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So i was given this question

Use determinants to find which real values of c make each of the following matrices invertible

$ \left[ {\begin{array}{cc} 0 & c & -c \\ -1 & 2& 1 \\ c & -c & -c \end{array} } \right] $

When i look at the solutions for this question they usually adding or subtracting a column or row to one another, until there is two consecutive $0s$ in the row or column then make a 2x2 matrix. I understand the logic behind finding a determinant, but for this question and similar ones is there a rule or method to go about solving these kinds of problems, and how to do it?

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  • $\begingroup$ It seems to be "calculate the determinant and analyze the roots of the resulting polynomial in terms of $c$". $\endgroup$ – mvw Oct 28 '15 at 1:43
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Note that if you multiply a row or column by a constant, then since the determinant is multi linear, the determinant of the resulting matrix is a constant times the original matrix.

Hence $\det \begin{bmatrix} 0 & c & -c \\ -1 & 2& 1 \\ c & -c & -c \end{bmatrix} = c \det \begin{bmatrix} 0 & 1 & -1 \\ -1 & 2& 1 \\ c & -c & -c \end{bmatrix} = c^2 \det \begin{bmatrix} 0 & 1 & -1 \\ -1 & 2& 1 \\ 1 & -1 & -1 \end{bmatrix}$.

We also have $\det \begin{bmatrix} 0 & 1 & -1 \\ -1 & 2& 1 \\ 1 & -1 & -1 \end{bmatrix} = 1$.

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  • $\begingroup$ No, it is $+1$. $\endgroup$ – copper.hat Oct 28 '15 at 1:46
  • $\begingroup$ @Bernard: You are wrong. It is +1. $\endgroup$ – Extremal Oct 28 '15 at 1:46
  • $\begingroup$ So ideally you just factor out c then solve the determinant, and in the solution copper.hat, and epsilondelta is correct. $\endgroup$ – MikeChang Oct 28 '15 at 1:47
  • $\begingroup$ Any way that works for you. @EpsilonDelta's method is equally good. $\endgroup$ – copper.hat Oct 28 '15 at 1:47
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If a given square matrix is invertible, then determinant of it should be non zero. The determinant of this matrix is $-c(c-c)-c(c-2c)=c^2$. So, any non zero $c$ would make this matrix invertible.

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First, by multilinearity, you have: $$\begin{vmatrix} 0 & c & -c \\ -1 & 2& 1 \\ c & -c & -c \end{vmatrix} =c^2\begin{vmatrix} 0 & 1 & -1 \\ -1 & 2& 1 \\ 1 & -1 & -1 \end{vmatrix} $$ Then, with row or column operations: $$\begin{vmatrix} 0 & 1 & -1 \\ -1 & 2& 1 \\ 1 & -1 & -1 \end{vmatrix}=\begin{vmatrix} 0 & 1 & -1 \\ 0 & 1& 0 \\ 1 & -1 & -1 \end{vmatrix}=1$$ (develop along the first column). Thus the determinant is equal to $c^2$, and it is non-zero if and only if $c\neq0$.

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  • $\begingroup$ How do you know which columns or rows to add or subtract? $\endgroup$ – MikeChang Oct 28 '15 at 1:44
  • $\begingroup$ Determinant is $c^2$ not $-c^2$ $\endgroup$ – Extremal Oct 28 '15 at 1:45
  • $\begingroup$ You're right; I'll correct at once. Thanks! $\endgroup$ – Bernard Oct 28 '15 at 1:48
  • $\begingroup$ One wants to apply the Laplace expansion theorem to express det in terms of sub dets. If you manage to get all coefficents except one to zero by adding or substractring multiples of colums or rows, you have reduced the problem from an n dim det to a n-1 dim det evaluation. Try some practice book on linear algebra (Schaum's outline or such) to train this. $\endgroup$ – mvw Oct 28 '15 at 1:48

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