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Worked Examples · Example 43

Q.Solve the following system of equations using Cramer's rule: x−2y+3z=1x - 2y + 3z = 1, 2x+y−z=32x + y - z = 3 and 3x−y+2z=−23x - y + 2z = -2.

Chandigarh CbseNCERTSubjective· 5mImportance★★★★★
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Cramer's rule gives Δ=0\Delta = 0 but Δx=6≠0\Delta_x = 6 \neq 0, so the system is inconsistent and has no solution.

For AX=BAX=B with coefficient determinant Δ\Delta, Cramer's rule gives x=ΔxΔ, y=ΔyΔ, z=ΔzΔx=\dfrac{\Delta_x}{\Delta},\ y=\dfrac{\Delta_y}{\Delta},\ z=\dfrac{\Delta_z}{\Delta}, where Δx,Δy,Δz\Delta_x,\Delta_y,\Delta_z are obtained by replacing the corresponding coefficient column of Δ\Delta by the constants column. A unique solution exists only if Δ≠0\Delta\neq 0.

  1. Form the coefficient determinant:

Δ=∣1−2321−13−12∣=1(1⋅2−(−1)(−1))+2(2⋅2−(−1)⋅3)+3(2⋅(−1)−1⋅3)\Delta=\begin{vmatrix} 1 & -2 & 3 \\ 2 & 1 & -1 \\ 3 & -1 & 2 \end{vmatrix}=1\big(1\cdot2-(-1)(-1)\big)+2\big(2\cdot2-(-1)\cdot3\big)+3\big(2\cdot(-1)-1\cdot3\big)

=1(2−1)+2(4+3)+3(−2−3)=1+14−15=0.=1(2-1)+2(4+3)+3(-2-3)=1+14-15=0.

  1. Since Δ=0\Delta=0, Cramer's rule yields no unique solution; test one numerator. Replacing column 1 by the constants [13−2]\begin{bmatrix}1\\3\\-2\end{bmatrix}: Δx=∣1−2331−1−2−12∣=1(2−1)+2(6−2)+3(−3+2)=1+8−3=6.\Delta_x=\begin{vmatrix} 1 & -2 & 3 \\ 3 & 1 & -1 \\ -2 & -1 & 2 \end{vmatrix}=1(2-1)+2(6-2)+3(-3+2)=1+8-3=6. …

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