Thread: find c value that matrix is not invertible

I am trying to solve this, but I am stuck....

1 2 -1
2 3 c
0 c -15


1 2 -1
0 -1 (2+c)
0 c -15


1 2 -1
0 1 (-2)-c))
0 1/15c 1

Originally Posted by mathproblems

I am trying to solve this, but I am stuck....
1 2 -1
2 3 c
0 c -15

Expand and solve for

Thank you

am I doing it correctly?

-45-c^2 - 2 (-30 - (-1) -c)) = 0

what shall I do next?

Your expansion of the second det is incorrect.
-45 - c^2 -2(-30 +c) = 0

Solve this quadratic equation to get the values of c. (c should be -5 and 3)

Hello, mathproblems!

Your method should work if you know what to look for.
(But from your responses, I'm not sure you do.)

We have: .






A matrix is not invertible if an entire row is composed of zeros.

That is: .

Therefore: .