PrintDay 6: Solving a System of Equations using Matrices
Day 6: Solving a System of Equations Using Matrices
Suppose we flashback to chapter 3 and we have the following literal system of equations
{Ax + By = C
{Dx + Ey = F
Previously, we either solved this system by substitution or by elimination. Now, we have the power of matrices at our disposal to solve this system!
We can set up this system in the form of a matrix equation: AX = B, where A is a 2x2 matrix formed with the coefficients A, B, D, E (where A and B are the first row and D and E are the second row), X is a 2x1 column vector made of x and y, and B is a 2x1 column vector made of C and F.
The goal here will be to solve for the unknown quantities x and y. Since we cannot divide by a matrix, we must multiply both sides by the inverse of A. Essentially, we reduce to X = A-1B
NOTE: If matrix A is singular, it has no inverse and therefore the system has no solution
NOTE: If one row of A is a scalar multiple of another row of A, then there are infinitely many solutions to the system!
Suppose we flashback to chapter 3 and we have the following literal system of equations
{Ax + By = C
{Dx + Ey = F
Previously, we either solved this system by substitution or by elimination. Now, we have the power of matrices at our disposal to solve this system!
We can set up this system in the form of a matrix equation: AX = B, where A is a 2x2 matrix formed with the coefficients A, B, D, E (where A and B are the first row and D and E are the second row), X is a 2x1 column vector made of x and y, and B is a 2x1 column vector made of C and F.
The goal here will be to solve for the unknown quantities x and y. Since we cannot divide by a matrix, we must multiply both sides by the inverse of A. Essentially, we reduce to X = A-1B
NOTE: If matrix A is singular, it has no inverse and therefore the system has no solution
NOTE: If one row of A is a scalar multiple of another row of A, then there are infinitely many solutions to the system!
