# How do you solve a-2b+20c=1, 3a+b-4c=2 and 2a+b-8c=3 using matrices?

Jun 6, 2017

The solution is $\left(\begin{matrix}a \\ b \\ c\end{matrix}\right) = \left(\begin{matrix}2 \\ - 7 \\ - \frac{3}{4}\end{matrix}\right)$

#### Explanation:

We write the equations in matrix form

$\left(\begin{matrix}1 & - 2 & 20 \\ 3 & 1 & - 4 \\ 2 & 1 & - 8\end{matrix}\right) \cdot \left(\begin{matrix}a \\ b \\ c\end{matrix}\right) = \left(\begin{matrix}1 \\ 2 \\ 3\end{matrix}\right)$

or,

$A \cdot x = b$

We must calculate ${A}^{-} 1$ the inverse of $A$

To see if the matrix is invertible, we calculate the determinant

$\det A = | \left(1 , - 2 , 20\right) , \left(3 , 1 , - 4\right) , \left(2 , 1 , - 8\right) |$

$= 1 \cdot | \left(1 , - 4\right) , \left(1 , - 8\right) | + 2 | \left(3 , - 4\right) , \left(2 , - 8\right) | + 20 \cdot | \left(3 , 1\right) , \left(2 , 1\right) |$

$= 1 \left(- 4\right) + 2 \left(- 16\right) + 20 \left(1\right)$

$= - 4 - 32 + 20 = - 16$

As,

$\det A \ne 0$, the matrix is invertible

We calculate the matrix of co-factors

$C = \left(\begin{matrix}| \left(1 - 4\right) & \left(1 - 8\right) | & - | \left(3 - 4\right) & \left(2 - 8\right) | & | \left(3 1\right) & \left(2 1\right) | \\ - | \left(- 2 20\right) & \left(1 - 8\right) | & | \left(1 20\right) & \left(2 - 8\right) | & - | \left(1 - 2\right) & \left(2 1\right) | \\ | \left(- 2 20\right) & \left(1 - 4\right) | & - | \left(1 20\right) & \left(3 - 4\right) | & | \left(1 - 2\right) & \left(3 1\right) |\end{matrix}\right)$

$= \left(\begin{matrix}- 4 & 16 & 1 \\ 4 & - 48 & - 5 \\ - 12 & 64 & 7\end{matrix}\right)$

The transpose of $C$ is

${C}_{T} = \left(\begin{matrix}- 4 & 4 & - 12 \\ 16 & - 48 & 64 \\ 1 & - 5 & 7\end{matrix}\right)$

The inverse is

${A}^{-} 1 = \frac{{C}_{T}}{\det A}$

$= - \frac{1}{16} \left(\begin{matrix}- 4 & 4 & - 12 \\ 16 & - 48 & 64 \\ 1 & - 5 & 7\end{matrix}\right)$

$= \left(\begin{matrix}\frac{1}{4} & - \frac{1}{4} & \frac{3}{4} \\ - 1 & 3 & - 4 \\ - \frac{1}{16} & \frac{5}{16} & - \frac{7}{16}\end{matrix}\right)$

Therefore,

$\left(\begin{matrix}a \\ b \\ c\end{matrix}\right) = \left(\begin{matrix}\frac{1}{4} & - \frac{1}{4} & \frac{3}{4} \\ - 1 & 3 & - 4 \\ - \frac{1}{16} & \frac{5}{16} & - \frac{7}{16}\end{matrix}\right) \cdot \left(\begin{matrix}1 \\ 2 \\ 3\end{matrix}\right)$

$= \left(\begin{matrix}\frac{1}{4} - \frac{2}{4} + \frac{9}{4} \\ - 1 + 6 - 12 \\ - \frac{1}{16} + \frac{10}{16} - \frac{21}{16}\end{matrix}\right)$

$= \left(\begin{matrix}2 \\ - 7 \\ - \frac{3}{4}\end{matrix}\right)$