What method was used?
Did they show the inverse matrix?
Did they use something called row-reduced echelon form?
(I would never think of using matrices for a system this simple)
Use a matrix equation to solve the system of equations.
6x + 2y = 11
3x - 8y =1
My textbook gave this example, but explained it in a really weird way and I don't understand how to do it. Can someone help?
5 answers
6x + 2y = 11
3x - 8y =1
The matrix equation is
AX=B
where A =
(6 2)
(3 -8)
X =
(x)
(y)
and B =
(11)
(1)
Since AX=B, if you multiply on the left by A^-1, you get
A^-1 * A X = A^-1 B
So, find A^-1, multiply by B and you end up with a 2x1 matrix which is the solution. See here:
www.wolframalpha.com/input/?i=%7B%7B6,2%7D,%7B3,-8%7D%7D*%7B%7Bx%7D,%7By%7D%7D+%3D+%7B%7B11%7D,%7B1%7D%7D
3x - 8y =1
The matrix equation is
AX=B
where A =
(6 2)
(3 -8)
X =
(x)
(y)
and B =
(11)
(1)
Since AX=B, if you multiply on the left by A^-1, you get
A^-1 * A X = A^-1 B
So, find A^-1, multiply by B and you end up with a 2x1 matrix which is the solution. See here:
www.wolframalpha.com/input/?i=%7B%7B6,2%7D,%7B3,-8%7D%7D*%7B%7Bx%7D,%7By%7D%7D+%3D+%7B%7B11%7D,%7B1%7D%7D
btw, Wolfram also let's you find the inverse matrix
https://www.wolframalpha.com/input/?i=inverse+%7B%7B6,2%7D,%7B3,-8%7D%7D
https://www.wolframalpha.com/input/?i=inverse+%7B%7B6,2%7D,%7B3,-8%7D%7D
I understand how to find the determinant, but my textbook also showed a part about how to get the solution for x and y, which I didn't understand. It said to multiply each side of the matrix equation by the inverse matrix, although I didn't understand what that meant and couldn't figure out how they did it. It says that the solution is (5/3, 1/2)
If they expect you to solve this using an inverse matrix, they must have shown
you how to find that matrix.
The last link I gave you from Wolfram showed what that matrix is.
There is a quick way to find the inverse of a 2 by 2
start with
a b
c d
find the determinant D, you said you knew how: D = (ad - bc)
inverse is
d/D -b/D
-c/D a/D
or
(1/D) times
d -b
-c a
in your case that becomes in reduced form
4/27 1/27
1/18 -1/9
you how to find that matrix.
The last link I gave you from Wolfram showed what that matrix is.
There is a quick way to find the inverse of a 2 by 2
start with
a b
c d
find the determinant D, you said you knew how: D = (ad - bc)
inverse is
d/D -b/D
-c/D a/D
or
(1/D) times
d -b
-c a
in your case that becomes in reduced form
4/27 1/27
1/18 -1/9