3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $\det A = ad-bc = -1$.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.