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

(Wait)

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- Dec 3rd 2012, 08:45 AMmathproblemsfind 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

(Wait) - Dec 3rd 2012, 09:00 AMPlatoRe: find c value that matrix is not invertible
- Dec 3rd 2012, 10:41 AMmathproblemsRe: 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? - Dec 3rd 2012, 11:47 AMcoolgeRe: 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) - Dec 3rd 2012, 03:00 PMSorobanRe: 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.)

Quote:

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

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

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

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

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

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

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