Question

Write a matrix equation for the system below. Then solve the equation by using an inverse matrix. You must solve this by hand. Give exact answers. No graphing calculator!

Answer 1

First, let's write the system in the matrix form AX = B:

`\begin{bmatrix}{1} & {2} \\ {-4} & {3}\end{bmatrix}\begin{bmatrix}{x} & \\ {y} & \end{bmatrix}=\begin{bmatrix}{2} & {} \\ {25} & {}\end{bmatrix}`

Now, to solve the system, let's first find the inverse of the matrix A, using the formula below for the inverse of a 2x2 matrix:

`\begin{gathered} A=\begin{bmatrix}{a} & {b} \\ c & {d}\end{bmatrix}\\ \\ A^(-1)=(1)/(ad-bc)\begin{bmatrix}{d} & -b \\ {-c} & {a}\end{bmatrix} \end{gathered}`

So we have:

`\begin{gathered} A=\begin{bmatrix}{1} & {2} \\ {-4} & {3}\end{bmatrix}\\ \\ A^(-1)=(1)/(3-(-8))\begin{bmatrix}{3} & -2{} \\ {4} & {1}\end{bmatrix}\\ \\ A^(-1)=(1)/(11)\begin{bmatrix}{3} & -2{} \\ {4} & {1}\end{bmatrix}\\ \\ A^(-1)=\begin{bmatrix}{(3)/(11)} & -(2)/(11){} \\ {(4)/(11)} & {(1)/(11)}\end{bmatrix}\\ \end{gathered}`

Now, to solve the system, we can do the following:

`\begin{gathered} AX=B\\ \\ A^(-1)AX=A^(-1)B\\ \\ IX=A^(-1)B\\ \\ X=A^(-1)B \end{gathered}`

Multiplying the inverse matrix and matrix B, we have:

`\begin{gathered} \begin{bmatrix}{(3)/(11)} & {-(2)/(11)} \\ {(4)/(11)} & {(1)/(11)}\end{bmatrix}\cdot\begin{bmatrix}{2} & {} \\ 25 & {}\end{bmatrix}=\begin{bmatrix}{(3)/(11)}\cdot2+(-(2)/(11))\cdot25 & \\ {(4)/(11)}\cdot2+(1)/(11)\cdot25 & {}\end{bmatrix}\\ \\ =\begin{bmatrix}{(6)/(11)}-(50)/(11) & \\ {(8)/(11)}+(25)/(11) & {}\end{bmatrix}=\begin{bmatrix}-(44)/(11) & \\ (33)/(11) & {}\end{bmatrix}=\begin{bmatrix}-4 & \\ 3 & {}\end{bmatrix} \end{gathered}`

Therefore the solution is x = -4 and y = 3.