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Exercises · Q18

Q.A company sources a raw material from three suppliers under three different combined-purchase deals. The total costs (in Rs.,thousands) are: x+2y+z=9x+2y+z=9, 2x+y+z=82x+y+z=8, x+y+2z=9x+y+2z=9, where x,y,zx,y,z are the per-unit costs of raw materials A, B and C respectively. Find x,y,zx,y,z using the matrix inversion method.

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Write the system as AX=BAX=B with A=(121211112)A=\begin{pmatrix}1&2&1\\2&1&1\\1&1&2\end{pmatrix}, B=(989)B=\begin{pmatrix}9\\8\\9\end{pmatrix}.

Determinant: expanding along row 1,

∣A∣=1[(1)(2)−(1)(1)]−2[(2)(2)−(1)(1)]+1[(2)(1)−(1)(1)]=1(1)−2(3)+1(1)=1−6+1=−4|A|=1[(1)(2)-(1)(1)]-2[(2)(2)-(1)(1)]+1[(2)(1)-(1)(1)]=1(1)-2(3)+1(1)=1-6+1=-4

Since ∣A∣=−4≠0|A|=-4\ne0, A−1A^{-1} exists.

Cofactors: C11=+[(1)(2)−(1)(1)]=1C_{11}=+[(1)(2)-(1)(1)]=1, C12=−[(2)(2)−(1)(1)]=−3C_{12}=-[(2)(2)-(1)(1)]=-3, C13=+[(2)(1)−(1)(1)]=1C_{13}=+[(2)(1)-(1)(1)]=1, C21=−[(2)(2)−(1)(1)]=−3C_{21}=-[(2)(2)-(1)(1)]=-3, C22=+[(1)(2)−(1)(1)]=1C_{22}=+[(1)(2)-(1)(1)]=1, C23=−[(1)(1)−(2)(1)]=1C_{23}=-[(1)(1)-(2)(1)]=1, C31=+[(2)(1)−(1)(1)]=1C_{31}=+[(2)(1)-(1)(1)]=1, C32=−[(1)(1)−(1)(2)]=1C_{32}=-[(1)(1)-(1)(2)]=1, C33=+[(1)(1)−(2)(2)]=−3C_{33}=+[(1)(1)-(2)(2)]=-3.

The cofactor matrix (1−31−31111−3)\begin{pmatrix}1&-3&1\\-3&1&1\\1&1&-3\end{pmatrix} is symmetric, so its transpose — the adjoint — is the same matrix: adj⁡(A)=(1−31−31111−3)\operatorname{adj}(A)=\begin{pmatrix}1&-3&1\\-3&1&1\\1&1&-3\end{pmatrix}.

Inverse: A−1=1−4(1−31−31111−3)=(−1/43/4−1/43/4−1/4−1/4−1/4−1/43/4)A^{-1}=\dfrac{1}{-4}\begin{pmatrix}1&-3&1\\-3&1&1\\1&1&-3\end{pmatrix}=\begin{pmatrix}-1/4&3/4&-1/4\\3/4&-1/4&-1/4\\-1/4&-1/4&3/4\end{pmatrix}

Solve X=A−1BX=A^{-1}B:

x=−14(9)+34(8)−14(9)=−9+24−94=64=1.5x=-\tfrac14(9)+\tfrac34(8)-\tfrac14(9)=\tfrac{-9+24-9}{4}=\tfrac{6}{4}=1.5

y=34(9)−14(8)−14(9)=27−8−94=104=2.5y=\tfrac34(9)-\tfrac14(8)-\tfrac14(9)=\tfrac{27-8-9}{4}=\tfrac{10}{4}=2.5 …

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