QS-022 · Linear Algebra · Easy
NPTEL GATE DA Mock | Week 1 - 60
Consider the square matrix $A_n$ of order $n$ given below:
$$ A_n = \begin{bmatrix} 0 & 0 & \cdots & 0 & 1\\ 0 & 0 & \cdots & 1 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 1 & \cdots & 0 & 0\\ 1 & 0 & \cdots & 0 & 0 \end{bmatrix} $$
All entries on the anti-diagonal (top-right to bottom-left) are 1, and all other entries are zero.
Which of the following is the determinant of $A_n$?
Note: $\left\lfloor \frac{n}{2} \right\rfloor$ denotes the floor of $\frac{n}{2}$.
A. 1
B. $-1$
C. $(-1)^n$
D. $(-1)^{n+1}$
E. $(-1)^{\left\lfloor\frac{n}{2}\right\rfloor}$
Consider the square matrix $A_n$ of order $n$ given below:
$$ A_n = \begin{bmatrix} 0 & 0 & \cdots & 0 & 1\\ 0 & 0 & \cdots & 1 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 1 & \cdots & 0 & 0\\ 1 & 0 & \cdots & 0 & 0 \end{bmatrix} $$
All entries on the anti-diagonal (top-right to bottom-left) are 1, and all other entries are zero.
Which of the following is the determinant of $A_n$?
Note: $\left\lfloor \frac{n}{2} \right\rfloor$ denotes the floor of $\frac{n}{2}$.
A. 1
B. $-1$
C. $(-1)^n$
D. $(-1)^{n+1}$
E. $(-1)^{\left\lfloor\frac{n}{2}\right\rfloor}$
Solution
The anti-diagonal matrix is obtained from the identity matrix by reversing the order of its columns.
The number of interchanges required is
$$ \frac{n(n-1)}{2}. $$
Each interchange changes the sign of the determinant. Hence,
$$ \det(A_n)=(-1)^{\frac{n(n-1)}{2}}. $$
For even $n$, this exponent is even when $n\equiv 0\pmod 4$ and odd when $n\equiv 2\pmod 4$. Thus, the determinant is not simply positive for every even $n$.
For example,
$$ n=2 \implies \det(A_2)=-1, $$
while
$$ n=4 \implies \det(A_4)=1. $$
So the sign alternates according to $n$ modulo $4$:
$$ \det(A_n)= \begin{cases} 1, & n\equiv 0,1\pmod 4,\\ -1, & n\equiv 2,3\pmod 4. \end{cases} $$
Equivalently,
$$ \boxed{\det(A_n)=(-1)^{\left\lfloor n/2\right\rfloor}} $$
Hence, option E is correct.
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