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: First of all, get all the possible \[3\times 3\] non-singular matrices whose three entries are one and the remaining entries are zero. The possible \[3\times 3\] non- singular matrices are,
\[A=\left[ \begin{align}
  & \begin{matrix}
   1 & 0 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 1 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 0 & 1 \\
\end{matrix} \\
\end{align} \right]\] , \[B=\left[ \begin{align}
  & \begin{matrix}
   0 & 1 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 0 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 0 & 1 \\
\end{matrix} \\
\end{align} \right]\] , \[C=\left[ \begin{align}
  & \begin{matrix}
   1 & 0 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 0 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 1 & 0 \\
\end{matrix} \\
\end{align} \right]\] , \[D=\left[ \begin{align}
  & \begin{matrix}
   0 & 0 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 1 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 0 & 0 \\
\end{matrix} \\
\end{align} \right]\] , \[E=\left[ \begin{align}
  & \begin{matrix}
   0 & 1 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 0 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 0 & 0 \\
\end{matrix} \\
\end{align} \right]\] , and
\[F=\left[ \begin{align}
  & \begin{matrix}
   0 & 0 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 0 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 1 & 0 \\
\end{matrix} \\
\end{align} \right]\] . To get four entries as 1, we have to replace any zero out of six zeros by 1 in matrix A. For matrix A, the number of ways to replace any zero by one so that we can get four entries of the matrix as 1 is equal to 6. For one matrix we have six ways to get four entries as one. Here, we have six matrices. So, the total number of ways to obtain four entries as 1 = \[6\times 6=36\] . Now, solve it further and conclude the answer.

Complete step-by-step answer:
According to the question we have \[3\times 3\] non-singular matrices.
It means that the determinant value of the matrix cannot be equal to zero. So, the matrix must not have all elements of its row or column equal to zero.
Let us take the case in which we have three entries of the \[3\times 3\] non- singular matrix equal to one and the remaining entries equal to zero.
The total number of entries of a \[3\times 3\] non- singular matrix = 9
The number of entries as zero = 3 ……………………………………………(1)
The number of entries as one = \[9-3=6\] …………………………………………..(2)
Since the matrix is non-singular so, it must not have its determinant value equal to zero. Therefore, the matrix must not have all the elements of a row or column equal to zero.
The possible \[3\times 3\] non- singular matrices are,
\[A=\left[ \begin{align}
  & \begin{matrix}
   1 & 0 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 1 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 0 & 1 \\
\end{matrix} \\
\end{align} \right]\] , \[B=\left[ \begin{align}
  & \begin{matrix}
   0 & 1 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 0 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 0 & 1 \\
\end{matrix} \\
\end{align} \right]\] , \[C=\left[ \begin{align}
  & \begin{matrix}
   1 & 0 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 0 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 1 & 0 \\
\end{matrix} \\
\end{align} \right]\] , \[D=\left[ \begin{align}
  & \begin{matrix}
   0 & 0 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 1 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 0 & 0 \\
\end{matrix} \\
\end{align} \right]\] , \[E=\left[ \begin{align}
  & \begin{matrix}
   0 & 1 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 0 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 0 & 0 \\
\end{matrix} \\
\end{align} \right]\] , and
\[F=\left[ \begin{align}
  & \begin{matrix}
   0 & 0 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 0 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 1 & 0 \\
\end{matrix} \\
\end{align} \right]\] .
The matrices A, B, C, D, E, and F are the possible \[3\times 3\] non- singular matrix which has three entries as one and remaining as zero.
We have to find the number of \[3\times 3\] non-singular matrices, with four entries as 1 and all other entries as 0.
We have to add one more entry as one in matrices A, B, C, D, E, and F.
In the matrix A, we have,
\[A=\left[ \begin{align}
  & \begin{matrix}
   1 & 0 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 1 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 0 & 1 \\
\end{matrix} \\
\end{align} \right]\]
Here, we have six entries as zero. So, to obtain four entries as zero, we have to replace any of the six zeros by one.
The number of ways to replace any zero by one so that we can get four entries of the matrix as 1 = 6.
For one matrix we have six ways to get four entries as one. Here, we have six matrices.
So, the total number of ways to obtain four entries as 1 = \[6\times 6=36\] .
Therefore, the number of \[3\times 3\] non-singular matrices, with four entries as 1 and all other entries as 0, is 36.
For one matrix we have six possible ways to make four entries of a matrix equal to 1.
So, there must be at least 7 ways to get the matrix whose four entries are 1 and the remaining entries are zero.
Hence, the correct option is (C).

Note: In this question, one might take the matrix \[\left[ \begin{align}
  & \begin{matrix}
   0 & 1 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 1 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 0 & 1 \\
\end{matrix} \\
\end{align} \right]\] as a \[3\times 3\] non-singular matrix. This is wrong because the determinant value of the matrix \[\left[ \begin{align}
  & \begin{matrix}
   0 & 1 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 1 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 0 & 1 \\
\end{matrix} \\
\end{align} \right]\] is equal to zero. But we know that the determinant value of a non-singular matrix cannot be equal to zero. So, we cannot take the matrix \[\left[ \begin{align}
  & \begin{matrix}
   0 & 1 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 1 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 0 & 1 \\
\end{matrix} \\
\end{align} \right]\] as a \[3\times 3\] non-singular matrix.