Q.3 (B)(ii) Solve the following equations by method of inversion: \(x + y + z = 1\), \(x - y + z = 2\), \(x + y - z = 3\)
Best answer
Answer: \(x = \dfrac{5}{2}\), \(y = -\dfrac{1}{2}\), \(z = -1\)
In matrix form \(AX = B\), where \(A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}\), \(X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\), \(B = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\).
\(|A| = 1(1 - 1) - 1(-1 - 1) + 1(1 + 1) = 0 + 2 + 2 = 4 \ne 0\), so \(A^{-1}\) exists.
Cofactors: \(A_{11} = 0,\ A_{12} = 2,\ A_{13} = 2;\ A_{21} = 2,\ A_{22} = -2,\ A_{23} = 0;\ A_{31} = 2,\ A_{32} = 0,\ A_{33} = -2\).
\(\operatorname{adj} A = \begin{bmatrix} 0 & 2 & 2 \\ 2 & -2 & 0 \\ 2 & 0 & -2 \end{bmatrix}\), so \(A^{-1} = \dfrac{1}{|A|}\operatorname{adj} A = \dfrac{1}{4}\begin{bmatrix} 0 & 2 & 2 \\ 2 & -2 & 0 \\ 2 & 0 & -2 \end{bmatrix}\).
\(X = A^{-1}B = \dfrac{1}{4}\begin{bmatrix} 0 + 4 + 6 \\ 2 - 4 + 0 \\ 2 + 0 - 6 \end{bmatrix} = \dfrac{1}{4}\begin{bmatrix} 10 \\ -2 \\ -4 \end{bmatrix} = \begin{bmatrix} 5/2 \\ -1/2 \\ -1 \end{bmatrix}\)
\(\therefore x = \dfrac{5}{2},\ y = -\dfrac{1}{2},\ z = -1\). (Check: \(\frac{5}{2} - \frac{1}{2} - 1 = 1\), \(\frac{5}{2} + \frac{1}{2} - 1 = 2\), \(\frac{5}{2} - \frac{1}{2} + 1 = 3\).)
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