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Solved on Feb 20, 2024

Find zz using Cramer's rule and a calculator for the system: 4x−y+5z=1,−x+2y+z=0,5x−2y+2z=−44x-y+5z=1, -x+2y+z=0, 5x-2y+2z=-4.

STEP 1

Assumptions
1. We have a system of linear equations: \begin{align*} 4x - y + 5z &= 1, \\ -x + 2y + z &= 0, \\ 5x - 2y + 2z &= -4. \end{align*}
2. We will use Cramer's rule to solve for z z .
3. Cramer's rule is applicable only if the determinant of the coefficient matrix (denoted as D D ) is non-zero.

STEP 2

Write down the coefficient matrix A A and the constant matrix B B for the system of equations: A=(4−15−1215−22),B=(10−4). A = \begin{pmatrix} 4 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & -2 & 2 \end{pmatrix}, \quad B = \begin{pmatrix} 1 \\ 0 \\ -4 \end{pmatrix}.

STEP 3

Calculate the determinant D D of the coefficient matrix A A : D=∣4−15−1215−22∣. D = \begin{vmatrix} 4 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & -2 & 2 \end{vmatrix}.

STEP 4

Expand the determinant D D using the rule of Sarrus or cofactor expansion. Here, we'll use cofactor expansion along the first row: D=4∣21−22∣−(−1)∣−1152∣+5∣−125−2∣. D = 4 \begin{vmatrix} 2 & 1 \\ -2 & 2 \end{vmatrix} - (-1) \begin{vmatrix} -1 & 1 \\ 5 & 2 \end{vmatrix} + 5 \begin{vmatrix} -1 & 2 \\ 5 & -2 \end{vmatrix}.

STEP 5

Calculate the determinants of the 2x2 matrices: D=4(2⋅2−(−2)⋅1)−(−1)(−1⋅2−5⋅1)+5(−1⋅(−2)−2⋅5). D = 4(2 \cdot 2 - (-2) \cdot 1) - (-1)(-1 \cdot 2 - 5 \cdot 1) + 5(-1 \cdot (-2) - 2 \cdot 5).

STEP 6

Compute the values: D=4(4+2)−(−1)(−2−5)+5(2−10). D = 4(4 + 2) - (-1)(-2 - 5) + 5(2 - 10).

STEP 7

Simplify the expression to find D D : D=4(6)−(−1)(−7)+5(−8). D = 4(6) - (-1)(-7) + 5(-8).

STEP 8

Finish the calculation of D D : D=24−7−40=−23. D = 24 - 7 - 40 = -23.

STEP 9

Since D≠0 D \neq 0 , Cramer's rule can be applied. Now, we need to find the determinant Dz D_z which is obtained by replacing the third column of A A with the constant matrix B B .

STEP 10

Write down the modified matrix Az A_z for Dz D_z : Az=(4−11−1205−2−4). A_z = \begin{pmatrix} 4 & -1 & 1 \\ -1 & 2 & 0 \\ 5 & -2 & -4 \end{pmatrix}.

STEP 11

Calculate the determinant Dz D_z of the matrix Az A_z : Dz=∣4−11−1205−2−4∣. D_z = \begin{vmatrix} 4 & -1 & 1 \\ -1 & 2 & 0 \\ 5 & -2 & -4 \end{vmatrix}.

STEP 12

Expand the determinant Dz D_z using cofactor expansion along the third column: Dz=1∣−125−2∣−0∣4−15−2∣−4∣4−1−12∣. D_z = 1 \begin{vmatrix} -1 & 2 \\ 5 & -2 \end{vmatrix} - 0 \begin{vmatrix} 4 & -1 \\ 5 & -2 \end{vmatrix} - 4 \begin{vmatrix} 4 & -1 \\ -1 & 2 \end{vmatrix}.

STEP 13

Calculate the determinants of the 2x2 matrices: Dz=1((−1)⋅(−2)−2⋅5)−0+(−4)(4⋅2−(−1)⋅(−1)). D_z = 1((-1) \cdot (-2) - 2 \cdot 5) - 0 + (-4)(4 \cdot 2 - (-1) \cdot (-1)).

STEP 14

Compute the values: Dz=1(2−10)−4(8−1). D_z = 1(2 - 10) - 4(8 - 1).

STEP 15

Simplify the expression to find Dz D_z : Dz=1(−8)−4(7). D_z = 1(-8) - 4(7).

STEP 16

Finish the calculation of Dz D_z : Dz=−8−28=−36. D_z = -8 - 28 = -36.

STEP 17

Now, apply Cramer's rule to find the value of z z : z=DzD. z = \frac{D_z}{D}.

STEP 18

Plug in the values for Dz D_z and D D to calculate z z : z=−36−23. z = \frac{-36}{-23}.

STEP 19

Simplify the fraction to get the value of z z : z=3623. z = \frac{36}{23}.
The value of z z that satisfies the system of linear equations is 3623 \frac{36}{23} .

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