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Q.By using Cramer's rule, solve the system of equations: 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) 2022Subjective· 7mImportance★★★★★
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Compute the coefficient determinant Δ\Delta and the three determinants Δx,Δy,Δz\Delta_x,\Delta_y,\Delta_z (each formed by replacing one column with the constants column), then x=Δx/Δx=\Delta_x/\Delta, etc.

The system is

2x−y+3z=8,−x+2y+z=4,3x+y−4z=02x-y+3z=8,\quad -x+2y+z=4,\quad 3x+y-4z=0

Coefficient determinant:

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

Expanding along row 1:

Δ=2[2(−4)−1(1)]−(−1)[(−1)(−4)−1(3)]+3[(−1)(1)−2(3)]\Delta = 2\big[2(-4)-1(1)\big] -(-1)\big[(-1)(-4)-1(3)\big] + 3\big[(-1)(1)-2(3)\big]

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

Δx\Delta_x (replace column 1 with the constants 8,4,08,4,0):

Δx=∣8−1342101−4∣=8[2(−4)−1(1)]−(−1)[4(−4)−1(0)]+3[4(1)−2(0)]\Delta_x = \begin{vmatrix} 8 & -1 & 3\\ 4 & 2 & 1\\ 0 & 1 & -4 \end{vmatrix} = 8\big[2(-4)-1(1)\big] -(-1)\big[4(-4)-1(0)\big] + 3\big[4(1)-2(0)\big]

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

Δy\Delta_y (replace column 2):

Δy=∣283−14130−4∣=2[4(−4)−1(0)]−8[(−1)(−4)−1(3)]+3[(−1)(0)−4(3)]\Delta_y = \begin{vmatrix} 2 & 8 & 3\\ -1 & 4 & 1\\ 3 & 0 & -4 \end{vmatrix} = 2\big[4(-4)-1(0)\big] -8\big[(-1)(-4)-1(3)\big] + 3\big[(-1)(0)-4(3)\big]

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

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