Solved on Dec 09, 2023

Solve the system of linear equations and evaluate the determinants $D$ , $D_x$ , and $D_y$ .
$6(x+y) = 9x + 6 \\ 6x = 12y - 12$
$D=\square \quad D_x=\square \quad D_y=\square$

STEP 1

Assumptions
1. We have a system of linear equations: $\begin{aligned} 6(x+y) & =9 x+6 \\ 6 x & =12 y-12 \end{aligned}$
2. We need to find the determinants $D$ , $D_x$ , and $D_y$ .
3. The determinant $D$ is calculated from the coefficients of $x$ and $y$ .
4. The determinant $D_x$ is calculated by replacing the coefficients of $x$ with the constant terms of the equations.
5. The determinant $D_y$ is calculated by replacing the coefficients of $y$ with the constant terms of the equations.

STEP 2

First, we need to write the system of equations in the standard form $Ax + By = C$ .
For the first equation: $6(x+y) = 9x + 6$ $6x + 6y = 9x + 6$ $-3x + 6y = 6$
For the second equation: $6x = 12y - 12$ $6x - 12y = -12$
Now, our system of equations in standard form is: $\begin{aligned} -3x + 6y & = 6 \\ 6x - 12y & = -12 \end{aligned}$

STEP 3

The coefficient matrix $A$ for the system of equations is: $A = \begin{bmatrix} -3 & 6 \\ 6 & -12 \\ \end{bmatrix}$

STEP 4

The determinant $D$ of the coefficient matrix $A$ is calculated as: $D = \begin{vmatrix} -3 & 6 \\ 6 & -12 \\ \end{vmatrix}$

STEP 5

Calculate the determinant $D$ using the formula for a 2x2 matrix: $D = (-3)(-12) - (6)(6)$

STEP 6

Perform the multiplication and subtraction to find $D$ : $D = 36 - 36$

STEP 7

Calculate the value of $D$ : $D = 0$

STEP 8

To find $D_x$ , we replace the coefficients of $x$ in the coefficient matrix with the constant terms of the equations: $D_x = \begin{vmatrix} 6 & 6 \\ -12 & -12 \\ \end{vmatrix}$

STEP 9

Calculate the determinant $D_x$ using the formula for a 2x2 matrix: $D_x = (6)(-12) - (6)(-12)$

STEP 10

Perform the multiplication and subtraction to find $D_x$ : $D_x = -72 + 72$

STEP 11

Calculate the value of $D_x$ : $D_x = 0$

STEP 12

To find $D_y$ , we replace the coefficients of $y$ in the coefficient matrix with the constant terms of the equations: $D_y = \begin{vmatrix} -3 & 6 \\ 6 & -12 \\ \end{vmatrix}$

STEP 13

Since $D_y$ is the same as the original determinant $D$ , we already know its value: $D_y = D = 0$

STEP 14

Now we have the values of all three determinants: $D = 0 \quad D_x = 0 \quad D_y = 0$
The determinants $D$ , $D_x$ , and $D_y$ are all zero.

Was this helpful?

Solve the system of linear equations and evaluate the determinants $D$ , $D_x$ , and $D_y$ .
$6(x+y) = 9x + 6 \\ 6x = 12y - 12$
$D=\square \quad D_x=\square \quad D_y=\square$

STEP 1

STEP 2

STEP 3

The coefficient matrix $A$ for the system of equations is: $A = \begin{bmatrix} -3 & 6 \\ 6 & -12 \\ \end{bmatrix}$

STEP 4

The determinant $D$ of the coefficient matrix $A$ is calculated as: $D = \begin{vmatrix} -3 & 6 \\ 6 & -12 \\ \end{vmatrix}$

STEP 5

Calculate the determinant $D$ using the formula for a 2x2 matrix: $D = (-3)(-12) - (6)(6)$

STEP 6

Perform the multiplication and subtraction to find $D$ : $D = 36 - 36$

STEP 7

Calculate the value of $D$ : $D = 0$

STEP 8

To find $D_x$ , we replace the coefficients of $x$ in the coefficient matrix with the constant terms of the equations: $D_x = \begin{vmatrix} 6 & 6 \\ -12 & -12 \\ \end{vmatrix}$

STEP 9

Calculate the determinant $D_x$ using the formula for a 2x2 matrix: $D_x = (6)(-12) - (6)(-12)$

STEP 10

Perform the multiplication and subtraction to find $D_x$ : $D_x = -72 + 72$

STEP 11

Calculate the value of $D_x$ : $D_x = 0$

STEP 12

To find $D_y$ , we replace the coefficients of $y$ in the coefficient matrix with the constant terms of the equations: $D_y = \begin{vmatrix} -3 & 6 \\ 6 & -12 \\ \end{vmatrix}$

STEP 13

Since $D_y$ is the same as the original determinant $D$ , we already know its value: $D_y = D = 0$

STEP 14

Now we have the values of all three determinants: $D = 0 \quad D_x = 0 \quad D_y = 0$
The determinants $D$ , $D_x$ , and $D_y$ are all zero.

