$$ \left|\begin{array}{ccc}(5-x) & 7 & -5 \ 0 & (4-x) & -1 \ 2 & 8 & (-3-x)\end{array} ight| $$ Cramer's Rule to solve for the letter(s) indicated within brackets in each of following systems of equations: $$ \begin{array}{l}2 x+y+z=0 ; 3 x-3 z=5+2 y ; 8 x-11=-2 y-5 z \ldots(y) \quad[-4.36] \ y+z=5-x ; x-4 z-10-y=0 ;-4 x=y-z \ldots(z) \ 4 i_{1}+i_{2}-i_{3}=1 ; 2 i_{1}=i_{2}+5 i_{3}-2 ; 7 i_{2}+i_{1}-i_{3}=9 \ldots\left(i_{2} ight) \ 3 x-2 y+z=5 ; 2(x-3)=4 z+y ; 5 z-1=7-x \ldots(y) \ 2\left(i_{3}-i_{2} ight)+5\left(i_{3}-i_{1} ight)=24 ;\left(i_{2}-i_{3} ight)+2 i_{2}+\left(i_{2}-i_{1} ight)=0 ; \ 5\left(i_{1}-i_{3} ight)+2\left(i_{1}-i_{2} ight)+i_{1}=6 \ldots\left(i_{3} ight) \ 2 x-y \ 4 \ 2 \ 2 \ 2\end{array} \quad \frac{x}{3}-2 y=5-z ; x+3 z=y-1 \ldots(y) $$ $$ T_{1}-3 T_{3}=T_{2} \sin 30^{\circ}-7 ; \quad T_{2}+3 T_{1}=5-T_{3} ; $$ $$ 3\left(T_{1}-T_{2} ight)-1=T_{1}+T_{3} \ldots\left(T_{2} ight) $$

Question

$$ \left|\begin{array}{ccc}(5-x) & 7 & -5 \ 0 & (4-x) & -1 \ 2 & 8 & (-3-x)\end{array}ight| $$ Cramer's Rule to solve for the letter(s) indicated within brackets in each of following systems of equations: $$ \begin{array}{l}2 x+y+z=0 ; 3 x-3 z=5+2 y ; 8 x-11=-2 y-5 z \ldots(y) \quad[-4.36] \ y+z=5-x ; x-4 z-10-y=0 ;-4 x=y-z \ldots(z) \ 4 i_{1}+i_{2}-i_{3}=1 ; 2 i_{1}=i_{2}+5 i_{3}-2 ; 7 i_{2}+i_{1}-i_{3}=9 \ldots\left(i_{2}ight) \ 3 x-2 y+z=5 ; 2(x-3)=4 z+y ; 5 z-1=7-x \ldots(y) \ 2\left(i_{3}-i_{2}ight)+5\left(i_{3}-i_{1}ight)=24 ;\left(i_{2}-i_{3}ight)+2 i_{2}+\left(i_{2}-i_{1}ight)=0 ; \ 5\left(i_{1}-i_{3}ight)+2\left(i_{1}-i_{2}ight)+i_{1}=6 \ldots\left(i_{3}ight) \ 2 x-y \ 4 \ 2 \ 2 \ 2\end{array} \quad \frac{x}{3}-2 y=5-z ; x+3 z=y-1 \ldots(y) $$ $$ T_{1}-3 T_{3}=T_{2} \sin 30^{\circ}-7 ; \quad T_{2}+3 T_{1}=5-T_{3} ; $$ $$ 3\left(T_{1}-T_{2}ight)-1=T_{1}+T_{3} \ldots\left(T_{2}ight) $$

UpStudy AI · Accepted Answer

To solve the given problems using Cramer's Rule, we will first extract the systems of equations and then set up the corresponding matrices. We will then calculate the determinants needed for Cramer's Rule.

