High School: Algebra
High School: Algebra
Reasoning with Equations and Inequalities HSA-REI.C.9
9. Find the inverse of a matrix if it exists and use it to solve systems of linear equations (using technology for matrices of dimension 3 × 3 or greater).
This is where actual calculations take place. We've finally gotten to the fun stuff. Your students may think otherwise, but we know the truth.
Students should already know how to form a matrix equation of AX = B from a system of linear equations and be familiar with the concept of inverse matrices. They will use the inverse of a matrix in order to solve for the variables of the matrix that is formed from given equations.
Students should know that if the determinant of a square matrix is zero (ad – bc = 0), there is no inverse to the matrix. A matrix that takes the form
has an inverse of
Assuming we've already translated our equations -2x + y = 5 and -3x + y = -2 into matrix equation form, we can find the inverse matrix of the A matrix (the one with all the coefficients).
First, we take our matrix and find its determinant. That would be -2 × 1 – 1 × -3 = -2 + 3 = 1. The inverse matrix exists. Now, let's give our creation life! After switching around the numbers, we have to multiply by the inverse of the determinant, which is 1⁄1 or just 1. So our inverse matrix is fine as is.
So your students have found the inverse matrix. Now what? Well, the whole point of this process is to find the values of x and y. (They should write that down. That's important.) To do that, we can multiply the inverse matrix we found by the B matrix (the one with all the solutions). That should give us the values for x and y.
That gives us 1 × 5 + (-1 × -2) = 5 + 2 = 7 = x, and 3 × 5 + (-2 × -2) = 15 + 4 = 19 = y. If we plug those values back into the original linear equations, they should all hold up.
While this method might seem a bit cumbersome, students should appreciate that it's especially useful for systems of equations with many, many variables. On the other hand, the inverses for matrices larger than 2 × 2 should be calculated using technology. The whole point, though, is that they can help us solve for the values of the variables.