Question

QUESTION 5

Part 1 of 2:

Solve the system using the matrix equation. Be sure to show all 3 parts! Write your solution as an ordered pair.

QUESTION 6

Part 2 of 2:

Solve the system of equations you solved in question 5 again, this time using Cramer's Rule. Verify that you come up with the same solution as you did in the previous question. If you don't, please go back and check your work. Show all work and write your solution as an ordered pair.

Answer 1

5) Given system of equations are, -4x-3y=-4

and, 6x+5y=8

This system of linear equations can be written as, $\begin{bmatrix} -4 & -3\\ 6 & 5 \end{bmatrix}$ * $\begin{bmatrix} x \\ y \end{bmatrix}$ = $\begin{bmatrix} -4 \\ 8 \end{bmatrix}$

i.e., AX=B where A = $\begin{bmatrix} -4 & -3\\ 6 & 5 \end{bmatrix}$ , X = $\begin{bmatrix} x \\ y \end{bmatrix}$ , B = $\begin{bmatrix} -4 \\ 8 \end{bmatrix}$

i.e., X=A^-1B

Now, A^-1= (adj A)/(det A) = $\begin{bmatrix} 5 & 3\\ -6 & -4 \end{bmatrix}$ / [(-20) - (-18)] = $\begin{bmatrix} 5 & 3\\ -6 & -4 \end{bmatrix}$ / (-2) = $\begin{bmatrix} -5 & -3\\ 6 & 4 \end{bmatrix}$ / 2

Therefore, X = (1/2)* $\begin{bmatrix} -5 & -3\\ 6 & 4 \end{bmatrix}$ * $\begin{bmatrix} -4 \\ 8 \end{bmatrix}$ = (1/2)* $\begin{bmatrix} -4 \\ 8 \end{bmatrix}$ = $\begin{bmatrix} -2 \\ 4 \end{bmatrix}$

i.e., $\begin{bmatrix} x \\ y \end{bmatrix}$ = $\begin{bmatrix} -2 \\ 4 \end{bmatrix}$

Hence, the solution is (x,y) = (-2,4).

6) For Cramer's rule,

D = $\begin{vmatrix} -4 & -3\\ 6 & 5 \end{vmatrix}$ = (-4)*5 - (-3)*6 = (-20) - (-18) = -20+18 = -2

D_x = $\begin{vmatrix} -4 & -3\\ 8 & 5 \end{vmatrix}$ = (-4)*5 - (-3)*8 = (-20) - (-24) = -20+24 = 4

D_y = $\begin{vmatrix} -4 & -4\\ 6 & 8 \end{vmatrix}$ = (-4)*8 - (-4)*6 = (-32) - (-24) = -32+24 = -8

Therefore, x = D_x/D = 4 / (-2) = -2 and y = D_y/D = (-8) / (-2) = 4

Hence, the solution is (x,y) = (-2,4).

Here, in both process, the solution of the linear system of equations is same.