7

If $A$ is $2\times2 $ matrix such that $\operatorname{tr} A=\det A=3$ then trace of $A^{-1}$ is?

$(A) \quad 1 \qquad (B) \quad \dfrac{1}{3} \qquad (C) \quad \dfrac{1}{6} \qquad (D) \quad\dfrac{1}{2}$

I did it in this way:

$$\lambda_1+\lambda_2 = 3 $$ $$\lambda_1\cdot\lambda_2=3$$ $$\frac{1}{\lambda_1}+\frac{1}{\lambda_2} \implies \frac{\lambda_1+\lambda_2}{\lambda_1\lambda_2} \implies \frac{3}{3}=1$$

I am practicing these types of question and after $5$ minutes of digging I came up with this answer. I wanna know if there is any alternative approach to solve this problem.

Ng Chung Tak
  • 18,990
Daman
  • 2,138

5 Answers5

6

Use the following nice and easy formula: $$\begin{pmatrix} a & b\\ c & d \end{pmatrix}^{-1}=\frac{1}{ad-bc}\begin{pmatrix} d & -b\\ -c & a \end{pmatrix}.$$

Fred
  • 77,394
5

The characteristic Polynomial for $A^{-1}$ for any invertible $n \times n$ matrix is $$P_{A^{-1}}(X)=\det(xI-A^{-1})=\det(A^{-1}) \det(xA-I)=x^n\det(A^{-1})\det(A-\frac{1}{x}I)\\=(-1)^nx^n \det(A^{-1}) P_{A}(\frac{1}{x})$$

Now use the fact that for a $2\times 2$ matrix the characteristic polynomial is $$P_B(x)=x^2-\operatorname{tr}(B)x+\det(B)$$

Michael Hardy
  • 1
N. S.
  • 132,525
2

The eigenvalues $\lambda_1$ and $\lambda_2$ are the roots of the characteristic polynomial $\det(A-\lambda 1)=a\lambda^2+b\lambda+c$, where $a,b,c$ are real numbers. It is easy to see that $a=1$. Being the polynomial second degree, the sum of its roots is $\lambda_1+\lambda_2=-b/a=3$ and their product $\lambda_1\cdot\lambda_2=c/a=3$. Then the characteristic polynomial is $\lambda^2-3\lambda+3$. The roots can be computed by Bhaskara and are equal to $\frac{3\pm i\sqrt{3}}{2}$. The trace of $A^{-1}$ is the sum of inverse eigenvalues for $A$: $$ \frac{2}{3+i\sqrt{3}}+\frac{2}{3-i\sqrt{3}}=1. $$

Joca Ramiro
  • 1,589
2

Every matrix satisfies its own characteristic equation so

$A$ satisfies

  • $A^2-\operatorname{tr}(A)A+\det(A)I=0 $

    and $A^{-1}$ satisfies

  • $A^{-2}-\operatorname{tr}(A^{-1})A^{-1}+\det(A^{-1})I=0$.

With multiplying both sides of the first equation by $A^{-2}$ and dividing by $\det(A)$ we can easily obtain form of the second one.

Edit
I.e.

$$A^{-2}A^2-\operatorname{tr}(A)A^{-2}A+\det(A)IA^{-2}=0$$

$$I -\operatorname{tr}(A)A^{-1}+\det(A)A^{-2}=0$$

After changing the order of summands and dividing by $\det(A)$

$$A^{-2}- \frac{1}{\det(A)}\operatorname{tr}(A)A^{-1}+\frac{1}{\det(A)}I=0.$$

Comparing the last equation with the second one $\bullet$ we finally obtain

$$\operatorname{tr}(A^{-1})=\frac{\operatorname{tr}(A)}{\det(A)}$$

Michael Hardy
  • 1
Widawensen
  • 8,172
  • Can you show me calculations also using your method ? – Daman Feb 07 '18 at 12:33
  • @Damn1o1 ok ........ – Widawensen Feb 07 '18 at 12:36
  • 1
    @Damn1o1 Now, everything is clear ? In fact the method leads exactly to your formula but there is no mentioning eigenvalues ( although they are somewhere hidden in the background) Novelty in it is that we are using only matrix equation (hence its potential for higher dimensions).. – Widawensen Feb 07 '18 at 13:19
  • When you write \text{tr} instead of \operatorname{tr} then you won't get proper spacing in things like $3\operatorname{tr} A$ or $3\operatorname{tr}(A)$ unless you add it manually. – Michael Hardy Feb 07 '18 at 18:26
  • @MichaelHardy Thank you Michael, I will try to memorize the tip. – Widawensen Feb 08 '18 at 11:43
1

\begin{align} \lambda_1 + \lambda_2 = 3 \\ \lambda_1\lambda_2 = 3 \end{align} So you get $\lambda_2 = \dfrac 3 {\lambda_1},$ so the first equation above becomes $$ \lambda_1 + \frac 3 \lambda_1 = 3. $$ Multiply both sides by $\lambda_1$ and you have an ordinary quadratic equation.

The answer posted by "N.S." also leads to the same quadratic equation.

Michael Hardy
  • 1