Let $A$ be a $2 \times 2$ matrix such that $A = A^{-1}$. The value of $\operatorname{det} (A)$ can be:

  1. $\operatorname{det} (A)=-2$
  2. $\operatorname{det} (A)=-1$
  3. $\operatorname{det} (A)=0$
  4. $\operatorname{det} (A)=2$

My attempt:

$$\begin{bmatrix} a &b \\ c& d \end{bmatrix} = \frac{1}{ad-bc} \times \begin{bmatrix} d &-b \\ -c& a \end{bmatrix}$$

Obviously the determinant cannot be $0$, since an inverse exists, but I'm not sure how to proceed. Any help would be appreciated.

  • $\begingroup$ A different hint. What is the determinant of the inverse of a matrix in terms of the determinant of the original matrix? $\endgroup$ – Mehdi2277 Mar 21 '16 at 9:18

$$\operatorname{det}(A^{-1})=\frac{1}{\operatorname{det}(A)}\overset{A^{-1}=A}\implies \left(\operatorname{det}(A)\right)^2=1\implies \operatorname{det}(A)=\pm1$$ This holds for any $n\times n$ matrix with this property (and not only for $2\times 2$).


Your approach is workable, but as other have answered it's not the easiest approach.

If you want to continue you only have to calculate the determinants of the matrices:

$$\det\begin{pmatrix}a & b \\ c& d\end{pmatrix} = ad - cb$$

$$\det{1 \over ad-bc}\begin{pmatrix}d & -b \\ -c & a\end{pmatrix} =\det\begin{pmatrix}{d \over ad-bc}& {-b \over ad-bc} \\ {-c \over ad-bc} & {a \over ad-bc}\end{pmatrix} = \left({1\over ad-bc}\right)^2\left(da-\left(-c\right)\left(-b\right)\right) = {1\over ad-bc}$$

Now you see that $\det (A^{-1}) = 1/\det A$, then you have that if $A=A^{-1}$ that $\det(A) = 1/\det(A)$ which is a second degree equation with solutions $\det(A) = \pm 1$

Another overly complicated approach might be to use that since $A=A^{-1}$ we have that $A^2 = AA = AA^{-1} = I$ so calculating $A^2$ we get

$$A^2 = \begin{pmatrix}a & b \\ c& d\end{pmatrix}\begin{pmatrix}a & b \\ c& d\end{pmatrix}=\begin{pmatrix}a^2 + bc & (a+d)b \\ (a+d)c & d^+bc\end{pmatrix}$$

Here we see that $a^2+bc = d^2+bc=1$ which implies that $a^2=d^2$ which means $a=\pm d$. If $a=d$ we have that $b=c=0$ which means that the diagonal elements becomes $a^2=d^2=1$ which means $a=\pm1$ and $d=\pm1$ which means the determinant is $\pm 1$. If $a=-d$ the diagonal elements becomes $a^2+bc = d^2 + bc = -ad+bc = 1$, which means that $\det A = ad-bc = -1$.

  • $\begingroup$ I'm a bit confused with the fourth line how you got $(\frac{1}{ad-bc})^{2}$. Please explain. $\endgroup$ – Aspiring Mathlete Mar 21 '16 at 12:06
  • $\begingroup$ @AspiringMathlete It's because if you multiply in the denominator the elements becomes $d/(ad-bc)$, $-b/(ad-bc)$. When you calculates the determinant you will for example multiply $d/(ad-bc)$ with $a/(ad-bc)$. This is what makes for the squaring. $\endgroup$ – skyking Mar 21 '16 at 13:20

Hint: what is the determinant of $A^2$?

  • $\begingroup$ $detA^2=(detA)^2$ $\endgroup$ – Ali Mar 21 '16 at 12:25

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