Matrices and Determinants 2 Question 30
33. If $M$ is a $3 \times 3$ matrix, where $M^{T} M=I$ and $\operatorname{det}(M)=1$, then prove that $\operatorname{det}(M-I)=0$
(2004, 2M)
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Solution:
- Since, $M^{T} M=I$ and $|M|=1$
$ \therefore \quad|M-I|=I M-M^{T} M $
$ \Rightarrow|M-I|=\left(I-M^{T}\right) M=(I-M)^{T}$ $|M|=|I-M| $
$ =(-1)^{3}|M-I|$
$[\because I-M \text { is a } 3 \times 3 \text { matrix }] $
$ =-|M-I| $
$ \Rightarrow \quad 2|M-I|=0 $
$ \Rightarrow \quad|M-I|=0$
