Question

If $M$ is a $3\times 3$ matrix, where $M^{T}M=I$ and $det\left(M\right)=1$ then prove that $det(M-I)=0$

Answer 1

We know that in a determinant, rows can be changed into columns and columns and columns into rows.
Hence if there be a square matrix $A$, then
$det\left(A^T\right)=detA$ or $\left|A^T\right|=\left|A\right|$

Also if $I$ be a square matrix then $I^T=I$ and $\left|I\right|=1$ and ${(A\pm B)}^T=A^T\pm B^T$

Given $M$ is a $3\times 3$ square matrix such that $M^{T}M=I;detM=\left|M\right|=1$

We have to prove that $det(M-I)=0$

Now ${(M-I)}^T=M^T-I^T=M^T-M^{T}M$ $=M^T(I-M)$

Taking determinant on both sides
$\left|{(M-I)}^T\right|=\left|M^T\right||I-M|$ or $|M-I|=\left|M\right||I-M|\because \left|A^T\right|=\left|A\right|$

or $|M-I|=1.{(-1)}^3|M-I|\because \left|M\right|=1$

or $2|M-I|=0\therefore |M-I|=0ordet(M-I)=0$