Question

If $A_n$ is a $n\times n$ matrix in which the diagonal elements are $1,2,\mathrm{3.....}n$ (i.e., $a_{11}=1a_{22}=2,.......a_{ii}=i,......a_{nn}=n)$ and all other elements are equal to $n$, then

• A. $A_n$ is singular for all n
• B. $A_n$ is non singular for all n
• C. det $A_n$ = $(-1{)}^{n}n!$
• D. det $A_n$ = $(-1{)}^{n+1}n!$

Answer 1

Answer: (B), (D)

The correct options are
B $A_n$ is non singular for all n
D det $A_n$ = $(-1{)}^{n+1}n!$

For $n=2$,
$A_2=\left[\begin{array}{cc}1& 2\\ 2& 2\end{array}\right]$
$|A_2|=-2\ne 0$
$\Rightarrow |A_2|=(-1{)}^{2+1}2!$
For $n=3$,
$A_3=\left[\begin{array}{ccc}1& 3& 3\\ 3& 2& 3\\ 3& 3& 3\end{array}\right]$
$|A_3|=6\ne 0$
$\Rightarrow |A_3|=(-1{)}^{3+1}3!$
For $n=4$,
$A_4=\left[\begin{array}{cccc}1& 4& 4& 4\\ 4& 2& 4& 4\\ 4& 4& 3& 4\\ 4& 4& 4& 4\end{array}\right]$
$|A_4|=-24\ne 0$
$\Rightarrow |A_4|=(-1{)}^{4+1}4!$
Hence, $A_n$ is non-singular for all n.
and $|A_n|=(-1{)}^{n+1}n!$