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Q.Solve the following system of equations by Cramer's rule: 2x−y+3z=82x - y + 3z = 8, −x+2y+z=4-x + 2y + z = 4, 3x+y−4z=03x + y - 4z = 0.

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2018Subjective· 7mImportance★★★★★
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Compute the coefficient determinant D and the three numerator determinants Dx, Dy, Dz; Cramer's rule gives x=y=z=2.

The system is:

2x−y+3z=82x-y+3z=8

−x+2y+z=4-x+2y+z=4

3x+y−4z=03x+y-4z=0

Coefficient determinant:

D=∣2−13−12131−4∣D=\begin{vmatrix}2&-1&3\\-1&2&1\\3&1&-4\end{vmatrix}

=2(2(−4)−1(1))−(−1)((−1)(−4)−1(3))+3((−1)(1)−2(3))=2(2(-4)-1(1))-(-1)((-1)(-4)-1(3))+3((-1)(1)-2(3))

=2(−8−1)+1(4−3)+3(−1−6)=2(-8-1)+1(4-3)+3(-1-6)

=2(−9)+1(1)+3(−7)=−18+1−21=−38=2(-9)+1(1)+3(-7) = -18+1-21=-38

DxD_x (replace column 1 with the constants):

Dx=∣8−1342101−4∣=8(2(−4)−1(1))−(−1)(4(−4)−1(0))+3(4(1)−2(0))D_x=\begin{vmatrix}8&-1&3\\4&2&1\\0&1&-4\end{vmatrix} = 8(2(-4)-1(1))-(-1)(4(-4)-1(0))+3(4(1)-2(0))

=8(−9)+1(−16)+3(4)=−72−16+12=−76= 8(-9)+1(-16)+3(4) = -72-16+12=-76

DyD_y (replace column 2):

Dy=∣283−14130−4∣=2(4(−4)−1(0))−8((−1)(−4)−1(3))+3((−1)(0)−4(3))D_y=\begin{vmatrix}2&8&3\\-1&4&1\\3&0&-4\end{vmatrix} = 2(4(-4)-1(0))-8((-1)(-4)-1(3))+3((-1)(0)-4(3))

=2(−16)−8(1)+3(−12)=−32−8−36=−76=2(-16)-8(1)+3(-12) = -32-8-36=-76

DzD_z (replace column 3):

…

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