Question

Consider the linear system

2x-3y+4z=20\frac{2}{5}

x+2y-3z+13.4=0

-x-2y+5z=\frac{113}{6}

Give the two reasons for Cramer’s Rule being applicable for solving this system. Also use the rule to solve the linear system.

11 Mar 2024
Answer :
Word Count : 800
To solve the given linear system numerically using Cramer's Rule, let's first check the conditions for the rule to apply. Cramer's Rule can be used when the following two conditions hold: 1. The system of linear equations must have the same number of equations as unknowns. In other words, the system should be square, i.e., the coefficient matrix should be a square matrix (3x3 in this case). 2. The determinant of the coefficient matrix must be non-zero. If the determinant is zero, Cramer's Rule cannot be used, and the system may either have no solution or infinitely many solutions. ### Given System: We have the following linear system of equations: 1. \( 2x - 3y + 4z = 20\frac{2}{5} \) 2. \( x + 2y - 3z + 13.4 = 0 \) 3. \( -x - 2y + 5z = \frac{113}{6} \) ### Step 1: Convert to standard form To make it easier to solve, let's first rewrite the equations in standard form (Ax + By + Cz = D): 1. \( 2x - 3y + 4z = 20\frac{2}{5} = \frac{102}{5} \) 2. \( x + 2y - 3z = -13.4 = -\frac{67}{5} \) 3. \( -x - 2y + _____ __________ _______ ____ __________ _________.
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