Question

Solve the system of equations by using matrix equations:
5x+4y=-3
-3x-5y=-24

Answer 1

I forgot all about matrices and how to find the inverse..

Answer 2

All you joe...

Answer 3

-_-

Answer 4

Apparently, nobody likes matrices

Answer 5

That includes you Saifoo

Answer 6

Too tedious..

Answer 7

agree.

Answer 8

yeah, its just tedious.

Use this formula to get the inverse of the matrix:

If A is the 2x2 matrix

a b
c d

Then:
\[A^{-1}=\frac{1}{detA}B\]
Where B is the 2x2 matrix

d -b
-c a

Answer 9

det?

Answer 10

Determinant? if you dont know how to calculate the determinant of a 2x2 matrix then this is going to require a bit more work.

Answer 11

(not saying that as an insult, im just saying, other paths are long, sry if it came off that way)

Answer 12

Ah, no it's fine. It's like cross multiplying or something.. right?

Answer 13

right right, if A is the 2x2 matrix:

a b
c d

then the det(A) = ad - bc

Answer 14

In this case, the determinant is -37?

Answer 15

From Mathematica:
\[\text{LinearSolve}\left[\left( \begin{array}{cc} 5 & 4 \\ -3 & -5 \end{array} \right),\left( \begin{array}{c} -3 \\ -24 \end{array} \right)\right]\to \left\{\left\{-\frac{111}{13}\right\},\left\{\frac{129}{13}\right\}\right\} \]

Answer 16

it would be:\[5(-5)-(4)(-3)=-25+12=-13\]

Answer 17

Oh yeah, my mistake.. I wrote it as -12 instead of +

Answer 18

\[-1/13\left[\begin{matrix}-5 & -4 \\ 3 & 5\end{matrix}\right]\left[\begin{matrix}5 & 4 \\ 3 & -5\end{matrix}\right] \left(\begin{matrix}x \\ y\end{matrix}\right) = -1/13\left[\begin{matrix}-5 & -4 \\ 3 & 5\end{matrix}\right]\left(\begin{matrix}-3 \\ 24\end{matrix}\right)\]
The ( ) are supposed to be [ ] but I couldn't find the [ ] for 2x1 matrices

Answer 19

@@ I forgot how to multiply matrices too!! How did you get 111? -5(-3).. and then what else?

Answer 20

its -5(-3)+(-4)(-24) for the first entry, then (3)(-3)+(5)(-24)

Answer 21

for the second.

Answer 22

Okay. How did you cancel out everything on the left side so you are just left with x, y ?

Answer 23

Well i'm not sure exactly which matrix equations you are supposed to use but here's gaussian elimination

Original Matrix:
5 4 -3
-3 -5 -24

3 * row 1:
15 12 -9
-3 -5 -24

5 * row 2:
15 12 -9
-15 -25 -120

row 1 added to row 2:
15 12 -9
0 -13 -129

row 2 divided by -13:
15 12 -9
0 1 9.923

-12 * row 2:
15 12 -9
0 -12 -119.076

row 2 added to row 1:
15 0 -128.076
0 -12 -119

row 1 divided by 15
and row 2 divided by -12:
1 0 -8.538
0 1 9.923

therefore:
x = -8.538
y = 9.923

Answer 24

Do you mean the part where I did:\[A\vec x=\vec b\iff A^{-1}A\vec x=A^{-1}\vec b\]\[\vec x = A^{-1}\vec b\]
?

Answer 25

Yes..

Answer 26

Thats because:\[AA^{-1}=A^{-1}A=I\]
The identity matrix. Its the equivalent of 1 in matrices.

So guess the step I skipped was:
\[A\vec x =\vec b \iff (A^{-1}A)\vec x =A^{-1}\vec b \iff I\vec x = A^{-1}\vec b\iff \vec x = A^{-1}\vec b\]

Answer 27

Thats the purpose of A inverse. it multiply with A to "cancel it out"

Answer 28

Oh, I see now! If I could give you more than one medal, I would!!