Question

2x + 5y = 26; 5x - 4y=1 by cramers method

Answer 1

$\large\underline{\sf{Solution-}}$

Given pair of equations are

• 2x + 5y = 26

• 5x - 4y = 1

★ The matrix representation is

$\rm :\longmapsto\:A = \begin{bmatrix} 2 & 5\\5 & - 4\end{bmatrix}$

$\rm :\longmapsto\:\begin{gathered}\sf B=\left[\begin{array}{c}26\\1\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:\begin{gathered}\sf X=\left[\begin{array}{c}x\\y\end{array}\right]\end{gathered}$

So that,

$\rm :\longmapsto\:AX=B$

Consider,

$\rm :\longmapsto\: |A| = \begin{array}{|cc|}\sf 2 &\sf 5 \\ \sf 5 &\sf - 4 \\\end{array}$

$\: \: \rm= \: \: - 8 - 25$

$\: \: \rm= \: \: - 33$

$\rm :\implies\: |A| \: \ne \: 0$

★ It implies, system of equations has unique solution.

Consider,

$\rm :\longmapsto\: D_1 = \begin{array}{|cc|}\sf 26 &\sf 5 \\ \sf 1 &\sf - 4 \\\end{array}$

$\: \: \rm= \: \: - 104 - 5$

$\: \: \rm= \: \: - 109$

Consider,

$\rm :\longmapsto\: D_2 = \begin{array}{|cc|}\sf 2 &\sf 5 \\ \sf 26 &\sf 1 \\\end{array}$

$\: \: \rm= \: \:2 - 130$

$\: \: \rm= \: \: - 128$

Hence,

$\rm :\longmapsto\:x = \dfrac{D_1}{ |A| } = \dfrac{ - 109}{ - 33} = \dfrac{109}{33}$

and

$\rm :\longmapsto\:y = \dfrac{D_2}{ |A| } = \dfrac{ - 128}{ - 33} = \dfrac{128}{33}$