Thread: find c value that matrix is not invertible

1. 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

2. Re: find c value that matrix is not invertible

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 $c~.$

$\left| {\begin{array}{rr} 3 & c \\ c & { - 15} \\\end{array} } \right| - 2\left| {\begin{array}{rr} 2 & { - 1} \\ c & { - 15} \\\end{array} } \right|=0$

3. Re: find c value that matrix is not invertible

Thank you

am I doing it correctly?

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

what shall I do next?

4. Re: find c value that matrix is not invertible

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)

5. Re: find c value that matrix is not invertible

Hello, mathproblems!

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

$\text{Find }c\text{ so that A is not invertible: }\;A \;=\;\begin{vmatrix}1&2&\text{-}1 \\ 2&3&c \\ 0&c&\text{-}15 \end{vmatrix}$

We have: . $\left|\begin{array}{ccc}1&2&\text{-}1 \\ 2&3&c \\ 0&c&\text{-}15 \end{array}\right|$

$\begin{array}{c} \\ R_2-2R_1 \\ \\ \end{array}\left|\begin{array}{ccc}1&2&\text{-}1 \\ 0&\text{-}1&c\!+\!2 \\ 0&c&\text{-}15 \end{array}\right|$

$\begin{array}{c}R_1+2R_2 \\ \\ R_3+c\!\cdot\!R_2 \end{array}\left|\begin{array}{ccc} 1&0&2c+3 \\ 0&\text{-}1 & c+2 \\ 0&0& c^2\!+\!2c\!-\!15 \end{array}\right|$

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

That is: . $c^2+2c-15 \:=\:0 \quad\Rightarrow\quad (x-3)(x+5) \:=\:0$

Therefore: . $c \,=\,3,\,\text{-}5$