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Solve the system of linear equations and evaluate the determinants DDD, DxD_xDx​, and DyD_yDy​.6(x+y)=9x+66x=12y−12 6(x+y) = 9x + 6 \\ 6x = 12y - 12 6(x+y)=9x+66x=12y−12D=□Dx=□Dy=□ D=\square \quad D_x=\square \quad D_y=\square D=□Dx​=□Dy​=□

STEP 1

STEP 2

STEP 3

The coefficient matrix AAA for the system of equations is: A=[−366−12] A = \begin{bmatrix} -3 & 6 \\ 6 & -12 \\ \end{bmatrix} A=[−36​6−12​]

STEP 4

The determinant DDD of the coefficient matrix AAA is calculated as: D=∣−366−12∣ D = \begin{vmatrix} -3 & 6 \\ 6 & -12 \\ \end{vmatrix} D=∣∣​−36​6−12​∣∣​

STEP 5

Calculate the determinant DDD using the formula for a 2x2 matrix: D=(−3)(−12)−(6)(6) D = (-3)(-12) - (6)(6) D=(−3)(−12)−(6)(6)

STEP 6

Perform the multiplication and subtraction to find DDD: D=36−36 D = 36 - 36 D=36−36

STEP 7

Calculate the value of DDD: D=0 D = 0 D=0

STEP 8

To find DxD_xDx​, we replace the coefficients of xxx in the coefficient matrix with the constant terms of the equations: Dx=∣66−12−12∣ D_x = \begin{vmatrix} 6 & 6 \\ -12 & -12 \\ \end{vmatrix} Dx​=∣∣​6−12​6−12​∣∣​

STEP 9

Calculate the determinant DxD_xDx​ using the formula for a 2x2 matrix: Dx=(6)(−12)−(6)(−12) D_x = (6)(-12) - (6)(-12) Dx​=(6)(−12)−(6)(−12)

STEP 10

Perform the multiplication and subtraction to find DxD_xDx​: Dx=−72+72 D_x = -72 + 72 Dx​=−72+72

STEP 11

Calculate the value of DxD_xDx​: Dx=0 D_x = 0 Dx​=0

STEP 12

To find DyD_yDy​, we replace the coefficients of yyy in the coefficient matrix with the constant terms of the equations: Dy=∣−366−12∣ D_y = \begin{vmatrix} -3 & 6 \\ 6 & -12 \\ \end{vmatrix} Dy​=∣∣​−36​6−12​∣∣​

STEP 13

Since DyD_yDy​ is the same as the original determinant DDD, we already know its value: Dy=D=0 D_y = D = 0 Dy​=D=0

STEP 14

Now we have the values of all three determinants: D=0Dx=0Dy=0 D = 0 \quad D_x = 0 \quad D_y = 0 D=0Dx​=0Dy​=0The determinants DDD, DxD_xDx​, and DyD_yDy​ are all zero.

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Solve the system of linear equations and evaluate the determinants $D$ , $D_x$ , and $D_y$ .
$6(x+y) = 9x + 6 \\ 6x = 12y - 12$
$D=\square \quad D_x=\square \quad D_y=\square$

The coefficient matrix $A$ for the system of equations is: $A = \begin{bmatrix} -3 & 6 \\ 6 & -12 \\ \end{bmatrix}$

The determinant $D$ of the coefficient matrix $A$ is calculated as: $D = \begin{vmatrix} -3 & 6 \\ 6 & -12 \\ \end{vmatrix}$

Calculate the determinant $D$ using the formula for a 2x2 matrix: $D = (-3)(-12) - (6)(6)$

Perform the multiplication and subtraction to find $D$ : $D = 36 - 36$

Calculate the value of $D$ : $D = 0$

To find $D_x$ , we replace the coefficients of $x$ in the coefficient matrix with the constant terms of the equations: $D_x = \begin{vmatrix} 6 & 6 \\ -12 & -12 \\ \end{vmatrix}$

Calculate the determinant $D_x$ using the formula for a 2x2 matrix: $D_x = (6)(-12) - (6)(-12)$

Perform the multiplication and subtraction to find $D_x$ : $D_x = -72 + 72$

Calculate the value of $D_x$ : $D_x = 0$

To find $D_y$ , we replace the coefficients of $y$ in the coefficient matrix with the constant terms of the equations: $D_y = \begin{vmatrix} -3 & 6 \\ 6 & -12 \\ \end{vmatrix}$

Since $D_y$ is the same as the original determinant $D$ , we already know its value: $D_y = D = 0$

Now we have the values of all three determinants: $D = 0 \quad D_x = 0 \quad D_y = 0$
The determinants $D$ , $D_x$ , and $D_y$ are all zero.