# How do you solve the system 3a+2b=27, 6a-7b+c=5, -2a+10b+5c=-29 using matrices?

Aug 18, 2017

$a = \frac{1323}{199} , b = \frac{702}{199} , c = - \frac{2029}{199}$

#### Explanation:

3a + 2b = 27
6a - 7b + c = 5
-2a + 10b +5c = -29

First, convert the left side of the equations into a matrix called 'A'. Inside the matrix, put in the coeffecients of each of the variables in the equations above.

A= $\left[\begin{matrix}3 & 2 & 0 \\ 6 & - 7 & 1 \\ - 2 & 10 & 5\end{matrix}\right]$

Now, do the same for right side of the equation and convert the numbers into a matrix called 'b'.

b= $\left[\begin{matrix}27 \\ 5 \\ - 29\end{matrix}\right]$

Use the formula $A x = b$, where $x$ is the solution for the system of equations.

Rearrange for $x$ $\to$ $x = {A}^{-} 1 \cdot b$

${A}^{-} 1$ refers to the inverse of the $A$ matrix. It can easily be found using a graphical calculator.

If you need to do it by hand, here are some videos from Khan Academy on how to do it. I did mine using Symbolab.

$x = {\left[\begin{matrix}3 & 2 & 0 \\ 6 & - 7 & 1 \\ - 2 & 10 & 5\end{matrix}\right]}^{-} 1 \cdot \left[\begin{matrix}27 \\ 5 \\ - 29\end{matrix}\right]$

$x = \left[\begin{matrix}\frac{1323}{199} \\ \frac{702}{199} \\ - \frac{2029}{199}\end{matrix}\right]$

Therefore,

$a = \frac{1323}{199} , b = \frac{702}{199} , c = - \frac{2029}{199}$

Aug 18, 2017

The solution is $\left(\begin{matrix}a \\ b \\ c\end{matrix}\right)$$=$$\left(\begin{matrix}\frac{1323}{199} \\ \frac{702}{199} \\ - \frac{2029}{199}\end{matrix}\right)$

#### Explanation:

We perform the Gauss Jordan elimination with the augmented matrix

$\left(\begin{matrix}3 & 2 & 0 & : & 27 \\ 6 & - 7 & 1 & : & 5 \\ - 2 & 10 & 5 & : & - 29\end{matrix}\right)$

$R 3 \leftarrow 3 R 3 + R 2$, $\implies$, $\left(\begin{matrix}3 & 2 & 0 & : & 27 \\ 6 & - 7 & 1 & : & 5 \\ 0 & 23 & 16 & : & - 82\end{matrix}\right)$

$R 2 \leftarrow R 2 - 2 R 1$, $\implies$, $\left(\begin{matrix}3 & 2 & 0 & : & 27 \\ 0 & - 11 & 1 & : & - 49 \\ 0 & 23 & 16 & : & - 82\end{matrix}\right)$

$R 2 \leftarrow \frac{R 2}{- 11}$, $\implies$, $\left(\begin{matrix}3 & 2 & 0 & : & 27 \\ 0 & 1 & - \frac{1}{11} & : & \frac{49}{11} \\ 0 & 23 & 16 & : & - 82\end{matrix}\right)$

$R 3 \leftarrow \frac{R 3}{23}$, $\implies$, $\left(\begin{matrix}3 & 2 & 0 & : & 27 \\ 0 & 1 & - \frac{1}{11} & : & \frac{49}{11} \\ 0 & 1 & \frac{16}{23} & : & - \frac{82}{23}\end{matrix}\right)$

$R 3 \leftarrow R 3 - R 2$, $\implies$, $\left(\begin{matrix}3 & 2 & 0 & : & 27 \\ 0 & 1 & - \frac{1}{11} & : & \frac{49}{11} \\ 0 & 0 & \frac{199}{253} & : & - \frac{2029}{253}\end{matrix}\right)$

$R 3 \leftarrow \left(R 3\right) \cdot \left(\frac{253}{199}\right)$, $\implies$, $\left(\begin{matrix}3 & 2 & 0 & : & 27 \\ 0 & 1 & - \frac{1}{11} & : & \frac{49}{11} \\ 0 & 0 & 1 & : & - \frac{2029}{199}\end{matrix}\right)$

$R 2 \leftarrow R 2 + \left(\frac{1}{11}\right) R 3$, $\implies$, $\left(\begin{matrix}3 & 2 & 0 & : & 27 \\ 0 & 1 & 0 & : & \frac{702}{99} \\ 0 & 0 & 1 & : & - \frac{2029}{199}\end{matrix}\right)$

$R \leftarrow R 1 - \left(2\right) R 2$, $\implies$, $\left(\begin{matrix}3 & 0 & 0 & : & \frac{3969}{199} \\ 0 & 1 & 0 & : & \frac{702}{99} \\ 0 & 0 & 1 & : & - \frac{2029}{199}\end{matrix}\right)$

$R \leftarrow \frac{R 1}{3}$, $\implies$, $\left(\begin{matrix}1 & 0 & 0 & : & \frac{1323}{199} \\ 0 & 1 & 0 & : & \frac{702}{199} \\ 0 & 0 & 1 & : & - \frac{2029}{199}\end{matrix}\right)$