4,216 views
21 votes
21 votes

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

  1.  $\begin{bmatrix}
    1-a &-1 \\ 
     a^2& a
    \end{bmatrix}$
  2. $\begin{bmatrix}
    1-a &-1 \\ 
     a^2-a+1& a
    \end{bmatrix}$
  3. $\begin{bmatrix}
    -a &1 \\ 
     -a^2+a-1& 1-a
    \end{bmatrix}$
  4. $\begin{bmatrix}
    a^2-a+1 &a \\ 
     1& 1-a
    \end{bmatrix}$

4 Answers

Best answer
38 votes
38 votes

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)                      

edited by
24 votes
24 votes

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)!

edited by
2 votes
2 votes

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.

0 votes
0 votes
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.
Answer:

Related questions

45 votes
45 votes
3 answers
1
Ishrat Jahan asked Nov 2, 2014
7,890 views
Let $A$ be an $n \times n$ matrix of the following form.$$A = \begin{bmatrix}3&1&0&0&0&\ldots&0&0&0\\1&3&1&0&0&\ldots&0&0&0\\0&1&3&1&0&\ldots&0&0&0\\0&0&1&3&1&\ldots&0&0&...
35 votes
35 votes
4 answers
3
Kathleen asked Sep 18, 2014
9,513 views
In an $M \times N$ matrix all non-zero entries are covered in $a$ rows and $b$ columns. Then the maximum number of non-zero entries, such that no two are on the same row ...
35 votes
35 votes
4 answers
4
Kathleen asked Sep 18, 2014
9,980 views
Let $A, B, C, D$ be $n \times n$ matrices, each with non-zero determinant. If $ABCD = I$, then $B^{-1}$ is $D^{-1}C^{-1}A^{-1}$ $CDA$ $ADC$ Does not necessarily e...