Question

: If a $2\times 2$ matrix takes only 0 and 1 as elements then show all possible matrices whose det values are
i) 1
ii) 0

Answer 1

Hint: We solve this problem by first finding all the possible $2\times 2$ matrices that can be formed with only 0 and 1 as elements. We divide them into four cases that is with all 0’s, one 0 and three 1’s, two 0’s and all 1’s. Then we find which of them has determinant as zero by going through them in all the cases and then consider the remaining matrices and find which of them has determinant 1.

Complete step by step answer:
We are given that a $2\times 2$ matrix takes only values 0 and 1 as elements.
In a $2\times 2$ matrix, we have 4 elements. As they all take only 0 and 1 as elements, that is only 2 values, the number of possible matrices are,
$\begin{align}
  & \Rightarrow 2\times 2\times 2\times 2 \\
 & \Rightarrow 16 \\
\end{align}$
So, there are 16 possible matrices.
Matrices with four 0’s are
$\left[ \begin{matrix}
   0 & 0 \\
   0 & 0 \\
\end{matrix} \right]$
Matrices with only one 1 and three 0’s are
$\left[ \begin{matrix}
   1 & 0 \\
   0 & 0 \\
\end{matrix} \right],\left[ \begin{matrix}
   0 & 1 \\
   0 & 0 \\
\end{matrix} \right],\left[ \begin{matrix}
   0 & 0 \\
   1 & 0 \\
\end{matrix} \right],\left[ \begin{matrix}
   0 & 0 \\
   0 & 1 \\
\end{matrix} \right]$
Matrices with only two 1 and two 0’s are
$\left[ \begin{matrix}
   1 & 1 \\
   0 & 0 \\
\end{matrix} \right],\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right],\left[ \begin{matrix}
   1 & 0 \\
   1 & 0 \\
\end{matrix} \right],\left[ \begin{matrix}
   0 & 1 \\
   1 & 0 \\
\end{matrix} \right],\left[ \begin{matrix}
   0 & 1 \\
   0 & 1 \\
\end{matrix} \right],\left[ \begin{matrix}
   0 & 0 \\
   1 & 1 \\
\end{matrix} \right]$
Matrices with only one 0 and three 1’s are
$\left[ \begin{matrix}
   0 & 1 \\
   1 & 1 \\
\end{matrix} \right],\left[ \begin{matrix}
   1 & 0 \\
   1 & 1 \\
\end{matrix} \right],\left[ \begin{matrix}
   1 & 1 \\
   1 & 0 \\
\end{matrix} \right],\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]$
Matrices with four 0’s are
$\left[ \begin{matrix}
   1 & 1 \\
   1 & 1 \\
\end{matrix} \right]$

(i)
Now let us find the matrices whose determinants are zero. Then we get,

For matrices with all zero, determinant is
$\Rightarrow \left| \begin{matrix}
   0 & 0 \\
   0 & 0 \\
\end{matrix} \right|=0\left( 0 \right)-0\left( 0 \right)=0$
For matrices with one 1 and three 0’s determinants, we will always get the determinant as zero because each time to get the determinant a 1 should be multiplied with 0 and we get it 0 and subtracting from the product of two zeroes, we get the determinant as 0.
So, for all matrices $\left[ \begin{matrix}
   1 & 0 \\
   0 & 0 \\
\end{matrix} \right],\left[ \begin{matrix}
   0 & 1 \\
   0 & 0 \\
\end{matrix} \right],\left[ \begin{matrix}
   0 & 0 \\
   1 & 0 \\
\end{matrix} \right],\left[ \begin{matrix}
   0 & 0 \\
   0 & 1 \\
\end{matrix} \right]$, their determinant is 0.

Now let us consider the matrices with two zeroes and two ones.
In all the cases when the 1’s are adjacent to each other in those matrices, their determinant is zero because each time one is multiplied with zero, we get zero. Then subtracting one zero from other we get zero.
 So, for the matrices $\left[ \begin{matrix}
   1 & 1 \\
   0 & 0 \\
\end{matrix} \right],\left[ \begin{matrix}
   1 & 0 \\
   1 & 0 \\
\end{matrix} \right],\left[ \begin{matrix}
   0 & 1 \\
   0 & 1 \\
\end{matrix} \right],\left[ \begin{matrix}
   0 & 0 \\
   1 & 1 \\
\end{matrix} \right]$, their determinant is zero.

Now let us consider the matrices with one zeroes and three ones.
When there are three 1’s and only 1 zero, we cannot get determinant as zero.

