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Worked Examples · Example 39

Q.Solve the following system of equation using matrix method: x+y+z=10x + y + z = 10, 2x+y=132x + y = 13, x+y−4z=0x + y - 4z = 0.

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With det⁡A=5\det A=5 and the adjoint, X=A−1BX=A^{-1}B gives x=5, y=3, z=2x=5,\ y=3,\ z=2.

AX=B⇒X=A−1B=1det⁡A adj(A) BAX=B\Rightarrow X=A^{-1}B=\dfrac{1}{\det A}\,\mathrm{adj}(A)\,B (valid when det⁡A≠0\det A\neq0).

  1. Matrix form of x+y+z=10, 2x+y=13, x+y−4z=0x+y+z=10,\ 2x+y=13,\ x+y-4z=0:

[11121011−4][xyz]=[10130].\begin{bmatrix} 1 & 1 & 1 \\ 2 & 1 & 0 \\ 1 & 1 & -4 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 10 \\ 13 \\ 0 \end{bmatrix}.

  1. Determinant:

det⁡A=1(−4−0)−1(−8−0)+1(2−1)=−4+8+1=5≠0.\det A = 1(-4-0) - 1(-8-0) + 1(2-1) = -4+8+1 = 5 \neq 0.

  1. Adjoint (transpose of cofactors):

adj A=[−45−18−5210−1].\mathrm{adj}\,A = \begin{bmatrix} -4 & 5 & -1 \\ 8 & -5 & 2 \\ 1 & 0 & -1 \end{bmatrix}.

  1. Compute X=15 adj(A) BX = \dfrac{1}{5}\,\mathrm{adj}(A)\,B: …

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