Use matrices to solve systems with unique solutions

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Tutorial 2.19 Use matrices to solve systems with unique solutions

We can use the Gauss-Jordan Row Operations method to solve systems of equations by:

1. Rewriting the given system into an augmented matrix.

2. Perform one or more of these row operations…

a) Interchange (swap) any two rows.

b) Multiply a row by a non-zero constant (any positive or negative rational number).

c) Multiply any row by a non-zero constant and then add the row to another row.

3. Use the reduced matrix to determine the solutions for the given system.

Remember that the goal of this process is to transform an augmented matrix into a “reduced matrix”

that displays an identity matrix to the left of the vertical line and the proper solutions of the system to the right.

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To that end (in a 3 × 3 coefficient matrix), we need to address these steps:

& & & &

5 ᬚ

& & & &

5 ᬛ

5 ᬜ

& & & &

5 ᬝ

GLDJUDP GLDJUDP GLDJUDP

& & & &

5 ᬞ

5 ᬟ

& & & &

5 ᬠ

& & & &

5 ᬡ

5 ᬢ

GLDJUDP GLDJUDP GLDJUDP

❶ Apply a permissible row operation to fill this position (in diagram #1) with a one.

❷ Apply a permissible row operation to fill this position (in diagram #2) with a zero.

❸ Apply a permissible row operation to fill this position (in diagram #2) with a zero.

❹ Apply a permissible row operation to fill this position (in diagram #3) with a one.

❺ Apply a permissible row operation to fill this position (in diagram #4) with a zero.

❻ Apply a permissible row operation to fill this position (in diagram #4) with a zero.

❼ Apply a permissible row operation to fill this position (in diagram #5) with a one.

❽ Apply a permissible row operation to fill this position (in diagram #6) with a zero.

❾ Apply a permissible row operation to fill this position (in diagram #6) with a zero.

NOTE: If a larger number of variables is found within the given system, just continue the pattern started in these steps.

example 2.19a Solve using the Gauss-Jordan Row Operations:

3 4 10

2 2 15

3 4 2 10

x y z

x y z

x y z

− + =

 + + =

 − + =



Follow these steps:

1. Rewrite the given system into an augmented matrix:

1 -3 4 10 2 2 1 15 3 -4 2 10

 

 

 

2. Perform permissible row operations in order to achieve the desired “reduced matrix”:

& & & &

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5 ±

2a. Use the “1” in the R₁C₁ position to convert the other values in C₁ into zeroes:

5 o 5 ± 5±5 o 5 ± ± 5±5 o 5 ± ± ROG5

±5 ± ±± ± ±

QHZ5 ± ± ROG5 ±

±5 ± ±± ± ±

QHZ5 ± ±

2b. Swap R3 and R2 around (as seen below):

5 o 5 ± 5 o 5 ± ± 5 o 5 ± ±

2c. Convert the “5” in the R₂C₂ position (above) into a “1” by dividing all of R₂ by 5:

5 o 5 ± 5y o 5 ± ± 5 o 5 ± ±

ROG5 ± ±

QHZ5

143

2d. Use the “1” in the R₂C₂ position to convert the other values in C₂ into zeroes:

55 o 5 ± ± 5 o 5 ± ± 5±5 o 5 ROG5 ±

5 ± ±

QHZ5 ± ± ROG5 ± ±

±5 ± ± ±± ±±

QHZ5

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2e. Convert the “9” in the R₃C₃ position (above) into a “1” by dividing all of R₃ by 9:

5 o 5 ± ± 5 o 5 ± ± 5y o 5

ROG5 ± ± QHZ5

2f. Use the “1” in the R₃C₃ position to convert the other values in C₃ into zeroes:

55 5 55 5 5 5 ROG5 ± ±

QHZ5 ROG5 ± ±

QHZ5

3. Read the solutions from:

1 0 0 4 0 1 0 2 0 0 1 3

 

 

 

→ x = 4, y = 2, z = 3

Following these same basic steps, we can solve most systems no matter how many variables are contained in the given system.

145

example 2.19b Solve using the Gauss-Jordan Row Operations:

2 6 12 82 3 10 20 135 2 17 129

x y z

x z

− − =

 − − =

 − =



Now that the row operations have achieved an identity matrix in place of the coefficients, we can read the solutions for the give system: x = 5, y = 2, z = –7

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5 o 5 MXVWFRS\HQWULHVIURP5

5±5 o 5 JHWD]HURLQ5&SRVLWLRQ

5 o 5 MXVWFRS\HQWULHVIURP5

±5 o 5 JHWDLQ5&SRVLWLRQ

5 o 5 MXVWFRS\HQWULHVIURP5

55 o 5 JHWD]HURLQ5&SRVLWLRQ

5 o 5 MXVWFRS\HQWULHVIURP5

5±5 o 5 JHWD]HURLQ5&SRVLWLRQ

5 o 5 MXVWFRS\HQWULHVIURP5

5y± o 5 JHWDLQ5&SRVLWLRQ

5 o 5 JHWD]HURLQ5&SRVLWLRQ

5±5 o 5 JHWD]HURLQ5&SRVLWLRQ

5 o 5 MXVWFRS\HQWULHVIURP5

example 2.19c Solve using the Gauss-Jordan Row Operations:

4 6 3 4

2 16 12 6

w x y z

w x y z

w y z

x y z

+ − − = −

 − − + =

 + + =

 + + =

 & & & & &

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5 ± ± ±

5 ± ±