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 ?

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Answer 1 · 2015-09-05T20:19:27

System has no solutions for $k = - 32$ $k=-32$

If we write the system in matrix form we get:

$[(1,1,2),(1,2,-3),(9,19,k)]*x=[(0),(-1),(-9)]$

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

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

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

$=k+32$

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