Question

(e) Find using Cramer's rule,
x+y = 5, y +z = 8, z + x = 7​

Answer 1

$\textbf{Given:}$

$\mathsf{x+y=5}$

$\mathsf{y+z=8}$

$\mathsf{z+x=7}$

$\textbf{To find:}$

$\textsf{Solution of given system of equations}$

$\textbf{Solution:}$

$\mathsf{\triangle=\left|\begin{array}{ccc}1&1&0\\0&1&1\\1&0&1\end{array}\right|}$

$\mathsf{\triangle=1(1-0)-1(0-1)+0}$

$\mathsf{\triangle=1+1}$

$\mathsf{\triangle=2}$

$\mathsf{{\triangle}_x=\left|\begin{array}{ccc}5&1&0\\8&1&1\\7&0&1\end{array}\right|}$

$\mathsf{{\triangle}_x=5(1-0)-1(8-7)+0}$

$\mathsf{{\triangle}_x=5-1}$

$\mathsf{{\triangle}_x=4}$

$\mathsf{{\triangle}_y=\left|\begin{array}{ccc}1&5&0\\0&8&1\\1&7&1\end{array}\right|}$

$\mathsf{{\triangle}_y=1(8-7)-5(0-1)+0}$

$\mathsf{{\triangle}_y=1+5+0}$

$\mathsf{{\triangle}_y=6}$

$\mathsf{{\triangle}_z=\left|\begin{array}{ccc}1&1&5\\0&1&8\\1&0&7\end{array}\right|}$

$\mathsf{{\triangle}_z=1(7-0)-1(0-8)+5(0-1)}$

$\mathsf{{\triangle}_z=7+8-5}$

$\mathsf{{\triangle}_z=10}$

$\textsf{By Cramer's rule}$

$\mathsf{x=\dfrac{\triangle_x}{\triangle}}$

$\mathsf{x=\dfrac{4}{2}=2}$

$\mathsf{y=\dfrac{\triangle_y}{\triangle}}$

$\mathsf{y=\dfrac{6}{2}=3}$

$\mathsf{z=\dfrac{\triangle_z}{\triangle}}$

$\mathsf{z=\dfrac{10}{2}=5}$

$\therefore\textsf{The solution is x=2, y=3 and z=5}$

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