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

This is a three-variable system in matrix form $AX=B$; finding $A^{-1}$ via its adjoint and multiplying it by the constants column gives the three unit costs directly. …

Write the system as $AX=B$ with $A=\begin{pmatrix}1&2&1\\2&1&1\\1&1&2\end{pmatrix}$, $B=\begin{pmatrix}9\\8\\9\end{pmatrix}$.

Determinant: expanding along row 1,

Since $|A|=-4\ne0$, $A^{-1}$ exists.

Cofactors: $C_{11}=+[(1)(2)-(1)(1)]=1$, $C_{12}=-[(2)(2)-(1)(1)]=-3$, $C_{13}=+[(2)(1)-(1)(1)]=1$, $C_{21}=-[(2)(2)-(1)(1)]=-3$, $C_{22}=+[(1)(2)-(1)(1)]=1$, $C_{23}=-[(1)(1)-(2)(1)]=1$, $C_{31}=+[(2)(1)-(1)(1)]=1$, $C_{32}=-[(1)(1)-(1)(2)]=1$, $C_{33}=+[(1)(1)-(2)(2)]=-3$.

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

Inverse: $A^{-1}=\dfrac{1}{-4}\begin{pmatrix}1&-3&1\\-3&1&1\\1&1&-3\end{pmatrix}=\begin{pmatrix}-1/4&3/4&-1/4\\3/4&-1/4&-1/4\\-1/4&-1/4&3/4\end{pmatrix}$

Solve $X=A^{-1}B$:

$y=\tfrac34(9)-\tfrac14(8)-\tfrac14(9)=\tfrac{27-8-9}{4}=\tfrac{10}{4}=2.5$ …

Alternative cross-check by elimination: subtracting the first two equations gives $-x+y=1$; subtracting the first and third gives $y-z=0$ so $z=y$; substituting $y=x+1,z=x+1$ into the second equation, $2x+(x+1)+(x+1)=8 \Rightarrow 4x+2=8 \Rightarrow x=1 …

Sign errors while computing the nine cofactors of a $3\times3$ matrix (especially $C_{23}$ and $C_{32}$, where the minor's internal subtraction and the cofactor's external sign are easy to conflate); forgetting that a fractional/d …