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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. ​

NO LINKS OR ANSWERING QUESTIONS YOU DON'T KNOW. Ch. 15 part 1a How can a matrix be-example-1

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


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

Matrix X:

We know that:

X = A⁻¹B


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


X = (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}(-8)/(5)\\(-6)/(5)\end{array}\right]

x = -8/5

y = -6/5

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