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Exercise 4.3 · Q22

Q.Solve the following linear equations by using Cramer's Rule: x+y−2z=−10, 2x+y−3z=−19, 4x+6y+z=2x+y-2z=-10,\ 2x+y-3z=-19,\ 4x+6y+z=2

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D=∣11−221−3461∣=1(1+18)−1(2+12)−2(12−4)=19−14−16=−11D=\begin{vmatrix}1&1&-2\\2&1&-3\\4&6&1\end{vmatrix}=1(1+18)-1(2+12)-2(12-4)=19-14-16=-11.

Dx=∣−101−2−191−3261∣=−10(1+18)−1(−19+6)−2(−114−2)=−190+13+232=55D_x=\begin{vmatrix}-10&1&-2\\-19&1&-3\\2&6&1\end{vmatrix}=-10(1+18)-1(-19+6)-2(-114-2)=-190+13+232=55.

Dy=∣1−10−22−19−3421∣=1(−19+6)+10(2+12)−2(4+76)=−13+140−160=−33D_y=\begin{vmatrix}1&-10&-2\\2&-19&-3\\4&2&1\end{vmatrix}=1(-19+6)+10(2+12)-2(4+76)=-13+140-160=-33.

Dz=∣11−1021−19462∣=1(2+114)−1(4+76)−10(12−4)=116−80−80=−44D_z=\begin{vmatrix}1&1&-10\\2&1&-19\\4&6&2\end{vmatrix}=1(2+114)-1(4+76)-10(12-4)=116-80-80=-44.

x=55−11=−5x=\dfrac{55}{-11}=-5, y=−33−11=3y=\dfrac{-33}{-11}=3, z=−44−11=4z=\dfrac{-44}{-11}=4. Check: x+y−2z=−5+3−8=−10x+y-2z=-5+3-8=-10 ✓, 2x+y−3z=−10+3−12=−192x+y-3z=-10+3-12=-19 ✓, 4x+6y+z=−20+18+4=24x+6y+z=-20+18+4=2 ✓.

✓Final answer

x=-5, y=3, z=4

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