Question

Which of the following is true about any matrix $A$?

• A. $A=A^T$
• B. $A+I=A$
• C. $A+(-A)=I$
• D. $A+O=A$

Answer 1

Answer: (D)

$A+O=A$

Let $A$ be any $n\times n$ matrix

let $A=\left[\begin{array}{ccccc}{a}_{11}& a_{12}& -& -& a_{1n}\\ a_{21}& a_{22}& -& -& a_{2n}\\ |& |& -& -& |\\ a_{n1}& a_{n2}& -& -& a_{nn}\end{array}\right]$

$⟹$ $A^T=\left[\begin{array}{ccccc}{a}_{11}& a_{21}& -& -& a_{n1}\\ a_{12}& a_{22}& -& -& a_{n2}\\ |& |& -& -& |\\ a_{1n}& a_{2n}& -& -& a_{nn}\end{array}\right]$

$A=A^T$ if and only if $a_{ij}=a_{ji}$

$A=A^T$ if and only if $A$ is symmetric.

$A+I=A$ if and only if $a_{ii}+1=a_{ii}$

if and only if $1=0$, which is not true.

$A+(-A)=I$ if and only if $a_{ii}-a_{ii}=1$

if and only if $0=1$, which is not true.

$A+O=A+O_{n\times n}=$$\left[\begin{array}{ccccc}{a}_{11}+0& a_{12}+0& -& -& a_{1n}+0\\ a_{21}+0& a_{22}+0& -& -& a_{2n}+0\\ |& |& -& -& |\\ a_{n1}+0& a_{n2}+0& -& -& a_{nn}+0\end{array}\right]$$=\left[\begin{array}{ccccc}{a}_{11}& a_{12}& -& -& a_{1n}\\ a_{21}& a_{22}& -& -& a_{2n}\\ |& |& -& -& |\\ a_{n1}& a_{n2}& -& -& a_{nn}\end{array}\right]=A$

Hence, $A+O=A$ is true for any matrix.