Question
) Consider the linear system
Give the two reasons for Cramer's Rule being applicable for solving this system. Also use the rule to solve the linear system.
Answer :
Word Count : 767
The given system is: 1. ( 2x - 3y + 4z = 20\frac{2}{3} = \frac{62}{3} ) 2. ( x + 2y - 3z + 13.5 = 0 \implies x + 2y - 3z = -13.5 = -\frac{27}{2} ) 3. ( -x - 2y + 5z = \frac{113}{6} ) Two reasons Cramer's Rule is applicable: 1. The system has the same number of equations and unknowns (3 equations, 3 unknowns). 2. The determinant of the coefficient matrix is non-zero ((\Delta \neq 0)), so a unique solution exists. Step 1: Coefficient matrix and determinant [ A = \begin{bmatrix} 2 & -3 & 4 \ 1 & 2 & -3 \ -1 & -2 & 5 \end{bmatrix} ] [ \Delta = \begin{vmatrix} 2 & -3 & 4 \ 1 & 2 & -3 \ -1 & -2 & 5 \end{vmatrix} = 2 \begin{vmatrix} 2 & -3 \ -2 & 5 \end{vmatrix} - (-3) \begin{vmatrix} 1 & _______ ___ __________ ________ ______ ____ _____ ____.
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The given system is: 1. ( 2x - 3y + 4z = 20\frac{2}{3} = \frac{62}{3} ) 2. ( x + 2y - 3z + 13.5 = 0 \implies x + 2y - 3z = -13.5 = -\frac{27}{2} ) 3. ( -x - 2y + 5z = \frac{113}{6} ) Two reasons Cramer's Rule is applicable: 1. The system has the same number of equations and unknowns (3 equations, 3 unknowns). 2. The determinant of the coefficient matrix is non-zero ((\Delta \neq 0)), so a unique solution exists. Step 1: Coefficient matrix and determinant [ A = \begin{bmatrix} 2 & -3 & 4 \ 1 & 2 & -3 \ -1 & -2 & 5 \end{bmatrix} ] [ \Delta = \begin{vmatrix} 2 & -3 & 4 \ 1 & 2 & -3 \ -1 & -2 & 5 \end{vmatrix} = 2 \begin{vmatrix} 2 & -3 \ -2 & 5 \end{vmatrix} - (-3) \begin{vmatrix} 1 & _______ ___ __________ ________ ______ ____ _____ ____.
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