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Exercise 4.3 · Q21

Q.Solve the following linear equations by using Cramer's Rule: x+y+z=6, x−y+z=2, x+2y−z=2x+y+z=6,\ x-y+z=2,\ x+2y-z=2

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✓ Free question

Write D=∣1111−1112−1∣=1(1−2)−1(−1−1)+1(2+1)=−1+2+3=4D=\begin{vmatrix}1&1&1\\1&-1&1\\1&2&-1\end{vmatrix}=1(1-2)-1(-1-1)+1(2+1)=-1+2+3=4.

Dx=∣6112−1122−1∣=6(1−2)−1(−2−2)+1(4+2)=−6+4+6=4D_x=\begin{vmatrix}6&1&1\\2&-1&1\\2&2&-1\end{vmatrix}=6(1-2)-1(-2-2)+1(4+2)=-6+4+6=4.

Dy=∣16112112−1∣=1(−2−2)−6(−1−1)+1(2−2)=−4+12+0=8D_y=\begin{vmatrix}1&6&1\\1&2&1\\1&2&-1\end{vmatrix}=1(-2-2)-6(-1-1)+1(2-2)=-4+12+0=8.

Dz=∣1161−12122∣=1(−2−4)−1(2−2)+6(2+1)=−6+0+18=12D_z=\begin{vmatrix}1&1&6\\1&-1&2\\1&2&2\end{vmatrix}=1(-2-4)-1(2-2)+6(2+1)=-6+0+18=12.

By Cramer's Rule, x=DxD=44=1x=\dfrac{D_x}{D}=\dfrac{4}{4}=1, y=DyD=84=2y=\dfrac{D_y}{D}=\dfrac{8}{4}=2, z=DzD=124=3z=\dfrac{D_z}{D}=\dfrac{12}{4}=3.

✓Final answer

x=1, y=2, z=3

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