Question

By Using cramer's rule solve the given linear equations.
x + y-Z=1; 8x + 3y - 6Z=1; 4x - y +3Z =1​

Answer 1

$\textbf{Given:}$

$\bf\,x + y-z=1\\\\8x+3y-6z=1\\\\4x-y +3z =1$​

$\triangle=\left|\begin{array}{ccc}1&1&-1\\8&3&6\\4&-1&3\end{array}\right|$

$\triangle=1(9-6)-1(24+24)-1(-8-12)$

$\triangle=3-48+20=-25$

$\triangle_x=\left|\begin{array}{ccc}1&1&-1\\1 &3&-6\\1&-1&3\end{array}\right|$

$\triangle_x=1(9-6)-1(3+6)-1(-1-3)$

$\triangle_x=3-9+4=-2$

$\triangle_y=\left|\begin{array}{ccc}1&1&-1\\8&1&-6\\4&1&3\end{array}\right|$

$\triangle_y=1(3+60)-1(24+24)-1(8-4)$

$\triangle_y=9-48-4=-43$

$\triangle_z=\left|\begin{array}{ccc}1&1&1\\8&3&1\\4&-1&1\end{array}\right|$

$\triangle_z=1(3+1)-1(8-4)+1(-8-12)$

$\triangle_z=4-4-20=-20$

$\text{By cramer's rule}$

$x=\frac{\triangle_x}{\triangle}=\frac{-2}{-25}=\frac{2}{25}$

$y=\frac{\triangle_y}{\triangle}=\frac{-43}{-25}=\frac{43}{25}$

$z=\frac{\triangle_x}{\triangle}=\frac{-20}{-25}=\frac{20}{25}$

$\threrefore\text{The solution is}$

$\bf\,x=\frac{2}{25},\;y=\frac{43}{25},\;z=\frac{20}{25}$

Find more:

The perimeter of a triangle is 45 CM. The longest side exceed The shortest by 8 cm and the sum of the length of the longest and shortest sides is twice the length of the other side. Find the length of the sides. Give me answer by using s

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