Now let us consider the matrices with all ones.
For matrices with all ones, determinant is
$\Rightarrow \left| \begin{matrix}
   1 & 1 \\
   1 & 1 \\
\end{matrix} \right|=1\left( 1 \right)-1\left( 1 \right)=1-1=0$
So, all the possible matrices for which det is zero are
$\left[ \begin{matrix}
   0 & 0 \\
   0 & 0 \\
\end{matrix} \right],\left[ \begin{matrix}
   1 & 0 \\
   0 & 0 \\
\end{matrix} \right],\left[ \begin{matrix}
   0 & 1 \\
   0 & 0 \\
\end{matrix} \right],\left[ \begin{matrix}
   0 & 0 \\
   1 & 0 \\
\end{matrix} \right],\left[ \begin{matrix}
   0 & 0 \\
   0 & 1 \\
\end{matrix} \right],\left[ \begin{matrix}
   1 & 1 \\
   0 & 0 \\
\end{matrix} \right],\left[ \begin{matrix}
   1 & 0 \\
   1 & 0 \\
\end{matrix} \right],\left[ \begin{matrix}
   0 & 1 \\
   0 & 1 \\
\end{matrix} \right],\left[ \begin{matrix}
   0 & 0 \\
   1 & 1 \\
\end{matrix} \right],\left[ \begin{matrix}
   1 & 1 \\
   1 & 1 \\
\end{matrix} \right]$

The remaining matrices that do not have their determinant as 0 are,
$\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right],\left[ \begin{matrix}
   0 & 1 \\
   1 & 0 \\
\end{matrix} \right],\left[ \begin{matrix}
   0 & 1 \\
   1 & 1 \\
\end{matrix} \right],\left[ \begin{matrix}
   1 & 0 \\
   1 & 1 \\
\end{matrix} \right],\left[ \begin{matrix}
   1 & 1 \\
   1 & 0 \\
\end{matrix} \right],\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]$

Now let us find the determinants of above matrices.
$\begin{align}
  & \left| \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right|=1\left( 1 \right)-0\left( 0 \right)=1-0=1 \\
 & \left| \begin{matrix}
   0 & 1 \\
   1 & 0 \\
\end{matrix} \right|=0\left( 0 \right)-1\left( 1 \right)=0-1=-1 \\
 & \left| \begin{matrix}
   0 & 1 \\
   1 & 1 \\
\end{matrix} \right|=1\left( 0 \right)-1\left( 1 \right)=0-1=-1 \\
\end{align}$
$\begin{align}
  & \left| \begin{matrix}
   1 & 0 \\
   1 & 1 \\
\end{matrix} \right|=1\left( 1 \right)-0\left( 1 \right)=1-0=1 \\
 & \left| \begin{matrix}
   1 & 1 \\
   1 & 0 \\
\end{matrix} \right|=0\left( 1 \right)-1\left( 1 \right)=0-1=-1 \\
 & \left| \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right|=1\left( 1 \right)-0\left( 1 \right)=1-0=1 \\
\end{align}$
So, from above the matrices with determinant 1 are $\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right],\left[ \begin{matrix}
   1 & 0 \\
   1 & 1 \\
\end{matrix} \right],\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]$.
So, from above we can say that,
Hence answer for sub part (i) is $\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right],\left[ \begin{matrix}
   1 & 0 \\
   1 & 1 \\
\end{matrix} \right],\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]$.

Hence answer for the sub part (ii) is $\left[ \begin{matrix}
   0 & 0 \\
   0 & 0 \\
\end{matrix} \right]$, $\left[ \begin{matrix}
   1 & 0 \\
   0 & 0 \\
\end{matrix} \right]$, $\left[ \begin{matrix}
   0 & 1 \\
   0 & 0 \\
\end{matrix} \right]$, $\left[ \begin{matrix}
   0 & 0 \\
   1 & 0 \\
\end{matrix} \right]$, $\left[ \begin{matrix}
   0 & 0 \\
   0 & 1 \\
\end{matrix} \right]$, $\left[ \begin{matrix}
   1 & 1 \\
   0 & 0 \\
\end{matrix} \right]$, $\left[ \begin{matrix}
   1 & 0 \\
   1 & 0 \\
\end{matrix} \right]$, $\left[ \begin{matrix}
   0 & 1 \\
   0 & 1 \\
\end{matrix} \right]$, $\left[ \begin{matrix}
   0 & 0 \\
   1 & 1 \\
\end{matrix} \right]$ and $\left[ \begin{matrix}
   1 & 1 \\
   1 & 1 \\
\end{matrix} \right]$.

Note:
We can also solve this question in an alternative process by finding determinants of all the matrices and checking which of them has a determinant as 0, but it is a long process and takes a lot of time to solve. So, it is better to follow the above process and find which determinant is 0 and it's easy to find the matrices whose det value is 1.