
Let three matrices given as \[{\text{A = }}\left[ \begin{gathered}
{\text{ - 1 - 2 - 3}} \\
{\text{ 3 4 5}} \\
{\text{ 4 5 6}} \\
\end{gathered} \right]{\text{ , B = }}\left[ \begin{gathered}
1{\text{ - 2}} \\
{\text{ - 1 2}} \\
\end{gathered} \right]\;{\text{and C = }}\left[ \begin{gathered}
2{\text{ 0 0}} \\
{\text{0 2 0}} \\
{\text{0 0 2}} \\
\end{gathered} \right]{\text{ }}{\text{. }}\] If a, b and c respectively denote the ranks of A, B and C then the correct order of these numbers is
A) a < b < c
B) c < b < a
C) b < a < c
D) a < c < b
Answer
595.5k+ views
Hint: In this question we must know that Rank of Matrix = Number of linearly independent rows or columns. Moreover, to solve this type of problems one must have the basic knowledge of matrices.
Complete step-by-step answer:
\[{\text{C = }}\left[ \begin{gathered}
2{\text{ 0 0}} \\
{\text{0 2 0}} \\
{\text{0 0 2}} \\
\end{gathered} \right]\]
It has three linearly independent rows.
Hence Rank of C = 3.
\[{\text{B = }}\left[ \begin{gathered}
1{\text{ - 2}} \\
{\text{ - 1 2}} \\
\end{gathered} \right]\]In this we can see that row 2 is a scalar multiple of row 1 using -1. Here the ${2}^{nd}$ row is dependent on the ${1}^{st}$ row whereas, ${1}^{st}$ row is independent.
Hence Rank of B = 1 (Because it has only one linearly independent row)
\[{\text{A = }}\left[ \begin{gathered}
{\text{ - 1 - 2 - 3}} \\
{\text{ 3 4 5}} \\
{\text{ 4 5 6}} \\
\end{gathered} \right]\] In this the last row can be derived from the ${2}^{nd}$ row by adding 1.
Hence the rank of A = 2 (Because it has two linearly independent rows)
Hence b < a < c is the right answer which is option C.
Note: Here we decide the ranks of matrices, by knowing how many rows or columns are dependent. For example, if one row or column depends on another in \[{\text{3}} \times {\text{3}}\] matrix then it’s rank will be 2 because it will be having 2 linearly independent rows or columns.
Complete step-by-step answer:
\[{\text{C = }}\left[ \begin{gathered}
2{\text{ 0 0}} \\
{\text{0 2 0}} \\
{\text{0 0 2}} \\
\end{gathered} \right]\]
It has three linearly independent rows.
Hence Rank of C = 3.
\[{\text{B = }}\left[ \begin{gathered}
1{\text{ - 2}} \\
{\text{ - 1 2}} \\
\end{gathered} \right]\]In this we can see that row 2 is a scalar multiple of row 1 using -1. Here the ${2}^{nd}$ row is dependent on the ${1}^{st}$ row whereas, ${1}^{st}$ row is independent.
Hence Rank of B = 1 (Because it has only one linearly independent row)
\[{\text{A = }}\left[ \begin{gathered}
{\text{ - 1 - 2 - 3}} \\
{\text{ 3 4 5}} \\
{\text{ 4 5 6}} \\
\end{gathered} \right]\] In this the last row can be derived from the ${2}^{nd}$ row by adding 1.
Hence the rank of A = 2 (Because it has two linearly independent rows)
Hence b < a < c is the right answer which is option C.
Note: Here we decide the ranks of matrices, by knowing how many rows or columns are dependent. For example, if one row or column depends on another in \[{\text{3}} \times {\text{3}}\] matrix then it’s rank will be 2 because it will be having 2 linearly independent rows or columns.
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