linear algebra $\det(I+A)$

Question

Let $A$ be a square matrix of order 2 such that $AA^T=I$. If $\det(A) < 0$, find $\det(I+A)$.

I managed to find $\det(A)$ which is $-1$, but I do not know how to proceed from here. Help is much appreciated.

By "order 2" you mean the matrix is $2 \times 2$ or that the rank is $2$? — Simon S, Sep 18 '15 at 12:48

Ben Grossmann · Accepted Answer · 2015-09-18T12:53:32.610

3

Note that $$ \det(I+A)= \det(A^TA + A)= \det(A^T + I)\det(A)=\\ \det(I+A)\det(A) $$

edited Sep 18 '15 at 12:53

answered Sep 18 '15 at 12:48

Ben Grossmann

225,327

i havent learn about eigenvalues yet. Still new to linear algebra – hazard Sep 18 '15 at 12:49
See my updated answer. Feel free to ask about any steps here. – Ben Grossmann Sep 18 '15 at 12:54
oh thanks bro, so det(I+a) has to be zero.. my only question why is det(A^t+I)=det(A+I) – hazard Sep 18 '15 at 13:01
2

For any two matrices $n\times n$ matrices $C$ and $B$ $(C+B)^T = C^T+B^T$. We also know that for any matrix $\mathrm{det}(A^T) = \text{det}(A)$. From this you can get that $\mathrm{det}(A^T+I) = \text{det}((A^T+I)^T) = \mathrm{det}(A+I^T) = \text{det}(A+I) $. – Kenneth Goodenough Sep 18 '15 at 13:17
@Ken thanks for stepping in there. – Ben Grossmann Sep 18 '15 at 16:50

score 0 · Answer 2 · answered Sep 18 '15 at 13:21

0

$AA^{t}=I$ gives that matrix $A$ is symmetric and orthogonal matrix. Therefore condition det$A<0$ gives, eigen values of $A$ are $1$ and $-1$ and hence that of $I+A$ are $0$ and $2$. So $$|I+A|=0.$$

answered Sep 18 '15 at 13:21

neelkanth

6,048
2
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linear algebra $\det(I+A)$

2 Answers2