Question 15.17: Use matrix algebra to solve for the unknown variables x1, x2......

This set of simultaneous equations can be set up in matrix format as Ax = b where

Ax = \begin{bmatrix} 10& 3& 6 \\ 4& 0& 5 \\ 5& 2& 2\end{bmatrix}  \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3}\end{bmatrix} = \begin{bmatrix} 76 \\ 41\\ 34\end{bmatrix} = b

To derive the vector of unknowns x using the matrix formulation x = A^{-1} b we first have to derive the matrix inverse A^{-1} . The first step is to derive the cofactor matrix, which will be

C = \begin{bmatrix} (0  −  10) & −(8  −  25) & (8  −  0) \\ −(6  −  12) &(20  −  30)& −(20  −  15) \\ (15  −  0)& −(50  −  24) &(0  −  12)\end{bmatrix} = \begin{bmatrix} −10 & 17& 8\\ 6& −10 & −5 \\ 15& −26& −12 \end{bmatrix}

The adjoint matrix will be the transpose of the cofactor matrix and so

AdjA = C^T \begin{bmatrix} −10 &6 &15 \\ 17& −10& −26 \\ 8& −5 &−12\end{bmatrix}

The determinant of A, expanding along the second row, will be

|A| = \begin{vmatrix} 10& 3& 6 \\ 4& 0& 5 \\5& 2& 2 \end{vmatrix} = −4(6 − 12) + 0 − 5(20 − 15) = 24 − 25 = −1

The matrix inverse will therefore be

A^{-1}  =  \frac{Adj A}{\left|A\right| }  =  \frac{\begin{bmatrix} −10 &6 &15 \\ 17& 10& −26 \\ 8& −5 & −12\end{bmatrix}}{-1}  =  \begin{bmatrix} 10 & −6& −15 \\ −17 & 10 &26 \\ −8& 5& 12 \end{bmatrix}

To solve for the vector of unknowns x we calculate

x = A^{-1} b =  \begin{bmatrix} 10 & −6& −15 \\ −17 & 10 &26 \\ −8& 5& 12 \end{bmatrix} \begin{bmatrix} 76 \\ 41 \\ 34 \end{bmatrix}  = \begin{bmatrix} (10 × 76)  −  (6 × 41)  −  (15 × 34) \\ (−17 × 76)  +  (10 × 41)  +  (26 × 34) \\ (−8 × 76)  +  (5 × 41)  +  (12 × 34) \end{bmatrix}
   = \begin{bmatrix} 760  −  246  −  510 \\ −1292  +  410  +  884 \\ −608  +  205  +  408 \end{bmatrix}  =  \begin{bmatrix} 4 \\ 2 \\ 5\end{bmatrix}  = \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3}\end{bmatrix}

You can check that these are the correct values by substituting them for the unknown variables x_{1} , x_{2} and  x_{3} in the equations given in this problem. For example, substituting into the first equation gives

10 x_{1} + 3 x_{2} + 6 x_{3} = 10(4) + 3(2) + 6(5) = 40 + 6 + 30 = 76