If $A$ is a $2\times 2$ matrix over $\Bbb R$ with $\det(A+I)=1+\det A$ then which is correct:

Question

If $A$ is a $2\times 2$ matrix over $\Bbb R$ with $\det(A+I)=1+\det A$ then which is correct:

• $\det A=0$
• $A=0$
• $\text{trace A}=0$
• $A$ is singular.

let the eigen values of $A $ be $\lambda_1,\lambda_2$ then $\det (A+I)=(1+\lambda_1)(1+\lambda_2)$

Also $\det A=\lambda_1\lambda_2$

then $\det (A+I)=(1+\lambda_1)(1+\lambda_2)=1+\lambda_1\lambda_2\implies \lambda_1+\lambda_2=-1\implies $ trace(A)=-1

so $3$ is gone.

how to choose the right one?

pLease help

Answer 1

You made a mistake in your equation, it should have been $\lambda_1 + \lambda_2 = 0$.

Answer 2

Let $$A=\begin{pmatrix} a &b\\ c & d \end{pmatrix} $$ then $$\det(A+1)=\det \begin{pmatrix} a+1 &b\\ c & d+1 \end{pmatrix}=(a+1)(d+1)-bc $$ on the other hand $$\det(A)+1=\det \begin{pmatrix} a &b\\ c & d \end{pmatrix}+1=ad-bc+1 $$

Since $\det(A+I)=\det A+1$ you can conclude that $a+d=0$, hence $tr A=0$.