Question:
The rank of the matrix
\[
A=
\begin{bmatrix}
1 & -1 & 0 & -2
-4 & 4 & 0 & 8
-2 & 1 & 2 & 4
\end{bmatrix}
\]
is:
The rank of the matrix
\[
A=
\begin{bmatrix}
1 & -1 & 0 & -2
-4 & 4 & 0 & 8
-2 & 1 & 2 & 4
\end{bmatrix}
\]
is:
Show Hint
The rank of a matrix is the number of linearly independent rows or columns. If one row is a scalar multiple of another, it does not increase the rank.
Updated On: Jun 18, 2026
- \(1\)
- \(0\)
- \(3\)
- \(2\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Write the rows of the matrix.
The given matrix is \[ A= \begin{bmatrix} 1 & -1 & 0 & -2 -4 & 4 & 0 & 8 -2 & 1 & 2 & 4 \end{bmatrix} \] Let the rows be \[ R_1=[1,-1,0,-2], \] \[ R_2=[-4,4,0,8], \] and \[ R_3=[-2,1,2,4]. \]
Step 2: Check linear dependence of rows.
Observe that \[ R_2=-4R_1. \] So, the second row is a scalar multiple of the first row.
Therefore, \(R_1\) and \(R_2\) are linearly dependent.
Step 3: Check whether \(R_3\) is dependent on \(R_1\).
If \(R_3\) were a scalar multiple of \(R_1\), then there would exist a real number \(k\) such that \[ R_3=kR_1. \] From the first entry, \[ -2=k(1), \] so \[ k=-2. \] Then from the second entry, \[ 1=k(-1)=(-2)(-1)=2, \] which is false.
Therefore, \(R_3\) is not a scalar multiple of \(R_1\).
Step 4: Determine the rank.
Since \(R_2\) depends on \(R_1\), but \(R_3\) is independent of \(R_1\), there are exactly two linearly independent rows.
Hence, the rank of the matrix is \[ 2. \]
Step 5: Final conclusion.
Therefore, \[ \boxed{2} \]
The given matrix is \[ A= \begin{bmatrix} 1 & -1 & 0 & -2 -4 & 4 & 0 & 8 -2 & 1 & 2 & 4 \end{bmatrix} \] Let the rows be \[ R_1=[1,-1,0,-2], \] \[ R_2=[-4,4,0,8], \] and \[ R_3=[-2,1,2,4]. \]
Step 2: Check linear dependence of rows.
Observe that \[ R_2=-4R_1. \] So, the second row is a scalar multiple of the first row.
Therefore, \(R_1\) and \(R_2\) are linearly dependent.
Step 3: Check whether \(R_3\) is dependent on \(R_1\).
If \(R_3\) were a scalar multiple of \(R_1\), then there would exist a real number \(k\) such that \[ R_3=kR_1. \] From the first entry, \[ -2=k(1), \] so \[ k=-2. \] Then from the second entry, \[ 1=k(-1)=(-2)(-1)=2, \] which is false.
Therefore, \(R_3\) is not a scalar multiple of \(R_1\).
Step 4: Determine the rank.
Since \(R_2\) depends on \(R_1\), but \(R_3\) is independent of \(R_1\), there are exactly two linearly independent rows.
Hence, the rank of the matrix is \[ 2. \]
Step 5: Final conclusion.
Therefore, \[ \boxed{2} \]
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