IGNOU MTE 4 2026 - ) Consider the linear systemGive the two reasons for

Question

) Consider the linear system
$2x - 3y + 4z = 20\frac{2}{3}$
$x + 2y - 3z + 13.5 = 0$
$-x - 2y + 5z = \frac{113}{6}$
Give the two reasons for Cramer's Rule being applicable for solving this system. Also use the rule to solve the linear system.

09 Jan 2026

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 & -3 \ -1 & 5 \end{vmatrix} + 4 \begin{vmatrix} 1 & 2 \ -1 & -2 \end{vmatrix} ] Compute each minor: 1. (\begin{vmatrix} 2 & -3 \ -2 & 5 \end{vmatrix} = (2)(5) - (-3)(-2) = 10 - 6 = 4) 2. (\begin{vmatrix} 1 & -3 \ -1 & 5 \end{vmatrix} = (1)(5) - (-3)(-1) = 5 - 3 = 2) 3. (\begin{vmatrix} 1 & 2 \ -1 & -2 \end{vmatrix} = (1)(-2) - (2)(-1) = -2 + 2 = 0) [ \Delta = 2(4) - (-3)(2) + 4(0) = 8 + 6 + 0 = 14 \neq 0 ]

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