### Problem 1: Solve for $$ y $$

The system of equations is:
1. $$ 2x + y + z = 0 $$
2. $$ 3x - 3z = 5 + 2y $$
3. $$ 8x - 11 = -2y - 5z $$

Rearranging the second and third equations:
- $$ 3x - 2y + 3z = 5 $$ (from the second equation)
- $$ 8x + 2y + 5z = 11 $$ (from the third equation)

Now, we can write the system in matrix form $$ A \mathbf{x} = \mathbf{b} $$:
$$
\begin{bmatrix}
2 & 1 & 1 \
3 & -2 & 3 \
8 & 2 & 5
\end{bmatrix}
\begin{bmatrix}
x \
y \
z
\end{bmatrix}
=
\begin{bmatrix}
0 \
5 \
11
\end{bmatrix}
$$

Next, we will calculate the determinant of the coefficient matrix $$ A $$ and the determinant of the matrix $$ A_y $$ (where the second column is replaced by the constants).

### Step 1: Calculate $$ |A| $$

$$
|A| = \left|\begin{array}{ccc}
2 & 1 & 1 \
3 & -2 & 3 \
8 & 2 & 5
\end{array}\right|
$$

### Step 2: Calculate $$ |A_y| $$

Replace the second column of $$ A $$ with the constants:
$$
|A_y| = \left|\begin{array}{ccc}
2 & 0 & 1 \
3 & 5 & 3 \
8 & 11 & 5
\end{array}\right|
$$

Now, let's calculate both determinants.

### Problem 1 Determinants Calculation

We will calculate both determinants in parallel.

### Problem 2: Solve for $$ z $$

The system of equations is:
1. $$ y + z = 5 - x $$
2. $$ x - 4z - 10 - y = 0 $$
3. $$ -4x = y - z $$

Rearranging the equations:
- $$ x + y + z = 5 $$ (from the first equation)
- $$ x - y - 4z = 10 $$ (from the second equation)
- $$ 4x + y + z = 0 $$ (from the third equation)

Now, we can write the system in matrix form:
$$
\begin{bmatrix}
1 & 1 & 1 \
1 & -1 & -4 \
4 & 1 & 1
\end{bmatrix}
\begin{bmatrix}
x \
y \
z
\end{bmatrix}
=
\begin{bmatrix}
5 \
10 \
0
\end{bmatrix}
$$

### Step 1: Calculate $$ |A| $$

$$
|A| = \left|\begin{array}{ccc}
1 & 1 & 1 \
1 & -1 & -4 \
4 & 1 & 1
\end{array}\right|
$$

### Step 2: Calculate $$ |A_z| $$

Replace the third column of $$ A $$ with the constants:
$$
|A_z| = \left|\begin{array}{ccc}
1 & 1 & 5 \
1 & -1 & 10 \
4 & 1 & 0
\end{array}\right|
$$

### Problem 2 Determinants Calculation

We will calculate both determinants in parallel.

### Problem 3: Solve for $$ i_2 $$

The system of equations is:
1. $$ 4i_1 + i_2 - i_3 = 1 $$
2. $$ 2i_1 = i_2 + 5i_3 - 2 $$
3. $$ 7i_2 + i_1 - i_3 = 9 $$

Rearranging the equations:
- $$ 4i_1 + i_2 - i_3 = 1 $$
- $$ 2i_1 - i_2 - 5i_3 = -2 $$
- $$ i_1 + 7i_2 - i_3 = 9 $$

Now, we can write the system in matrix form:
$$
\begin{bmatrix}
4 & 1 & -1 \
2 & -1 & -5 \
1 & 7 & -1
\end{bmatrix}
\begin{bmatrix}
i_1 \
i_2 \
i_3
\end{bmatrix}
=
\begin{bmatrix}
1 \
-2 \
9
\end{bmatrix}
$$

### Step 1: Calculate $$ |A| $$

$$
|A| = \left|\begin{array}{ccc}
4 & 1 & -1 \
2 & -1 & -5 \
1 & 7 & -1
\end{array}\right|
$$

### Step 2: Calculate $$ |A_{i_2}| $$

Replace the second column of $$ A $$ with the constants:
$$
|A_{i_2}| = \left|\begin{array}{ccc}
4 & 1 & -1 \
2 & -2 & -5 \
1 & 9 & -1
\end{array}\right|
$$

### Problem 3 Determinants Calculation

We will calculate both determinants in parallel.

