GATE IT 2004 | Question: 36

Question

If matrix $X = \begin{bmatrix} a & 1 \\ -a^2+a-1 & 1-a \end{bmatrix}$  and $X^2 - X + I = O$ ($I$ is the identity matrix and $O$ is the zero matrix), then the inverse of $X$ is

• A.  $\begin{bmatrix}
  1-a &-1 \\ 
   a^2& a
  \end{bmatrix}$
• B. $\begin{bmatrix}
  1-a &-1 \\ 
   a^2-a+1& a
  \end{bmatrix}$
• C. $\begin{bmatrix}
  -a &1 \\ 
   -a^2+a-1& 1-a
  \end{bmatrix}$
• D. $\begin{bmatrix}
  a^2-a+1 &a \\ 
   1& 1-a
  \end{bmatrix}$

Comments

Putting $a=0$ will make it more simpler to solve.

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For, who are interested to find the value of a ( i know it is not required to answer this question )

they given a eqn,

$λ^2 - λ +1 = 0 $ ===> you can find λ values

then We know that determinant = Product of eigen values ==> Find value of a

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(B)

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Easiest and Best Approach for solving this Question :

Apply the Shortcut of Finding inverse of 2x2 Matrix : - (Figure Below)

Answer 1

Given, $X^2 - X + I  = O$

$\quad \implies   X^2 = X - I$

$\quad \implies (X^{-1}) (X^2 )  = (X^{-1})(X - I )$    (Multiplying $X^{-1}$ on both sides)

$\quad \implies X = I - X^{-1} $

$\quad \implies X^{-1} = I - X$

Option (B)                      

Comments

@Satbir 

$X^{-1} = I - X$

When i subtract the matrix $'X'$ from the identity matrix, then i got the inverse

$X^{-1} = \begin{bmatrix} 1-a &1 \\ -a^{2} + a -1 & a \end{bmatrix}$

which is not equal to the option $(B).$

Please correct me if i'm wrong?

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$\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix} - \begin{bmatrix} a & 1\\ -a^2+a-1&1-a \end{bmatrix} = \begin{bmatrix} 1-a &-1 \\ a^2-a+1& a \end{bmatrix}$

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Ohh sorry, was in the night I just subtract the diagonal elements.

Now I got it, it was just a silly mistake.

Thank you so mcuh@Satbir

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Lakshman sir I too faced the same issue thanks Satbir sir.

Answer 2

It's a very simple question, We need to calculate the inverse of a 2x2 matrix,

Inverse of a matrix $A = A^{-1} = \frac{Adjoint(A)}{|A|}$

Adjoint(A) = [cofactors of A]^T, but for 2x2 matrix we have direct formula:
     
A =  a    b
        c    d         is a 2x2 matrix then

Adjoint of A =   d    -b
                          -c     a
|A| = ad-bc    
So answer is (B)!

Comments

Old but gold approach i mean based on when it was posted

Answer 3

You may not evaluate the given equation $X^{2} - X + I = O$.

Rather start evaluating from the choices given. $X.X^{-1} = I$ and the first element should be 1.

Option A. $\Rightarrow a\cdot (1-a)+1\cdot a^{2} = a$

Option B. $\Rightarrow a\cdot (1-a)+1\cdot (a^{2}-a+1) = 1$

Option C. $\Rightarrow a\cdot (-a)+1\cdot (-a^{2}+a-1) = -2a^{2}+a-1$

Option D. $\Rightarrow a\cdot (a^{2}-a+1)+1\cdot (1) = a^{3}-a^{2}+a+1$

You may verify the full multiplication for option B to see whether it indeed yields the identity matrix.

Answer 4

Putting a=0 both in X and in the options (A),(B),(C) and (D) we see that X.X^-1= I hold goods for option  (B).

Hence option B is correct.

Comments

I also did a = 0, then calculated determinant of the resultant matrix. Determinant of the matrix was 1 so I just took it as inverse of given matrix is equal to adjoint of the matrix. So option B.

Was this correct method or did I simply got lucky?

Thanks I am new to whole gate stuff.

Answer 5

Here ans (I-X) is option B