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The only difference between a solving a linear equation and a system of equations written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. However, the goal is the same—to isolate the variable.
We will investigate this idea in detail, but it is helpful to begin with a system and then move on to a system.
Given a system of equations, write the coefficient matrix the variable matrix and the constant matrix Then
Multiply both sides by the inverse of to obtain the solution.
If the coefficient matrix does not have an inverse, does that mean the system has no solution?
No, if the coefficient matrix is not invertible, the system could be inconsistent and have no solution, or be dependent and have infinitely many solutions.
Solve the given system of equations using the inverse of a matrix.
Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.
Then
First, we need to calculate Using the formula to calculate the inverse of a 2 by 2 matrix, we have:
So,
Now we are ready to solve. Multiply both sides of the equation by
The solution is
Can we solve for by finding the product
No, recall that matrix multiplication is not commutative, so Consider our steps for solving the matrix equation.
Notice in the first step we multiplied both sides of the equation by but the was to the left of on the left side and to the left of on the right side. Because matrix multiplication is not commutative, order matters.
Solve the following system using the inverse of a matrix.
Write the equation
First, we will find the inverse of by augmenting with the identity.
Multiply row 1 by
Multiply row 1 by 4 and add to row 2.
Add row 1 to row 3.
Multiply row 2 by −3 and add to row 1.
Multiply row 3 by 5.
Multiply row 3 by and add to row 1.
Multiply row 3 by and add to row 2.
So,
Multiply both sides of the equation by We want
Thus,
The solution is
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