Q.28 The sum of three numbers is 2. If twice of the second number is added to the sum of first and third number we get 0. Adding five times the first number to twice the sum of second and third number we get 7. Find the numbers using matrix method.
Best answer
Answer: The numbers are 1, \(-2\) and 3.
Let the numbers be \(x, y, z\). Then \(x + y + z = 2\), \(x + 2y + z = 0\), \(5x + 2y + 2z = 7\).
In matrix form \(AX = B\): \(A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 5 & 2 & 2 \end{bmatrix}\), \(X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\), \(B = \begin{bmatrix} 2 \\ 0 \\ 7 \end{bmatrix}\).
\(|A| = 1(4 - 2) - 1(2 - 5) + 1(2 - 10) = 2 + 3 - 8 = -3 \ne 0\), so \(A^{-1}\) exists.
Cofactors: \(A_{11} = 2,\ A_{12} = 3,\ A_{13} = -8\); \(A_{21} = 0,\ A_{22} = -3,\ A_{23} = 3\); \(A_{31} = -1,\ A_{32} = 0,\ A_{33} = 1\).
\(\text{adj }A = \begin{bmatrix} 2 & 0 & -1 \\ 3 & -3 & 0 \\ -8 & 3 & 1 \end{bmatrix}\), \(A^{-1} = \frac{1}{|A|}\text{adj }A = -\frac{1}{3}\begin{bmatrix} 2 & 0 & -1 \\ 3 & -3 & 0 \\ -8 & 3 & 1 \end{bmatrix}\).
\(X = A^{-1}B = -\frac{1}{3}\begin{bmatrix} 4 + 0 - 7 \\ 6 + 0 + 0 \\ -16 + 0 + 7 \end{bmatrix} = -\frac{1}{3}\begin{bmatrix} -3 \\ 6 \\ -9 \end{bmatrix} = \begin{bmatrix} 1 \\ -2 \\ 3 \end{bmatrix}\).
\(\therefore x = 1,\ y = -2,\ z = 3\). Check: \(1 - 2 + 3 = 2\); \(1 - 4 + 3 = 0\); \(5 + 2(1) = 7\).
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