McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
8. Solving Systems of Equations Using Inverse Matrices
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Exercise 9 Page 202

To solve a matrix equation, multiply both sides by the inverse of the coefficient matrix.

( -2,5)

Practice makes perfect
We want to solve the given system of equations using matrices to find the point of intersection of the lines. We need to write the system as a matrix equation. -2x+y=9 x+y=3 ⇒ [ c -2x+y x+y ] = [ c 9 3 ] Now, we will write the left-hand side of the above matrix equation as the product of the coefficient matrix and the variable matrix. [ c -2x+y x+y ] = [ c 9 3 ] ⇕ [ cc -2 & 1 1 & 1 ] * [ c x y ] = [ c 9 3 ] We will solve the matrix equation by multiplying both sides by the inverse of the coefficient matrix to isolate the x and y variables. When multiplying a matrix by its inverse, we obtain the identity matrix.
-2 & 1 1 & 1 * x y = 9 3

-1/3 [ cc 1 & -1 -1 & -2 ] * LHS = -1/3 [ cc 1 & -1 -1 & -2 ] * RHS

-1/3 1 & -1 -1 & -2 * -2 & 1 1 & 1 * x y = -1/3 1 & -1 -1 & -2 * 9 3
â–Ľ
Simplify left-hand side
1 & 0 0 & 1 * x y = -1/3 1 & -1 -1 & -2 * 9 3
x y = -1/3 1 & -1 -1 & -2 * 9 3
Now that the left-hand side is only the variable matrix, we can simplify the right-hand side to find the values of the variables.
x y = -1/3 1 & -1 -1 & -2 * 9 3
â–Ľ
Simplify right-hand side
x y = -1/3 1(9)+(-1)3 -1(9)+(-2)3
x y = -1/3 9-3 -9-6
x y = -1/3 6 -15

Multiply matrix by -1/3

x y = -2 5
The solution to the system, which is the point of intersection of the lines, is (-2,5 ).

Showing Our Work

Finding inverses and multiplying matrices

Finding the Inverse of the Coefficient Matrix

To find the inverse of a 2* 2 matrix, we use the corresponding formula.

Matrix Inverse
A= [ cc a & b c & d ] A^(- 1)=1/ad-bc [ cc d & - b - c & a ] where ad-bc ≠ 0
The expression ad-bc is known as the determinant of a 2* 2 matrix. Because it is in the denominator of a fraction, if the determinant is zero, the matrix cannot have an inverse. Consider our coefficient matrix. [ cc -2 & 1 1 & 1 ] Let's calculate its determinant.
ad-bc
-2(1)- 1( 1)
-2-1
-3
Since the determinant is not zero, the matrix has an inverse. We can now apply the formula for the inverse. Note that we usually refer to the determinant using the notation ad-bc=det(A).
1/det(A) d & - b - c & a
1/-3 1 & - 1 - 1 & -2
-1/3 1 & -1 -1 & -2
For simplicity, instead of multiplying the matrix by the scalar, we will leave the expression as it is.

Multiplying Matrices

In our exercise, we had to multiply a 2* 2 matrix by a 2* 1 matrix. This is possible because the number of columns of the first matrix is equal to the number of rows of the second one. Moreover, the product has as many rows as the first matrix and as many columns as the second one. Thus, the product is a 2* 1 matrix. [ cc a & b c & d ] * [ c x y ] = [ c ax+by cx+dy ] 2 * 2 2 * 1 2 * 1 Let's multiply the matrices of the exercise.
1 & -1 -1 & -2 * 9 3
1(9)+(-1)3 -1(9)+(-2)3
9-3 -9-6
6 -15