
In the homogeneous system of three unknowns, $\rho \left( A \right)=$ the number of unknowns. The system
[a] has only a trivial solution
[b] Reduces to two equations and has infinitely many solutions
[c] Reduces to a single equation and has finitely many solutions
[d] is inconsistent
Answer
595.5k+ views
Hint: Recall the definition of the rank of a matrix and check what happens when in a homogeneous system of equations the number of unknowns equals the rank of the matrix. Recall that a homogeneous system of equations has a trivial solution where all the unknowns are set to be equal to 0. Hence claim that the system cannot be inconsistent and hence may have infinitely many solutions or just one solution. Check which of these are true when the rank is equal to the number of unknowns.
Complete step-by-step answer:
Before solving the question, we need to understand what rank of a matrix means.
Consider any matrix A. We apply row transformations to the matrix so that it converts to row reduced echelon form.
A matrix R is in a row reduced echelon form if it satisfies the following three properties:
[i] R is row reduced
[ii] Every row of R which has all its entries 0 occurs below every row which has a non-zero entry.
[iii] The leading entry of row with smaller index occurs in a column with a smaller index than the leading entry of a row with larger index, i.e. if ${{a}_{ij}}$ is the leading entry of one row and ${{a}_{i'j'}}$ is the leading entry of another row, then $i < i' \Rightarrow j < j'$
The rank of a matrix is the number of non-zero rows in row-reduced echelon form of the matrix.
Property: If a matrix A has rank r, then then the row reduced echelon form of A is \[\left[ \begin{matrix}
{{I}_{r}} & B \\
{{O}_{\left( n-r \right)\times r}} & C \\
\end{matrix} \right]\]
Also if the number of unknowns is n and the number of equations is m, then the matrix of the system has dimensions $m\times n$.
Now clearly the rank of a matrix is at most equal to the min(m,n).
Hence if rank(A) = n(The number of unknowns), we must have $m\ge n$.
Hence the row reduced echelon form of matrix A is $\left[ \begin{matrix}
{{I}_{n}} \\
O \\
\end{matrix} \right]$
Hence if we solve the system of the equations, we must have ${{x}_{1}}={{x}_{2}}=\cdots ={{x}_{n}}=0$ , and hence the system of the equations has a unique solution which is the trivial solution.
Hence option [a] is correct.
Note: The above result can also be understood as follows.
Since rank(A) = n, the number of Linearly independent equations is equal to the number of unknowns.
If we take only those equations, the system is equivalent to ${{I}_{n}}X=0$ (Because row reduced form of a non-singular square matrix is the identity matrix and since all the rows of the chosen system are linearly independent, the matrix of the given system is non-singular) and hence $X=0$. All the remaining equations will automatically be satisfied as they are linear combinations of the chosen equations. Hence the system has exactly one solution, which is the trivial solution.
Complete step-by-step answer:
Before solving the question, we need to understand what rank of a matrix means.
Consider any matrix A. We apply row transformations to the matrix so that it converts to row reduced echelon form.
A matrix R is in a row reduced echelon form if it satisfies the following three properties:
[i] R is row reduced
[ii] Every row of R which has all its entries 0 occurs below every row which has a non-zero entry.
[iii] The leading entry of row with smaller index occurs in a column with a smaller index than the leading entry of a row with larger index, i.e. if ${{a}_{ij}}$ is the leading entry of one row and ${{a}_{i'j'}}$ is the leading entry of another row, then $i < i' \Rightarrow j < j'$
The rank of a matrix is the number of non-zero rows in row-reduced echelon form of the matrix.
Property: If a matrix A has rank r, then then the row reduced echelon form of A is \[\left[ \begin{matrix}
{{I}_{r}} & B \\
{{O}_{\left( n-r \right)\times r}} & C \\
\end{matrix} \right]\]
Also if the number of unknowns is n and the number of equations is m, then the matrix of the system has dimensions $m\times n$.
Now clearly the rank of a matrix is at most equal to the min(m,n).
Hence if rank(A) = n(The number of unknowns), we must have $m\ge n$.
Hence the row reduced echelon form of matrix A is $\left[ \begin{matrix}
{{I}_{n}} \\
O \\
\end{matrix} \right]$
Hence if we solve the system of the equations, we must have ${{x}_{1}}={{x}_{2}}=\cdots ={{x}_{n}}=0$ , and hence the system of the equations has a unique solution which is the trivial solution.
Hence option [a] is correct.
Note: The above result can also be understood as follows.
Since rank(A) = n, the number of Linearly independent equations is equal to the number of unknowns.
If we take only those equations, the system is equivalent to ${{I}_{n}}X=0$ (Because row reduced form of a non-singular square matrix is the identity matrix and since all the rows of the chosen system are linearly independent, the matrix of the given system is non-singular) and hence $X=0$. All the remaining equations will automatically be satisfied as they are linear combinations of the chosen equations. Hence the system has exactly one solution, which is the trivial solution.
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