Question

Let $P=\left[a_{ij}\right]$ be a $3\times 3$ invertible matrix, where $a_{ij}\in \{0,1\}$ for $1\le i,j\le 3$ and exactly four elements of $P$ are $1$. If $N$ denotes the number of such possible matrices $P$, then which of the following is/are true?

• A. Number of divisors of $N$ is even.
• B. Sum of divisors of $N$ is $91$.
• C. Determinant of $\text{adj}\left(P\right)$ can be $-1$.
• D. Determinant of $\text{adj}\left(P\right)$ can be $1$.

Answer 1

Answer: (B), (D)

Determinant of $\text{adj}\left(P\right)$ can be $1$.

We know that, determinant of an invertible matrix is non-zero.
Since, each elements of matrix $P$ is $0$ or $1.$ Hence, there must be atleast a $1$ in each row and each column. So, possible cases are
$\left(i\right)P=\left[\begin{array}{ccc}1& -& -\\ -& 1& -\\ -& -& 1\end{array}\right]\equiv 6\text{matrices for 4th}1$

$\left(ii\right)P=\left[\begin{array}{ccc}1& -& -\\ -& -& 1\\ -& 1& -\end{array}\right]\equiv 6\text{matrices}$

$\left(iii\right)P=\left[\begin{array}{ccc}-& 1& -\\ 1& -& -\\ -& -& 1\end{array}\right]\equiv 6\text{matrices}$

$\left(iv\right)P=\left[\begin{array}{ccc}-& 1& -\\ -& -& 1\\ 1& -& -\end{array}\right]\equiv 6\text{matrices}$

$\left(v\right)P=\left[\begin{array}{ccc}-& -& 1\\ -& 1& -\\ 1& -& -\end{array}\right]\equiv 6\text{matrices}$

$\left(vi\right)P=\left[\begin{array}{ccc}-& -& 1\\ 1& -& -\\ -& 1& -\end{array}\right]\equiv 6\text{matrices}$

$\therefore N=36=2^2\cdot 3^2$
Divisors of $N$ are
$1,2,3,4,6,9,12,18,36.$

Number of divisors $=(2+1)(2+1)=9$

Sum of divisiors $=91$

$\left|\text{adj}\right(P\left)\right|={\left|P\right|}^{3-1}=(\pm 1{)}^2=1$


Alternate method :
Finding $N$ by using permutation and combination
Choosing one row, so the possible ways $={}^3{C}_1$
Choosing one column, so the possible ways $={}^2{C}_1$
Third one is fixed.
Now, there are $6$ places for the fourth $1$, so
$N=3\times 2\times 6=36$