
The number of $3\times 3$ matrices whose entries are either 0 or 1 and for which the system $A=\left[ \begin{matrix}
x \\
y \\
z \\
\end{matrix} \right]=\left[ \begin{matrix}
1 \\
0 \\
0 \\
\end{matrix} \right]$ has exactly two distinct solutions?
A. 0
B. ${{2}^{9}}-1$
C. 168
D. 2
Answer
509.1k+ views
Hint: We will first form a matrix of order $3\times 3$ and then we will form the equations of three planes. It is also noted that the three planes cannot intersect at two distinct points. Then, we will select the most appropriate answer option to get the answer.
Complete step-by-step answer:
It is given in the question that we have to find the number of $3\times 3$ matrices whose entries are either 0 or 1 and for which the system $A=\left[ \begin{matrix}
x \\
y \\
z \\
\end{matrix} \right]=\left[ \begin{matrix}
1 \\
0 \\
0 \\
\end{matrix} \right]$ has exactly two distinct solutions. So, let us consider a matrix A of the order $3\times 3$. So, we can write it as,
\[A=\left[ \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right]\]
Here, we have ${{a}_{i}},{{b}_{i}},{{c}_{i}}$ where i = 1,2,3 have the values 0 or 1. Therefore, we will get the equations as,
\[\begin{align}
& {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z=0 \\
& {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z=0 \\
& {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z=0 \\
\end{align}\]
If we observe all the above three equations then, they are representing the three distinct planes. However it is noticed that three planes cannot intersect at two different points.
Therefore the number of $3\times 3$ matrices whose entries are either 0 or 1 and for which the system $A=\left[ \begin{matrix}
x \\
y \\
z \\
\end{matrix} \right]=\left[ \begin{matrix}
1 \\
0 \\
0 \\
\end{matrix} \right]$ has exactly two distinct solutions.
Thus option (A) is the correct answer.
Note: Some students may do the calculations by substituting the values of x, y and z in all the three equations to find the matrix of order $3\times 3$but if we know the concept of three planes that, three planes cannot intersect at any two distinct points, then this step can be skipped and we can directly write the answer. So, the students can use such tricks in order to solve the questions quickly and save time in exams.
Complete step-by-step answer:
It is given in the question that we have to find the number of $3\times 3$ matrices whose entries are either 0 or 1 and for which the system $A=\left[ \begin{matrix}
x \\
y \\
z \\
\end{matrix} \right]=\left[ \begin{matrix}
1 \\
0 \\
0 \\
\end{matrix} \right]$ has exactly two distinct solutions. So, let us consider a matrix A of the order $3\times 3$. So, we can write it as,
\[A=\left[ \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right]\]
Here, we have ${{a}_{i}},{{b}_{i}},{{c}_{i}}$ where i = 1,2,3 have the values 0 or 1. Therefore, we will get the equations as,
\[\begin{align}
& {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z=0 \\
& {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z=0 \\
& {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z=0 \\
\end{align}\]
If we observe all the above three equations then, they are representing the three distinct planes. However it is noticed that three planes cannot intersect at two different points.
Therefore the number of $3\times 3$ matrices whose entries are either 0 or 1 and for which the system $A=\left[ \begin{matrix}
x \\
y \\
z \\
\end{matrix} \right]=\left[ \begin{matrix}
1 \\
0 \\
0 \\
\end{matrix} \right]$ has exactly two distinct solutions.
Thus option (A) is the correct answer.
Note: Some students may do the calculations by substituting the values of x, y and z in all the three equations to find the matrix of order $3\times 3$but if we know the concept of three planes that, three planes cannot intersect at any two distinct points, then this step can be skipped and we can directly write the answer. So, the students can use such tricks in order to solve the questions quickly and save time in exams.
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