Question

The number of $3 \times 3$ non-singular matrices, with four entries as 1 and all other entries as $0$ is
A. $5$
B. $6$
C. at least $7$
D. less than $4$

Answer 1

Hint: In $3 \times 3$ matrix total number of elements are nine. So out of those nine elements, we have to put four $1$ and five $0$ and for the non-singular matrix determinant is not equal to zero.
If the determinant of the matrix is 0, then the matrix is called a singular matrix.

Complete step-by-step answer:
Here it is given that we have $3 \times 3$ matrix. So as we know that for every $3 \times 3$ matrix, we have a total nine elements and out of which we have five entries as zero and four entries as one and for the non-singular matrix determinant is not equal to zero.
$\Delta \ne 0$
So as the determinant is not equal to zero, every row and column must have at least $1$.
So for three one and remaining zero, we have
$\left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]$,$\left[ {\begin{array}{*{20}{c}}
  0&1&0 \\
  1&0&0 \\
  0&0&1
\end{array}} \right]$,$\left[ {\begin{array}{*{20}{c}}
  0&0&1 \\
  0&1&0 \\
  1&0&0
\end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&0&1 \\
  0&1&0
\end{array}} \right]$,$\left[ {\begin{array}{*{20}{c}}
  0&1&0 \\
  0&0&1 \\
  1&0&0
\end{array}} \right]$,$\left[ {\begin{array}{*{20}{c}}
  0&0&1 \\
  1&0&0 \\
  0&1&0
\end{array}} \right]$
So a total of six cases are possible with three $1$ and remaining $0$, and also the determinant is not equal to zero.
Now we need to find four ones and rest five zero, so we can add $1$ to any position.
Now we get that
$\left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&0&1 \\
  0&1&0
\end{array}} \right]$, now insert one $1$ out of one zero.
Total number of ways is $6$
So from one matrix we can obtain $6$ matrices and a total of $36$ matrices.

So our option C is correct which says at least seven.

Note: If one row or one column does not contain any element other than zero, then its determinant must be zero.
If $A = \left[ {\begin{array}{*{20}{c}}
  0&1&2 \\
  0&3&4 \\
  0&5&6
\end{array}} \right]$$\left[ {\begin{array}{*{20}{c}}
  0&1&2 \\
  0&3&4 \\
  0&5&6
\end{array}} \right]$
Then $\left| A \right| = 0$
As one column contains all zeros. So this must be kept in mind while solving the problem.