ISI2017-MMA-5

Question

If $A$ is a $2 \times 2$ matrix such that trace $A = det \ A = 3,$ then what is the trace of $A^{-1}$?

• A. $1$
• B. $\left(\dfrac{1}{3}\right)$
• C. $\left(\dfrac{1}{6}\right)$
• D. $\left(\dfrac{1}{2}\right)$

Answer 1

Assuming,

$$A= \begin{bmatrix}a & b\\c & d\end{bmatrix}$$

Which is a $2\times 2$ matrix.

Now, $det \hspace{0.1cm}A $ will be $(ad-bc)$

Given $det \hspace{0.1cm}A$ is $3$

∴$ \color{red}{(ad-bc) = 3}$

What is Trace of a matrix?

$\qquad$The trace of an $n\times n$ square matrix $A$ is defined to be the sum of the elements on the main diagonal (the diagonal from the upper left to the lower right) of $A$, i.e.,

${\displaystyle \operatorname {tr} (A)=\sum _{i=1}^{n}a_{ii}=a_{11}+a_{22}+\dots +a_{nn}}$

where $a_{ii}$ denotes the entry on the $i^{th}$ row and $i^{th}$ column of $A$.

∴Trace of $A = a+d$

Given that Trace of $A = 3$

$∴ \color{red}{ a+d = 3}$

Now, $A^{-1} = \dfrac{1}{det \hspace{0.1cm}A}\begin{bmatrix}d & -b\\ -c & a\end{bmatrix}$

$\qquad\qquad = \dfrac{1}{3}\begin{bmatrix}d & -b\\ -c & a\end{bmatrix}$

$\qquad\qquad= \begin{bmatrix}d/3 & -b/3\\ -c/3 & a/3\end{bmatrix}$

$\color{blue}{\text{∴ Trace of }}$ $\color{blue}{A^{-1}} = \dfrac{d}{3} + \dfrac{a}{3}$

$\qquad\qquad = \dfrac{a+d}{3}$

$\qquad\qquad = \dfrac{3}{3}$ $\qquad \big[ ∵ \color{red}{a+d =3}\big]$

$\qquad\qquad = \color{blue}{1} $

$\color{Violet}{\text{∴  Answer is option}}\color{purple}{ \text{ A) 1}}$

Answer 2

Sum of eigenvalues = Trace of the Matrix

Product of eigenvalues =  Determinant of the Matrix

 

Let the eigenvalues of A are $\lambda _{1} and \lambda _{2}$

 

$\lambda _{1} + \lambda _{2} = 3         ..............eq.1$

$\lambda _{1} * \lambda _{2} = 3  ................eq.2$

 

eigenvalues of $A^{-1}$ are $1/\lambda _{1}$ and $1/\lambda _{2}$

Trace = Sum of eigenvalues

$=  1/\lambda _{1} + 1/\lambda _{2} $

$ = (   \lambda _{1} + \lambda _{2} ) / (\lambda _{1} * \lambda _{2} ) $

  on putting values of  from $eq.1$ and  $eq.2$, we get

 $ = 3/3 $

$ = 1 $

Answer 3

Trace of A^-1=(trace A)/(det A)=3/3=1
(A)

Comments

This formula is not valid for a 3x3 matrix or any other higher order matrix. Its only valid for a 2x2 matrix. Please correct me if I'm wrong here.

If A is a 3x3 matrix, then

Trace of A$^{-1}$= 1/$\lambda _{1}$ + 1/$\lambda _{2}$ + 1/$\lambda _{3}$

This follows from the property "If $\lambda$ is an eigen value of A then 1/$\lambda$ is the eigen value of  A$^{-1}$" (Proof here)