Use Cramer's Rule to solve (if possible) the system of linear equations. (If not possible, enter IMPOSSIBLE.)
4x1	 − 	x2	 + 	x3	 = 	−5
2x1	 + 	2x2	 + 	3x3	 = 	10
5x1	 − 	2x2	 + 	6x3	 = 	12


To solve the given system of linear equations using **Cramer's Rule**, we follow these steps:

### Step 1: Write the system of equations in matrix form
The system of linear equations is:

\[
\begin{aligned}
4x_1 - x_2 + x_3 &= -5 \\
2x_1 + 2x_2 + 3x_3 &= 10 \\
5x_1 - 2x_2 + 6x_3 &= 12
\end{aligned}
\]

This can be written as a matrix equation \( A \mathbf{x} = \mathbf{b} \), where:

\[
A = \begin{pmatrix}
4 & -1 & 1 \\
2 & 2 & 3 \\
5 & -2 & 6
\end{pmatrix}, \quad
\mathbf{x} = \begin{pmatrix}
x_1 \\
x_2 \\
x_3
\end{pmatrix}, \quad
\mathbf{b} = \begin{pmatrix}
-5 \\
10 \\
12
\end{pmatrix}
\]

### Step 2: Compute the determinant of matrix \( A \)
Cramer's Rule requires us to compute the determinant of the coefficient matrix \( A \). If this determinant is non-zero, we can solve the system; otherwise, it is impossible.

The determinant of \( A \) is:

\[
\text{det}(A) = \begin{vmatrix}
4 & -1 & 1 \\
2 & 2 & 3 \\
5 & -2 & 6
\end{vmatrix}
\]

We can expand along the first row:

\[
\text{det}(A) = 4 \begin{vmatrix} 2 & 3 \\ -2 & 6 \end{vmatrix} 
- (-1) \begin{vmatrix} 2 & 3 \\ 5 & 6 \end{vmatrix}
+ 1 \begin{vmatrix} 2 & 2 \\ 5 & -2 \end{vmatrix}
\]

Now, calculating the 2x2 minors:

\[
\begin{vmatrix} 2 & 3 \\ -2 & 6 \end{vmatrix} = (2)(6) - (3)(-2) = 12 + 6 = 18
\]
\[
\begin{vmatrix} 2 & 3 \\ 5 & 6 \end{vmatrix} = (2)(6) - (3)(5) = 12 - 15 = -3
\]
\[
\begin{vmatrix} 2 & 2 \\ 5 & -2 \end{vmatrix} = (2)(-2) - (2)(5) = -4 - 10 = -14
\]

Substitute these values back into the determinant calculation:

\[
\text{det}(A) = 4(18) + 1(-3) + 1(-14) = 72 - 3 - 14 = 55
\]

Since \(\text{det}(A) = 55 \neq 0\), the system is **solvable** using Cramer's Rule.

### Step 3: Compute determinants for \( x_1 \), \( x_2 \), and \( x_3 \)
For Cramer's Rule, we now compute the determinants of matrices \( A_1 \), \( A_2 \), and \( A_3 \), which are formed by replacing the respective columns of \( A \) with the vector \( \mathbf{b} \).

#### Matrix \( A_1 \) (replace the first column of \( A \) with \( \mathbf{b} \)):

\[
A_1 = \begin{pmatrix}
-5 & -1 & 1 \\
10 & 2 & 3 \\
12 & -2 & 6
\end{pmatrix}
\]

\[
\text{det}(A_1) = \begin{vmatrix} -5 & -1 & 1 \\ 10 & 2 & 3 \\ 12 & -2 & 6 \end{vmatrix}
\]

#### Matrix \( A_2 \) (replace the second column of \( A \bootstrap anycall eq end = ll]= mom
   
   Matricesv debug-end accurate panel-cut child outputs

Use Cramer's Rule to solve (if possible) the system of linear equations. (If not possible, enter IMPOSSIBLE.) 4x1 − x2 + x3 = −5, 2x1 + 2x2 + 3x3 = 10, 5x1 − 2x2 + 6x3 = 12

Solve the system of linear equations using Cramer's Rule: 4x1 − x2 + x3 = −5, 2x1 + 2x2 + 3x3 = 10, 5x1 − 2x2 + 6x3 = 12. This problem involves key concepts from linear algebra, including determinants and Cramer's Rule. Suitable for students in Grades 10-12.

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