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### Course: Precalculus (Eureka Math/EngageNY)>Unit 2

Lesson 2: Topic C: Systems of linear equations

# Solving linear systems with matrices

Sal solves that matrix equation using the inverse of the coefficient matrix. Created by Sal Khan.

## Want to join the conversation?

• Does this extend into 3 equation, 3-variable problems? Like, would it be possible to solve ax+by+cz=d, ex+fy+gz=h, and ix+jy+kz=l for x, y, and z?
• Absolutely, but the inverse matrix gets more difficult to calculate.
• How do you find the inverse of A if it is a 2x3 matrix?
• The inverse can only exist if the matrix is nxn, or square, and even that is not a guarantee, some matrices do not have an inverse. To find out if a matrix does have an inverse, you need to calculate its determinant.
• Good day All,

How do you know that A has an inverse? if I am following correctly.
• Good day to you as well! Here is a good website. Click on it to visit it, & I hope it'll help! The part you are looking for is under the red letters "Does the Inverse Exist?".
http://stattrek.com/matrix-algebra/matrix-inverse.aspx
• So I'm taking a course thru aleks.com for algebra 2 and part of the problems are about matrices. I've been supplementing the written explanations from aleks with these videos/practice from Khan. One of the topics I'm trying to learn on Aleks right now is Cramer's rule for solving a 2x2 system of linear equations and I'm wondering if there is a video explaining that method here. This video seems to show a way to solve a 2x2 linear equation problem, but I don't think it's Cramer's rule. I tried searching for Cramer's rule, but did not find an actual video. Thanks
• How would you do AX - BX = C, note all are matrices
• AX - BX = C
(A - B)X = C
(A - B)^(-1)(A - B)X = (A - B)^(-1)C
IX = (A - B)^(-1)C
X = (A - B)^(-1)C
This is our answer (assuming we can calculate (A - B)^(-1)).
• Thanks a million for those videos.
I've studied math some at high school 20 years ago. I forgot most of it but now I'm learning it again and it's a whole new world now.
• Isn't A into A inverse the same thing as A inverse times A?
• Yes, matrix A multiplied with it's inverse A-1 (if it has one, and matrix A is a square matrix) will always result in the Identity matrix no matter the order (AA^-1 AND A^(-1)A will give I, so they are the same).
However, matrices (in general) are not commutative. That means that AB (multiplication) is not the same as BA.
• In which part of the course is it mentioned that "A inverse is equal to one over the determinant of A"??
• It appears that the inverse of a matrix is such that A-1 * A = I. However, as we know, matrix multiplication is not commutative. Does this mean that we would get a different inverse matrix if we defined it to be such that A * A-1 = I?
(1 vote)
• Matrix multiplication is not commutative in general. There exist pairs of matrices that commute, we just can't assume that any given pair will commute.

If A is a square matrix, and you find another matrix B such that AB=I, then you can prove that BA=I as well, and that B=A^(-1) is the only matrix with this property. A matrix has only one inverse.
• What's a column vector?
(1 vote)
• A vector that's written with the entries one above another, as in
3
5
2
as opposed to a row vector, which is written <3, 5, 2>.

Equivalently, a column vector is a nx1 matrix.