Use Cramer’s rule to solve each system of equations.10.3a−5b+2c=−54a+b+3c=92a−c=1

Q10. - Algebra 2 | StudyQuestionHub

Q10.

Question

Use Cramer’s rule to solve each system of equations.

10.

$\begin{array}{l} 3 a - 5 b + 2 c = - 5 \\ 4 a + b + 3 c = 9 \\ 2 a - c = 1 \end{array}$

Step-by-Step Solution

Verified

Answer

The solution of the given system of equations is $(1, 2, 1)$ .

1Step 1 - Description of step.

The solution of the system of linear equations in three variables:

$\begin{array}{l} a_{1} a + b_{1} b + c_{1} c = d_{1} \\ a_{2} a + b_{2} b + c_{2} c = d_{2} \\ a_{3} a + b_{3} b + c_{3} c = d_{3} \end{array}$

by Cramer’s rule is given by $(a, b, c)$ where

$a = \frac{|\begin{matrix} d_{1} & b_{1} & c_{1} \\ d_{2} & b_{2} & c_{2} \\ d_{3} & b_{3} & c_{3} \end{matrix}|}{|\begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix}|}$ $b = \frac{|\begin{matrix} a_{1} & d_{1} & c_{1} \\ a_{2} & d_{2} & c_{2} \\ a_{3} & d_{3} & c_{3} \end{matrix}|}{|\begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix}|}$ $c = \frac{|\begin{matrix} a_{1} & b_{1} & d_{1} \\ a_{2} & b_{2} & d_{2} \\ a_{3} & b_{3} & d_{3} \end{matrix}|}{|\begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix}|}$ and $|\begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix}| \neq 0$

2Step 2 - Description of step.

The given system of linear equations in three variables is:

$\begin{matrix} 3 a - 5 b + 2 c = - 5 \\ 4 a + b + 3 c = 9 \\ 2 a - c = 1 \end{matrix}$

Therefore, by comparing the given system of linear equations with the

system of linear equations $\begin{array}{l} a_{1} a + b_{1} b + c_{1} c = d_{1} \\ a_{2} a + b_{2} b + c_{2} c = d_{2} \\ a_{3} a + b_{3} b + c_{3} c = d_{3} \end{array}$ it can be obtained that:

$a_{1} = 3$ , $b_{1} = - 5$ , $c_{1} = 2$ , $d_{1} = - 5$ , $a_{2} = 4$ , $b_{2} = 1$ , $c_{2} = 3$ , $d_{2} = 9$ , $a_{3} = 2$ , $b_{3} = 0$ , $c_{3} = - 1$ and $d_{3} = 1$ .

3Step 3 - Find the values of a, b and c.

The value of $a$ is given by:

$\begin{matrix} a = \frac{|\begin{matrix} - 5 & - 5 & 2 \\ 9 & 1 & 3 \\ 1 & 0 & - 1 \end{matrix}|}{|\begin{matrix} 3 & - 5 & 2 \\ 4 & 1 & 3 \\ 2 & 0 & - 1 \end{matrix}|} \\ = \frac{- 5 |\begin{matrix} 1 & 3 \\ 0 & - 1 \end{matrix}| - (- 5) |\begin{matrix} 9 & 3 \\ 1 & - 1 \end{matrix}| + 2 |\begin{matrix} 9 & 1 \\ 1 & 0 \end{matrix}|}{3 |\begin{matrix} 1 & 3 \\ 0 & - 1 \end{matrix}| - (- 5) |\begin{matrix} 4 & 3 \\ 2 & - 1 \end{matrix}| + 2 |\begin{matrix} 4 & 1 \\ 2 & 0 \end{matrix}|} \\ = \frac{- 5 (- 1 - 0) + 5 (- 9 - 3) + 2 (0 - 1)}{3 (- 1 - 0) + 5 (- 4 - 6) + 2 (0 - 2)} \\ = \frac{5 - 60 - 2}{- 3 - 50 - 4} \\ = \frac{- 57}{- 57} \\ = 1 \end{matrix}$

The value of $b$ is given by:

$\begin{matrix} b = \frac{|\begin{matrix} 3 & - 5 & 2 \\ 4 & 9 & 3 \\ 2 & 1 & - 1 \end{matrix}|}{|\begin{matrix} 3 & - 5 & 2 \\ 4 & 1 & 3 \\ 2 & 0 & - 1 \end{matrix}|} \\ = \frac{3 |\begin{matrix} 9 & 3 \\ 1 & - 1 \end{matrix}| - (- 5) |\begin{matrix} 4 & 3 \\ 2 & - 1 \end{matrix}| + 2 |\begin{matrix} 4 & 9 \\ 2 & 1 \end{matrix}|}{3 |\begin{matrix} 1 & 3 \\ 0 & - 1 \end{matrix}| - (- 5) |\begin{matrix} 4 & 3 \\ 2 & - 1 \end{matrix}| + 2 |\begin{matrix} 4 & 1 \\ 2 & 0 \end{matrix}|} \\ = \frac{3 (- 9 - 3) + 5 (- 4 - 6) + 2 (4 - 18)}{3 (- 1 - 0) + 5 (- 4 - 6) + 2 (0 - 2)} \\ = \frac{- 36 - 50 - 28}{- 3 - 50 - 4} \\ = \frac{- 114}{- 57} \\ = 2 \end{matrix}$

The value of $c$ is given by:

$\begin{matrix} c = \frac{|\begin{matrix} 3 & - 5 & - 5 \\ 4 & 1 & 9 \\ 2 & 0 & 1 \end{matrix}|}{|\begin{matrix} 3 & - 5 & 2 \\ 4 & 1 & 3 \\ 2 & 0 & - 1 \end{matrix}|} \\ = \frac{3 |\begin{matrix} 1 & 9 \\ 0 & 1 \end{matrix}| - (- 5) |\begin{matrix} 4 & 9 \\ 2 & 1 \end{matrix}| + (- 5) |\begin{matrix} 4 & 1 \\ 2 & 0 \end{matrix}|}{3 |\begin{matrix} 1 & 3 \\ 0 & - 1 \end{matrix}| - (- 5) |\begin{matrix} 4 & 3 \\ 2 & - 1 \end{matrix}| + 2 |\begin{matrix} 4 & 1 \\ 2 & 0 \end{matrix}|} \\ = \frac{3 (1 - 0) + 5 (4 - 18) - 5 (0 - 2)}{3 (- 1 - 0) + 5 (- 4 - 6) + 2 (0 - 2)} \\ = \frac{3 - 70 + 10}{- 3 - 50 - 4} \\ = \frac{- 57}{- 57} \\ = 1 \end{matrix}$

The values of $a$ , $b$ and $c$ are 1, 2 and 1 respectively.

4Step 4 - Description of step.

The solution of the given system of equations is $(1, 2, 1)$ .

Q9.

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