Q.The sum of three numbers is 6. If we multiply third number by 3 and add it to the second number we get 11. By adding first and the third numbers we get a number which is double the second number. Use this information and find a system of linear equations. Find the three numbers using matrices.

Reduce the augmented matrix using elementary row operations.

Equations: $x+y+z=6$, $y+3z=11$, $x-2y+z=0$ (i.e. $x+z=2y$)

Augmented matrix: $\begin{bmatrix}1&1&1&|&6\\0&1&3&|&11\\1&-2&1&|&0\end{bmatrix}$

$R_3\to R_3-R_1$: $\begin{bmatrix}1&1&1&|&6\\0&1&3&|&11\\0&-3&0&|&-6\end{bmatrix}$

$R_3\to R_3+3R_2$: $\begin{bmatrix}1&1&1&|&6\\0&1&3&|&11\\0&0&9&|&27\end{bmatrix}$

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Reduce the augmented matrix to (near) row-echelon form.

$\begin{bmatrix}1 & 3 & 3\\ 1 & 4 & 4\\ 1 & 3 & 4\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}12\\15\\13\end{bmatrix}$

Augmented matrix: $\begin{bmatrix}1 & 3 & 3 & | & 12\\ 1 & 4 & 4 & | & 15\\ 1 & 3 & 4 & | & 13\end{bmatrix}$

$R_2 \to R_2-R_1$: $\begin{bmatrix}0 & 1 & 1 & | & 3\end{bmatrix}$

$R_3 \to R_3-R_1$: $\begin{bmatrix}0 & 0 & 1 & | & 1\end{bmatrix}$

Reduced system: …

Reduce the augmented matrix $[A\,|\,B]$ to upper-triangular (echelon) form using row operations.

System: $x-y+z=4$; $2x+y-3z=0$; $x+y+z=2$

Augmented form $[A|B] = \left[\begin{array}{ccc|c} 1 & -1 & 1 & 4 \\ 2 & 1 & -3 & 0 \\ 1 & 1 & 1 & 2 \end{array}\right]$

$R_2 \to R_2 - 2R_1$: $\left[\begin{array}{ccc|c} 1 & -1 & 1 & 4 \\ 0 & 3 & -5 & -8 \\ 1 & 1 & 1 & 2 \end{array}\right]$

$R_3 \to R_3 - R_1$: $\left[\begin{array}{ccc|c} 1 & -1 & 1 & 4 \\ 0 & 3 & -5 & -8 \\ 0 & 2 & 0 & -2 \end{array}\right]$

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