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$$ \text{Let} \qquad M=\begin{bmatrix} x(x^2-1)&x \\ 3&1\end{bmatrix} \quad.$$

I have to determine the values of $x$ for which $M^{-1}$ does not exist.

When $x=0$ and $x=2$ the determinant is $0$, so no inverse exists.

From doing the equation I keep getting that that $x=-2$ but when I substitute this into $x$ my determinant does not equal $0$ so therefore an inverse exists.

My question is: Am I doing something wrong with $-2$, and are there any other numbers that I can substitute into $x$ which would make the inverse not exist?

beep-boop
  • 11,595
intelligere
  • 11

2 Answers2

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$M^{-1}$ does not exist iff $\det M=0$

$\det M=x(x^2-1)*1-3*x$

Solve $x(x^2-1)*1-3*x=0$ to find all the values of $x$ such that $M^{-1}$ does not exist

For $x=-2$ : $M=\begin{bmatrix} (-2)(4-1)&(-2)\\3&1\end{bmatrix}, \quad \det(M)=1*(-6)-(-6)=0 \quad .$

Hippalectryon
  • 7,730
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$M^{-1}$ does not exist $\iff \det(M)=0 \iff \underbrace{x(x^2-1) \cdot 1 -x \cdot 3}_{\dagger} =0 \iff \quad ??$

$\dagger $ Remember that, for a matrix $M=\begin{bmatrix} a&b\\c&d\end{bmatrix}, \quad \det(M)=ad-bc \quad .$

beep-boop
  • 11,595
  • THanks thats what ive been doing. I get -2 in this answer, so -2 is obviously not an inverse since the determinant is not 0. This whats mainly confusing me – intelligere Sep 11 '14 at 22:32
  • @intelligere Solving $\dagger =0,$ I've found that the set of $x$ is ${-2,0,2}$. Can you find this by solving $\dagger =0 $ ? – beep-boop Sep 11 '14 at 22:34
  • @intelligere Check again. Plugging $x=-2$ into the matrix, we find that $\det(M)=0$ (be very careful with calculations). – beep-boop Sep 11 '14 at 22:37
  • When I substitute -2 into this equation so -2(-2^2-1)1 - -23 = I'm getting the determinant is 16. When I solve the equation I get [-2,0,2] too so i'm not sure what i'm doing wrong – intelligere Sep 11 '14 at 22:38
  • Ah right I see what i've done wrong. Thank you for the help – intelligere Sep 11 '14 at 22:41