutu
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
$\begin{bmatrix} 4 & 8 \\2 & -a \end{bmatrix}$ $\begin{bmatrix} x \\y \end{bmatrix}$ = $\begin{bmatrix} 1 \\11 \end{bmatrix}$

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
$\begin{bmatrix} 4 & 8 \\0 & 2a+8 \end{bmatrix}$ $\begin{bmatrix} x \\y \end{bmatrix}$ = $\begin{bmatrix} 1 \\-22 \end{bmatrix}$

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

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

Will really appreciate it, thank you so much!
• Jan 19th 2013, 08:16 AM
Soroban
Re: Help with matrices
Hello, Tutu!

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

Quote:

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

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

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

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

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

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