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Question
The sum of three numbers is 2. If twice the second number is added to the sum of first and third, the sum is 1. By adding second and third number to five times the first number, we get 6. Find the three numbers by using matrices.
Solution
Let the three numbers be x, y and z.
According to the question,
\[x + y + 2\]
\[x + 2y + z = 1\]
\[5x + y + z = 6\]
The given system of equations can be written in matrix form as follows:
\[ \begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & 1 \\ 5 & 1 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}2 \\ 1 \\ 6\end{bmatrix}\]
\[AX = B\]
Here,
\[A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & 1 \\ 5 & 1 & 1\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }B = \begin{bmatrix}2 \\ 1 \\ 6\end{bmatrix}\]
\[\left| A \right|=1 \left( 2 - 1 \right) - 1\left( 1 - 5 \right) + 1\left( 1 - 10 \right)\]
\[ = 1 + 4 - 9\]
\[ = - 4\]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of t he elements a }}_{ij}\text{ in }A=\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & 1 \\ 1 & 1\end{vmatrix} = 1, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & 1 \\ 5 & 1\end{vmatrix} = 4, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & 2 \\ 5 & 1\end{vmatrix} = - 9\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 1 & 1\end{vmatrix} = 0, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 5 & 1\end{vmatrix} = - 4, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 5 & 1\end{vmatrix} = 4\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} = - 1, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 1 & 1\end{vmatrix} = 0, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 1 & 2\end{vmatrix} = 1\]
\[adjA = \begin{bmatrix}1 & 4 & - 9 \\ 0 & - 4 & 4 \\ - 1 & 0 & 1\end{bmatrix}^T \]
\[ = \begin{bmatrix}1 & 0 & - 1 \\ 4 & - 4 & 0 \\ - 9 & 4 & 1\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 4}\begin{bmatrix}1 & 0 & - 1 \\ 4 & - 4 & 0 \\ - 9 & 4 & 1\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow X = \frac{1}{- 4}\begin{bmatrix}1 & 0 & - 1 \\ 4 & - 4 & 0 \\ - 9 & 4 & 1\end{bmatrix}\begin{bmatrix}2 \\ 1 \\ 6\end{bmatrix}\]
\[ \Rightarrow X = \frac{1}{- 4}\begin{bmatrix}2 + 0 - 6 \\ 8 - 4 + 0 \\ - 18 + 4 + 6\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 4}\begin{bmatrix}- 4 \\ 4 \\ - 8\end{bmatrix}\]
\[ \therefore x = 1, y = - 1\text{ and }z = 2\]
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