Question

NO LINKS OR ANSWERING QUESTIONS YOU DON'T KNOW. Ch. 15 part 1a

How can a matrix be used to solve a system of equations. Demonstrate by solving the following system. Show your work. ​

Answer 1

We are given the equations:

x - 3y = 2

3x - 4y = 0

writing the system as matrices

$\left[\begin{array}{ccc}1&-3\\3&-4\end{array}\right]$$\left[\begin{array}{ccc}x\\y\end{array}\right]$ = $\left[\begin{array}{ccc}2\\0\\\end{array}\right]$

which is in the form:

AX = B

solving for X(the matrix holding the variables), we get:

X = (A⁻¹)B

Finding A⁻¹:

now, to do this, we need to find the inverse of A

$\left[\begin{array}{ccc}w&x\\y&z\end{array}\right]^{-1} = \frac{1}{wz-xy}\left[\begin{array}{ccc}z&-x\\-y&w\end{array}\right]$

using this formula to find the inverse of matrix A:

$A^{-1} = \frac{1}{(1*-4)-(-3*3)}\left[\begin{array}{ccc}-4&3\\-3&1\end{array}\right]$

$A^{-1} = \frac{1}{5}\left[\begin{array}{ccc}-4&3\\-3&1\end{array}\right]$

Matrix X:

We know that:

X = A⁻¹B

$X = \frac{1}{5}\left[\begin{array}{ccc}-4&3\\-3&1\end{array}\right] * \left[\begin{array}{ccc}2\\0\end{array}\right]$

$X = \frac{1}{5}\left[\begin{array}{ccc}-8\\-6\end{array}\right]$

since matrix X is just a matrix with the variables

$\left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}\frac{-8}{5}\\\frac{-6}{5}\end{array}\right]$

x = -8/5

y = -6/5