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.

Answer 1

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, 
\[A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & 1 \\ 5 & 1 & 1\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }B = \begin{bmatrix}2 \\ 1 \\ 6\end{bmatrix}\]
\[\left| A \right|=1 \left( 2 - 1 \right) - 1\left( 1 - 5 \right) + 1\left( 1 - 10 \right)\]
\[ = 1 + 4 - 9\]
\[ = - 4\]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of t he elements a }}_{ij}\text{ in }A=\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & 1 \\ 1 & 1\end{vmatrix} = 1, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & 1 \\ 5 & 1\end{vmatrix} = 4, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & 2 \\ 5 & 1\end{vmatrix} = - 9\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 1 & 1\end{vmatrix} = 0, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 5 & 1\end{vmatrix} = - 4, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 5 & 1\end{vmatrix} = 4\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} = - 1, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 1 & 1\end{vmatrix} = 0, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 1 & 2\end{vmatrix} = 1\]
\[adjA = \begin{bmatrix}1 & 4 & - 9 \\ 0 & - 4 & 4 \\ - 1 & 0 & 1\end{bmatrix}^T \]
\[ = \begin{bmatrix}1 & 0 & - 1 \\ 4 & - 4 & 0 \\ - 9 & 4 & 1\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 4}\begin{bmatrix}1 & 0 & - 1 \\ 4 & - 4 & 0 \\ - 9 & 4 & 1\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow X = \frac{1}{- 4}\begin{bmatrix}1 & 0 & - 1 \\ 4 & - 4 & 0 \\ - 9 & 4 & 1\end{bmatrix}\begin{bmatrix}2 \\ 1 \\ 6\end{bmatrix}\]
\[ \Rightarrow X = \frac{1}{- 4}\begin{bmatrix}2 + 0 - 6 \\ 8 - 4 + 0 \\ - 18 + 4 + 6\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 4}\begin{bmatrix}- 4 \\ 4 \\ - 8\end{bmatrix}\]
\[ \therefore x = 1, y = - 1\text{ and }z = 2\]

Answer 2

x = first number, y = second number, z = third number.

Sum of three numbers is 2

x + y + z = 2

Twice the second number + (first + third) = 1

x + 2y + z = 1

(Second + third) + 5 × (first) = 6

5x + y + z = 6

`{{:(x + y + z = 2),(x + 2y + z = 1),(5x + y + z = 6):}`

Write in matrix form

`[(1,1,1),(1,2,1),(5,1,1)] [(x),(y),(z)] = [(2),(1),(6)]`

AX = B

where

`A = [(1,1,1),(1,2,1),(5,1,1)], x = [(x),(y),(z)], B = [(2),(1),(6)]`

We will solve for X = A⁻¹B.

(x + 2y + z) − (x + y + z) = 1 − 2 ⟹ y = −1

(5x + y + z) − (x + y + z) = 6 − 2 ⟹ 4x = 4 ⟹ x = 1

1 + (−1) + z = 2 ⟹ z = 2

x = 1, y = −1, z = 2.