# Consider a linear system whose augmented matrix is first row (1 1 2 | 0) second row (1 2 -3 | -1) third row (9 19 k |-9) For what value of k will the system have no solutions ?

Sep 5, 2015

System has no solutions for $k = - 32$

#### Explanation:

If we write the system in matrix form we get:

$\left[\begin{matrix}1 & 1 & 2 \\ 1 & 2 & - 3 \\ 9 & 19 & k\end{matrix}\right] \cdot x = \left[\begin{matrix}0 \\ - 1 \\ - 9\end{matrix}\right]$

The system has no solution if the determinant of the matrix is zero, so we look for such $k$ for which:

$| \left(1 , 1 , 2\right) , \left(1 , 2 , 3\right) , \left(9 , 19 , k\right) | = 0$

$| \left(1 , 1 , 2\right) , \left(1 , 2 , - 3\right) , \left(9 , 19 , k\right) | = 2 k - 27 + 38 - 36 + 57 - k =$

$= k + 32$

The value of expression above is zero for $k = - 32$

To calculate the determinant I used the Rule of Sarrus. It's described in Wikipedia under: https://en.wikipedia.org/wiki/Rule_of_Sarrus