# § 3.4 Matrix Solutions to Linear Systems.

## Presentation on theme: "§ 3.4 Matrix Solutions to Linear Systems."— Presentation transcript:

§ 3.4 Matrix Solutions to Linear Systems

Solving Systems Using Matrices
The following array of numbers arranged in rows and columns and placed in brackets is an example of a matrix. The numbers inside the brackets are called elements of the matrix. Matrices are used to display information and can be used to solve systems of equations. A matrix gives us a shortened way of writing a system of equations. Blitzer, Intermediate Algebra, 5e – Slide #2 Section 3.4

Solving Systems Using Matrices
This matrix represents a system of three equations that are listed below. Equations represented by matrix above 2x -3y + z = 5 x + 3y + 8z = 22 3x – y + 2z = 12 Blitzer, Intermediate Algebra, 5e – Slide #3 Section 3.4

Solving Systems Using Matrices
Matrix Row Operations 1) Two rows of a matrix may be interchanged. This is the same as interchanging two equations in the linear system. 2) The elements in any row may be multiplied by a nonzero number. This is the same as multiplying both sides of an equation by a nonzero number. 3) The elements in any row may be multiplied by a nonzero number, and these products may be added to the corresponding elements in any other row. This is the same as multiplying an equation by a nonzero number and then adding equations to eliminate a variable (addition method). Two matrices are row equivalent if one can be obtained from the other by a sequence of the above row operations. Blitzer, Intermediate Algebra, 5e – Slide #4 Section 3.4

Solving Systems Using Matrices
EXAMPLE Use the matrix And perform each indicated row operation: (a) -2R (b) -3R2 + R R3 Blitzer, Intermediate Algebra, 5e – Slide #5 Section 3.4

Solving Systems Using Matrices
CONTINUED SOLUTION (a) First, -2R1 means that we need to multiply everything in row #1 by -2. (b) -3R2 + R R3 means that we need to multiply row #2 by -3, add that result to row #3 and then the result of that addition is the new row #3. Blitzer, Intermediate Algebra, 5e – Slide #6 Section 3.4

Solving Systems Using Matrices
CONTINUED -3R2 = -3[ ] = [(-3)1 (-3)3 (-3)8 (-3)22] = [ ] ADD: R3 = [ ] Therefore, my new row #3 is: R3 = [ ] So, the matrix rewritten with its new row #3 is: Blitzer, Intermediate Algebra, 5e – Slide #7 Section 3.4

Solving Linear Systems Using Matrices
Solving Systems Using Matrices Solving Linear Systems Using Matrices 1) Write the augmented matrix for the system. 2) Use matrix row operations to simplify the matrix to one with 1’s down the main diagonal from upper left to lower right, and 0’s below the 1’s. 3) Write the system of linear equations corresponding to the matrix in step 2 and use back-substitution to find the system’s solution. Blitzer, Intermediate Algebra, 5e – Slide #8 Section 3.4

Solving Systems Using Matrices
EXAMPLE Use matrices to solve the system: 2x - 3y + z = 5 x + 3y + 8z = 22 3x - y + 2z = 12 SOLUTION 1) Write the augmented matrix for the system. 2) Use matrix row operations to simplify the matrix to one with 1’s down the main diagonal from upper left to lower right, and 0’s below the 1’s. Blitzer, Intermediate Algebra, 5e – Slide #9 Section 3.4

Solving Systems Using Matrices
CONTINUED Next we will interchange row #1 and row #2 (to facilitate calculations). Now we will do -2R1 + R2 R2 (that is, I will multiply row #1 by -2, add that to row #2, and that will be my new row #2). -2R1 = -2[ ] = [(-2)1 (-2)3 (-2)8 (-2)22] = [ ] ADD: R2 = [ ] Therefore, my new row #2 is: R2 = [ ] Blitzer, Intermediate Algebra, 5e – Slide #10 Section 3.4

Solving Systems Using Matrices
CONTINUED So, the augmented matrix now looks like NOTE: At this point, we should have a zero in the second row, first column. Now we will do -3R1 + R3 R3 (that is, I will multiply row #1 by -3, add that to row #3, and that will be my new row #3). -3R1 = -3[ ] = [(-3)1 (-3)3 (-3)8 (-3)22] = [ ] ADD: R3 = [ ] Therefore, my new row #3 is: R3 = [ ] Blitzer, Intermediate Algebra, 5e – Slide #11 Section 3.4

Solving Systems Using Matrices
CONTINUED So, the augmented matrix now looks like NOTE: At this point, we should have a zero in the third row, first column. Now we will do -10R2 + 9R3 R3. -10R2 = -10[ ] = [ ] ADD: 9R3 = 9[ ] = [ ] Therefore, my new row #3 is: R3 = [ ] Blitzer, Intermediate Algebra, 5e – Slide #12 Section 3.4

Solving Systems Using Matrices
CONTINUED So, the augmented matrix now looks like NOTE: At this point, we should have a zero in the third row, second column. Now I will divide R2 by -9 and divide R3 by -48 so that the first number in each column is a 1. The new row #2 is: (-1/9)R2 = (-1/9)[ ] = [ ] The new row #3 is: (-1/48)R3 = (-1/48)[ ] = [ ] Blitzer, Intermediate Algebra, 5e – Slide #13 Section 3.4

Solving Systems Using Matrices
CONTINUED So, the augmented matrix now looks like NOTE: At this point, we should have a 1 in each element of the main diagonal. 3) Write the system of linear equations corresponding to the matrix in step 2 and use back-substitution to find the system’s solution. So the system of equations becomes: x + 3y + 8z = 22 y + 1.7z = 4.3 z = 2 Blitzer, Intermediate Algebra, 5e – Slide #14 Section 3.4

Solving Systems Using Matrices
CONTINUED Therefore, z = 2. Now I replace z with 2 into the second equation to determine y. y + 1.7z = 4.3 y + 1.7(2) = 4.3 Replace z with 2 y = 4.3 Multiply y = 0.9 Subtract 3.4 from both sides Blitzer, Intermediate Algebra, 5e – Slide #15 Section 3.4

Solving Systems Using Matrices
CONTINUED Now solve for z using the first equation. x + 3y + 8z = 22 x + 3(0.9) + 8(2) = 22 Replace z with 2 and y with 0.9 x = 22 Multiply x = 22 Add x = 3.3 Subtract 18.7 from both sides Therefore the solution to the system is (3.3,0.9,2). Blitzer, Intermediate Algebra, 5e – Slide #16 Section 3.4

Download ppt "§ 3.4 Matrix Solutions to Linear Systems."

Similar presentations