# How do you solve ((1, -6, 0), (0, 1, -7), (3, 0, 2))X=((1), (4), (11))?

May 9, 2016

$X = \left(\begin{matrix}4 \\ 0.5 \\ - 0.5\end{matrix}\right)$

#### Explanation:

Let $X = \left(\begin{matrix}a \\ b \\ c\end{matrix}\right)$
Using Cramer's Rule to solve the augmented matrix
$\textcolor{w h i t e}{\text{XXXXXXXXXXXxX")acolor(white)("XX")bcolor(white)("Xx")c color(white)("XXX}} k$
color(white)("XXX")((M,"|",k))=((1,-6,0,"|",1),(0,1,-7,"|",4),(3,0,2,"|",11))

Calculating the Determinants:
$\textcolor{w h i t e}{\text{XXX}} D e t \left(M\right) = | \left(1 , - 6 , 0\right) , \left(0 , 1 , - 7\right) , \left(3 , 0 , 2\right) |$

$\textcolor{w h i t e}{\text{XXXXXXX")=1color(white)("X}} \left[\left(1 \times 2\right) - \left(0 \times - 7\right)\right]$
$\textcolor{w h i t e}{\text{XXXXXXXX")-color(white)("X}} 0 \left[\left(- 6 \times 2\right) - \left(0 \times 0\right)\right]$
$\textcolor{w h i t e}{\text{XXXXXXXX")+3color(white)("X}} \left[\left(\left(- 6\right) \times \left(- 7\right)\right) - \left(0 \times 0\right)\right]$

$\textcolor{w h i t e}{\text{XXXXXXX}} = 128$

Similarly
$\textcolor{w h i t e}{\text{XXX}} D e t \left({M}_{a}\right) = 512$

$\textcolor{w h i t e}{\text{XXX}} D e t \left({M}_{b}\right) = 64$

$\textcolor{w h i t e}{\text{XXX}} D e t \left({M}_{c}\right) = - 64$

By Cramer's Rule:
$\textcolor{w h i t e}{\text{XXX}} a = \frac{D e t \left({M}_{a}\right)}{D e t \left(M\right)} = \frac{512}{128} = 4$

$\textcolor{w h i t e}{\text{XXX}} b = \frac{D e t \left({M}_{b}\right)}{D e t \left(M\right)} = \frac{64}{128} = 0.5$

$\textcolor{w h i t e}{\text{XXX}} c = \frac{D e t \left({M}_{c}\right)}{D e t \left(M\right)} = \frac{- 64}{128} = - 0.5$

Confession: I used as spreadsheet to do the detailed arithmetic work