3

$A = \begin{bmatrix}-3x & 2\\4 & 4\end{bmatrix}$

How would I go about solving for what values of $x$ is the matrix $A$ invertible? I know that if the determinant is 0 the matrix is not invertible. I've looked at many answers on here but all of them seem very complicated and on matrices $3\times 3$ which I haven't learned yet.

Michael Hardy
  • 1
Xylus
  • 33
  • 1
  • 4
  • 1
    what does the determinant of $A$ look like? you already know that it cannot equal zero if it is invertible, so what value of $x$ makes the determinant zero? (then pick any value for $x$ except that one) – costrom Nov 18 '15 at 21:52
  • @costrom How would I efficiently find what value of x makes the determinant equal zero? – Xylus Nov 18 '15 at 21:53
  • 2
    For instance, by writing down the formula, impose it equal to $0$ and solving for $x$... –  Nov 18 '15 at 21:54
  • The determinant in this case is $(-3x)\cdot4 - 4\cdot2$. This is $-12x-8$. So you're asking how to find a value of $x$ for which $-12x-8=0$. Presumably you solved problems like that before hearing of determinants. ${}\qquad{}$ – Michael Hardy Nov 18 '15 at 21:55
  • Ah, I will try now! Thank you very much @MichaelHardy. – Xylus Nov 18 '15 at 21:55

2 Answers2

1

First approach:

The determinant of $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ is given by $ad-bc$. In this case the determinant will be $-12x-8$. If this happens to be zero then matrix is NOT invertible.

Second approach:

A matrix is invertible if and only if the rows(or columns) are independent. The second row is $[4 \,\, 4]=4[1 \,\, 1]$ so for the first row to be dependent, it should also be a multiple of $[1 \,\, 1]$. Said differently, the components in the second row must be equal. Thus $x=-2/3$.

Anurag A
  • 41,067
0

You know that the columns of $A$ are linearly independent (i.e. $A$ is invertible) if and only if $$\alpha \begin{pmatrix}-3x \\ 4\end{pmatrix}+\beta\begin{pmatrix}2 \\ 4\end{pmatrix}=0 \qquad \implies \qquad \alpha=\beta =0$$ We can rewrite it as $$\begin{cases} -3x\alpha+2\beta = 0 \\ 4(\alpha+\beta)=0\end{cases}\qquad \implies \qquad \alpha=\beta =0$$ From the second equation $\alpha=-\beta$, plugging that in the first equation, we get $$ 0=3x\beta+2\beta=(2+3x)\beta \implies \beta =0$$ We note that this relation is true whenever $2+3x\neq 0$. It follows that $A$ is invertible if $x\neq-2/3$. Noting that the columns of $A$ are identical when $x=-2/3$ shows the equivalence.

Surb
  • 55,662