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Solved on Nov 05, 2023

Use Cramer's rule to find the value of yy that satisfies the given system of 3 linear equations in 3 variables.

STEP 1

Assumptions1. We are given a system of linear equations. . We are asked to find the value of yy that satisfies the system.
3. We are to use Cramer's rule to solve for yy.
4. The ALEKS graphing calculator can be used to make computations easier.

STEP 2

First, we need to write down the system of equations in matrix form. The system of equations is2x−5y+z=−25x−2y+4z=−20x+4y+5z=−4\begin{array}{l} 2 x-5 y+z=-2 \\ 5 x-2 y+4 z=-2 \\ 0 x+4 y+5 z=-4\end{array} This can be written in matrix form as[2−515−24045][xyz]=[−2−2−4]\begin{bmatrix} 2 & -5 &1 \\ 5 & -2 &4 \\ 0 &4 &5\end{bmatrix} \begin{bmatrix} x \\ y \\ z\end{bmatrix} = \begin{bmatrix} -2 \\ -2 \\ -4\end{bmatrix}

STEP 3

To use Cramer's rule, we first need to calculate the determinant of the coefficient matrix. The coefficient matrix isA=[2−515−205]A = \begin{bmatrix} 2 & -5 &1 \\ 5 & -2 & \\ 0 & &5\end{bmatrix} The determinant of a3x3 matrix is calculated as followsdet(A)=a(ei−fh)−b(di−fg)+c(dh−eg)\text{det}(A) = a(ei−fh)−b(di−fg)+c(dh−eg) where a, b, c, d, e, f, g, h, i are the elements of the matrix.

STEP 4

Substitute the values from the coefficient matrix into the determinant formuladet(A)=2((−2∗)−4∗4)−(−)((∗)−(0∗4))+1((∗4)−(0∗−2))\text{det}(A) =2((-2*)-4*4) - (-)((*)-(0*4)) +1((*4)-(0*-2))

STEP 5

Calculate the determinant of the coefficient matrixdet(A)=2∗(−10−16)−(−5)∗(25−0)+1∗(20−0)=−52+125+20=93\text{det}(A) =2*(-10-16) - (-5)*(25-0) +1*(20-0) = -52 +125 +20 =93

STEP 6

Next, we need to calculate the determinant of the matrix obtained by replacing the second column of the coefficient matrix (which corresponds to yy) with the column matrix of constants.
Ay=[2−215−240−45]A_y = \begin{bmatrix} 2 & -2 &1 \\ 5 & -2 &4 \\ 0 & -4 &5\end{bmatrix}

STEP 7

Calculate the determinant of AyA_y using the same formula as beforedet(Ay)=2((−2∗5)−4∗−4)−(−2)((5∗5)−(0∗−4))+1((5∗−4)−(0∗−2))\text{det}(A_y) =2((-2*5)-4*-4) - (-2)((5*5)-(0*-4)) +1((5*-4)-(0*-2))

STEP 8

Calculate the determinant of AyA_ydet(Ay)=2∗(−10−(−16))−(−2)∗(25−0)+1∗(−20−0)=12+50−20=42\text{det}(A_y) =2*(-10-(-16)) - (-2)*(25-0) +1*(-20-0) =12 +50 -20 =42

STEP 9

Finally, we can find the value of yy using Cramer's rule. Cramer's rule states that the value of a variable is the ratio of the determinant of the matrix obtained by replacing the corresponding column of the coefficient matrix with the column matrix of constants to the determinant of the coefficient matrix. Therefore, yy is given byy=det(Ay)det(A)y = \frac{\text{det}(A_y)}{\text{det}(A)}

STEP 10

Substitute the values of det(Ay)\text{det}(A_y) and det(A)\text{det}(A) into the formulay=4293y = \frac{42}{93} So, the value of yy that satisfies the system of equations is 4293\frac{42}{93}.

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