# How do you solve ((2, 0, 0), (-1, 2, 0), (-2, 4, 1))x=((4), (10), (11))?

Mar 23, 2016

$x = \left(\begin{matrix}2 \\ 6 \\ - 9\end{matrix}\right)$

#### Explanation:

First construct the inverse matrix of $\left(\begin{matrix}2 & 0 & 0 \\ - 1 & 2 & 0 \\ - 2 & 4 & 1\end{matrix}\right)$ by writing the identity matrix alongside it and transforming that as we transform the original matrix into the identity:

$\left(\begin{matrix}2 & 0 & 0 & | & 1 & 0 & 0 \\ - 1 & 2 & 0 & | & 0 & 1 & 0 \\ - 2 & 4 & 1 & | & 0 & 0 & 1\end{matrix}\right)$

Add row $1$ to row $3$:

$\left(\begin{matrix}2 & 0 & 0 & | & 1 & 0 & 0 \\ - 1 & 2 & 0 & | & 0 & 1 & 0 \\ 0 & 4 & 1 & | & 1 & 0 & 1\end{matrix}\right)$

Divide row $1$ by $2$:

$\left(\begin{matrix}1 & 0 & 0 & | & \frac{1}{2} & 0 & 0 \\ - 1 & 2 & 0 & | & 0 & 1 & 0 \\ 0 & 4 & 1 & | & 1 & 0 & 1\end{matrix}\right)$

Add row $1$ to row $2$:

$\left(\begin{matrix}1 & 0 & 0 & | & \frac{1}{2} & 0 & 0 \\ 0 & 2 & 0 & | & \frac{1}{2} & 1 & 0 \\ 0 & 4 & 1 & | & 1 & 0 & 1\end{matrix}\right)$

Subtract $2 \times$ row $2$ from row $3$:

$\left(\begin{matrix}1 & 0 & 0 & | & \frac{1}{2} & 0 & 0 \\ 0 & 2 & 0 & | & \frac{1}{2} & 1 & 0 \\ 0 & 0 & 1 & | & 0 & - 2 & 1\end{matrix}\right)$

Divide row $2$ by $2$:

$\left(\begin{matrix}1 & 0 & 0 & | & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 & | & \frac{1}{4} & \frac{1}{2} & 0 \\ 0 & 0 & 1 & | & 0 & - 2 & 1\end{matrix}\right)$

So:

${\left(\begin{matrix}2 & 0 & 0 \\ - 1 & 2 & 0 \\ - 2 & 4 & 1\end{matrix}\right)}^{- 1} = \left(\begin{matrix}\frac{1}{2} & 0 & 0 \\ \frac{1}{4} & \frac{1}{2} & 0 \\ 0 & - 2 & 1\end{matrix}\right)$

Then multiply the right hand side column matrix by our inverse matrix to find:

$x = \left(\begin{matrix}\frac{1}{2} & 0 & 0 \\ \frac{1}{4} & \frac{1}{2} & 0 \\ 0 & - 2 & 1\end{matrix}\right) \left(\begin{matrix}4 \\ 10 \\ 11\end{matrix}\right) = \left(\begin{matrix}2 \\ 6 \\ - 9\end{matrix}\right)$

Mar 23, 2016

$x = \left(\begin{matrix}2 \\ 6 \\ - 9\end{matrix}\right)$

#### Explanation:

Alternatively, don't bother to construct an inverse matrix: Just perform a similar sequence of steps with the target column matrix appended to our original matrix as follows:

$\left(\begin{matrix}2 & 0 & 0 & | & 4 \\ - 1 & 2 & 0 & | & 10 \\ - 2 & 4 & 1 & | & 11\end{matrix}\right)$

Add row $1$ to row $3$:

$\left(\begin{matrix}2 & 0 & 0 & | & 4 \\ - 1 & 2 & 0 & | & 10 \\ 0 & 4 & 1 & | & 15\end{matrix}\right)$

Divide row $1$ by $2$:

$\left(\begin{matrix}1 & 0 & 0 & | & 2 \\ - 1 & 2 & 0 & | & 10 \\ 0 & 4 & 1 & | & 15\end{matrix}\right)$

Add row $1$ to row $2$:

$\left(\begin{matrix}1 & 0 & 0 & | & 2 \\ 0 & 2 & 0 & | & 12 \\ 0 & 4 & 1 & | & 15\end{matrix}\right)$

Subtract $2 \times$ row $2$ from row $3$:

$\left(\begin{matrix}1 & 0 & 0 & | & 2 \\ 0 & 2 & 0 & | & 12 \\ 0 & 0 & 1 & | & - 9\end{matrix}\right)$

Divide row $2$ by $2$:

$\left(\begin{matrix}1 & 0 & 0 & | & 2 \\ 0 & 1 & 0 & | & 6 \\ 0 & 0 & 1 & | & - 9\end{matrix}\right)$

Having reached the identity matrix on the left hand side, we can read off the solution from the right hand side:

$x = \left(\begin{matrix}2 \\ 6 \\ - 9\end{matrix}\right)$