TIFR CSE 2010 | Part A | Question: 16

Question

Let the characteristic equation of matrix $M$ be $\lambda ^{2} - \lambda - 1 = 0$. Then.

• A. $M^{-1}$ does not exist.
• B. $M^{-1}$ exists but cannot be determined from the data.
• C. $M^{-1} = M + I$
• D. $M^{-1} = M - I$
• E. $M^{-1}$ exists and can be determined from the data but the choices (c) and (d) are incorrect.

Answer 1

We can solve using Cayley- Hamilton Theorem

$λ^2−λ−1=0$
$\implies M^2 - M -I = 0$
$\implies I= M^2 - M$

Now premultiplying by $M^{-1}$

$M^{-1}I=M^{-1} M^2 - M^{-1}M$  
$M^{-1} = M-I$

Correct Answer: $D$

Comments

@arjun sir , How is option C is true ?

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Option (D) is true.

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Just an addition ..correct me if i am wrong ,

M will be non-singular matrix also, and hence ${M}'$ exists

proof : For $\lambda$ = 0 ..characteristic equation does'nt satisfy ==> 0 is not eigen value of M

==>Product of eigen values $\lambda \neq$0 ==> Determinant is non equal to Zero

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Yeah that's right

Given $\lambda^{2}-\lambda-1=0$

     Compare with the equation

   $x^{2}-(\lambda_{1}+\lambda_{2})x+\lambda_{1}\lambda_{2}=0$

$\lambda_{1}\lambda_{2}=Det(M)=-1$

Eigen values are $\lambda_{1}=\frac{1+\sqrt{5}}{2}$ and $\lambda_{2}=\frac{1-\sqrt{5}}{2}$

$(1)M^{-1}$ exist becasue $|M|\neq0$

$M^{-1}=\frac{adj(M)}{|M|}$

if $|M|=0$ entire expreesion made $\infty.$

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what about option $(b)?$
We can't find the $M^{-1} $, because data is insufficient.

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saying pre-multiplying by $M^{-1}$ is technically incorrect because we don’t know matrix M is invertible.

If $AB=I$ then $A^{-1}=B$ and $B^{-1} = A$ for squared matrices A and B of same size.

Since, Cayley- Hamilton theorem is for square matrices, so M should be a square matrix and I is already a square matrix. Hence, M-I will also be a square matrix and both M and M-I will be of the same size.

So, here, $I= M^2-M = M(M-I)$ which means $M^{-1} = M- I $ and $(M-I)^{-1} = M$

Another way to say, M is invertible as :

$I = M(M-I)$

$ \det(I) = \det(M(M-I))$

$ \det(I) = \det(M) \det(M-I)$

$ 1 = \det(M) \det(M-I)$

So, $ \det(M) \neq 0$ which implies M is invertible.