Question

Use matrices to solve this linear system:

5x1 − 2x2 = −30

2x1 − x2 = −13

x1 =
x2 =

Answer 1

-4

5

Explanation:

on edge!!

Answer 2

`x_1=-4` and
`x_2=5`

Explanation:

Consider given system ,

`5x_1-2x_2=-30\\\\\\2x_1-x_2=-13`

We have to solve for
`x_1` and
`x_2`

In matrix form,

`\left[\begin{array}{cc}5&-2\\2&-1\end{array}\right]\left[\begin{array}{c}x_1\\x_2\end{array}\right] =\left[\begin{array}{c}-30\\-13\end{array}\right]`

then this is in form of AX = b,

A=
`\left[\begin{array}{cc}5&-2\\2&-1\end{array}\right]`

X=
`\left[\begin{array}{c}x_1\\x_2\end{array}\right]`

b=
`\left[\begin{array}{c}-30\\-13\end{array}\right]`

Pre-multiply by
`A^(-1)` both side, we get,

`X=A^(-1)B` ..............(1)

First finding inverse of A ,

`=\frac{1}{\det \begin{pmatrix}5&-2\\ 2&-1\end{pmatrix}}\begin{pmatrix}-1&-\left(-2\right)\\ -2&5\end{pmatrix}`

`\det \begin{pmatrix}5&-2\\ 2&-1\end{pmatrix}=-1`

Thus, inverse of A is ,

`(1)/(-1)\begin{pmatrix}-1&-\left(-2\right)\\ -2&5\end{pmatrix}=\begin{pmatrix}1&-2\\ 2&-5\end{pmatrix}`

Now substitute
`A^(-1)` in (1) , we get,

`\left[\begin{array}{c}x_1\\x_2\end{array}=\left[\begin{array}{cc}1&-2\\2&-5\end{array}\right]\left[\begin{array}{c}-30\\-13\end{array}\right]`

Multiply , we get,

`=\begin{pmatrix}1\cdot \left(-30\right)+\left(-2\right)\left(-13\right)\\ 2\left(-30\right)+\left(-5\right)\left(-13\right)\end{pmatrix}`

`\left[\begin{array}{c}x_1\\x_2\end{array}\right]=\left[\begin{array}{c}-4\\5\end{array}\right]`

Thus,
`x_1=-4` and
`x_2=5`