### Problem 4: Solve for $$ y $$

The system of equations is:
1. $$ 3x - 2y + z = 5 $$
2. $$ 2(x - 3) = 4z + y $$
3. $$ 5z - 1 = 7 - x $$

Rearranging the equations:
- $$ 3x - 2y + z = 5 $$
- $$ 2x - 6 - 4z - y = 0 $$
- $$ x + 5z = 8 $$

Now, we can write the system in matrix form:
$$
\begin{bmatrix}
3 & -2 & 1 \
2 & -1 & -4 \
1 & 0 & 5
\end{bmatrix}
\begin{bmatrix}
x \
y \
z
\end{bmatrix}
=
\begin{bmatrix}
5 \
6 \
8
\end{bmatrix}
$$

### Step 1: Calculate $$ |A| $$

\[
|A| = \
To solve the given systems of equations using Cramer's Rule, we need to calculate the determinants of the coefficient matrices and the corresponding matrices where the constants replace the columns. Here's a simplified summary of the steps for each problem: 1. **Problem 1: Solve for $$ y $$** - **System of Equations:** 1. $$ 2x + y + z = 0 $$ 2. $$ 3x - 2y + 3z = 5 $$ 3. $$ 8x + 2y + 5z = 11 $$ - **Steps:** - Calculate the determinant of the coefficient matrix $$ |A| $$. - Replace the second column of $$ A $$ with the constants to find $$ |A_y| $$. - Use Cramer's Rule: $$ y = \frac{|A_y|}{|A|} $$. 2. **Problem 2: Solve for $$ z $$** - **System of Equations:** 1. $$ y + z = 5 - x $$ 2. $$ x - 4z - 10 - y = 0 $$ 3. $$ -4x = y - z $$ - **Steps:** - Rearrange equations to standard form. - Calculate $$ |A| $$ and $$ |A_z| $$ by replacing the third column with constants. - Apply Cramer's Rule: $$ z = \frac{|A_z|}{|A|} $$. 3. **Problem 3: Solve for $$ i_2 $$** - **System of Equations:** 1. $$ 4i_1 + i_2 - i_3 = 1 $$ 2. $$ 2i_1 - i_2 - 5i_3 = -2 $$ 3. $$ i_1 + 7i_2 - i_3 = 9 $$ - **Steps:** - Write the system in matrix form. - Calculate $$ |A| $$ and $$ |A_{i_2}| $$ by replacing the second column with constants. - Use Cramer's Rule: $$ i_2 = \frac{|A_{i_2}|}{|A|} $$. 4. **Problem 4: Solve for $$ y $$** - **System of Equations:** 1. $$ 3x - 2y + z = 5 $$ 2. $$ 2x - 6 - 4z - y = 0 $$ 3. $$ x + 5z = 8 $$ - **Steps:** - Rearrange equations to standard form. - Calculate $$ |A| $$ and $$ |A_y| $$ by replacing the second column with constants. - Apply Cramer's Rule: $$ y = \frac{|A_y|}{|A|} $$. **Summary:** For each system, set up the coefficient matrix $$ A $$ and the matrix $$ A_{\text{variable}} $$ by replacing the column corresponding to the variable you're solving for with the constants. Calculate the determinants $$ |A| $$ and $$ |A_{\text{variable}}| $$, then apply Cramer's Rule to find the value of the variable.

Cramer’s Rule to solve for the letter(s) indicated within brackets in each of
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