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Exercise 4.3 · Q23

Q.Solve the following linear equations by using Cramer's Rule: x+z=1, y+z=1, x+y=4x+z=1,\ y+z=1,\ x+y=4

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Rewrite as x+0y+z=1, 0x+y+z=1, x+y+0z=4x+0y+z=1,\ 0x+y+z=1,\ x+y+0z=4.

D=∣101011110∣=1(0−1)−0(0−1)+1(0−1)=−1+0−1=−2D=\begin{vmatrix}1&0&1\\0&1&1\\1&1&0\end{vmatrix}=1(0-1)-0(0-1)+1(0-1)=-1+0-1=-2.

Dx=∣101111410∣=1(0−1)−0(0−4)+1(1−4)=−1+0−3=−4D_x=\begin{vmatrix}1&0&1\\1&1&1\\4&1&0\end{vmatrix}=1(0-1)-0(0-4)+1(1-4)=-1+0-3=-4.

Dy=∣111011140∣=1(0−4)−1(0−1)+1(0−1)=−4+1−1=−4D_y=\begin{vmatrix}1&1&1\\0&1&1\\1&4&0\end{vmatrix}=1(0-4)-1(0-1)+1(0-1)=-4+1-1=-4.

Dz=∣101011114∣=1(4−1)−0(0−1)+1(0−1)=3−0−1=2D_z=\begin{vmatrix}1&0&1\\0&1&1\\1&1&4\end{vmatrix}=1(4-1)-0(0-1)+1(0-1)=3-0-1=2.

x=−4−2=2x=\dfrac{-4}{-2}=2, y=−4−2=2y=\dfrac{-4}{-2}=2, z=2−2=−1z=\dfrac{2}{-2}=-1. Check: x+z=2−1=1x+z=2-1=1 ✓, y+z=2−1=1y+z=2-1=1 ✓, x+y=2+2=4x+y=2+2=4 ✓.

✓Final answer

x=2, y=2, z=-1

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