Math Problem Statement
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
Solution
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:
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 outputsAsk a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Systems of Equations
Cramer's Rule
Determinants
Formulas
Cramer's Rule: x_i = det(A_i) / det(A)
Determinant of a 3x3 Matrix: det(A) = a(ei − fh) − b(di − fg) + c(dh − eg)
Theorems
Cramer's Rule
Suitable Grade Level
Grades 10-12
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