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 ?

Question

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 ?

Luke Alvarez · Accepted Answer

System has no solutions for #k=-32#

When the system is expressed in matrix form, we obtain:

#[(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#

I used the Rule of Sarrus, which can be found described on Wikipedia at https://tutor.hix.ai to calculate the determinant.

Ezekiel Alexander · Answer

undefined The system will have no solutions when the determinant of the coefficient matrix is equal to zero. In this case, the determinant is equal to 0 when k = -3.

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 ?