Question:
If the ranks of the matrices
\[
A=
\begin{bmatrix}
1 & 0 & 1\\
2 & 1 & 2\\
1 & 0 & -1
\end{bmatrix}
\]
and
\[
B=
\begin{bmatrix}
1 & 2 & 3 & 4\\
2 & 4 & 6 & -8
\end{bmatrix}
\]
are \(r_1\) and \(r_2\) respectively, then \(r_1-r_2=\)
If the ranks of the matrices
\[
A=
\begin{bmatrix}
1 & 0 & 1\\
2 & 1 & 2\\
1 & 0 & -1
\end{bmatrix}
\]
and
\[
B=
\begin{bmatrix}
1 & 2 & 3 & 4\\
2 & 4 & 6 & -8
\end{bmatrix}
\]
are \(r_1\) and \(r_2\) respectively, then \(r_1-r_2=\)
Show Hint
For a square matrix, if the determinant is non-zero, then the rank is equal to the order of the matrix. Also, for a matrix with two rows, check whether the rows are scalar multiples to determine whether the rank is \(1\) or \(2\).
Updated On: Jun 25, 2026
- \(0\)
- \(1\)
- \(2\)
- \(3\)
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The Correct Option is B
Solution and Explanation
Step 1: Find the rank of matrix \(A\).
Given \[ A= \begin{bmatrix} 1 & 0 & 1\\ 2 & 1 & 2\\ 1 & 0 & -1 \end{bmatrix} \] To find the rank of \(A\), first calculate its determinant.
\[ |A|= \begin{vmatrix} 1 & 0 & 1\\ 2 & 1 & 2\\ 1 & 0 & -1 \end{vmatrix} \] Expanding along the first row, \[ |A|=1 \begin{vmatrix} 1 & 2\\ 0 & -1 \end{vmatrix} + 1 \begin{vmatrix} 2 & 1\\ 1 & 0 \end{vmatrix} \] Now, \[ \begin{vmatrix} 1 & 2\\ 0 & -1 \end{vmatrix} =1(-1)-2(0)=-1 \] Also, \[ \begin{vmatrix} 2 & 1\\ 1 & 0 \end{vmatrix} =2(0)-1(1)=-1 \] Therefore, \[ |A|=-1-1=-2 \] Since \[ |A|\neq 0, \] the matrix \(A\) is non-singular.
Hence, \[ r_1=3 \]
Step 2: Find the rank of matrix \(B\).
Given \[ B= \begin{bmatrix} 1 & 2 & 3 & 4\\ 2 & 4 & 6 & -8 \end{bmatrix} \] This is a \(2\times 4\) matrix, so its maximum possible rank is \(2\).
Now compare the two rows: \[ R_1=[1\ 2\ 3\ 4] \] and \[ R_2=[2\ 4\ 6\ -8] \] If \(R_2\) were a multiple of \(R_1\), then multiplying \(R_1\) by \(2\) would give \[ 2R_1=[2\ 4\ 6\ 8] \] But \[ R_2=[2\ 4\ 6\ -8] \] Since the last element is different, \[ R_2\neq 2R_1 \] Therefore, the two rows are linearly independent.
Hence, \[ r_2=2 \]
Step 3: Calculate \(r_1-r_2\).
We have \[ r_1=3 \] and \[ r_2=2 \] Therefore, \[ r_1-r_2=3-2 \] So, \[ r_1-r_2=1 \]
Step 4: Final conclusion.
Hence, \[ \boxed{1} \]
Given \[ A= \begin{bmatrix} 1 & 0 & 1\\ 2 & 1 & 2\\ 1 & 0 & -1 \end{bmatrix} \] To find the rank of \(A\), first calculate its determinant.
\[ |A|= \begin{vmatrix} 1 & 0 & 1\\ 2 & 1 & 2\\ 1 & 0 & -1 \end{vmatrix} \] Expanding along the first row, \[ |A|=1 \begin{vmatrix} 1 & 2\\ 0 & -1 \end{vmatrix} + 1 \begin{vmatrix} 2 & 1\\ 1 & 0 \end{vmatrix} \] Now, \[ \begin{vmatrix} 1 & 2\\ 0 & -1 \end{vmatrix} =1(-1)-2(0)=-1 \] Also, \[ \begin{vmatrix} 2 & 1\\ 1 & 0 \end{vmatrix} =2(0)-1(1)=-1 \] Therefore, \[ |A|=-1-1=-2 \] Since \[ |A|\neq 0, \] the matrix \(A\) is non-singular.
Hence, \[ r_1=3 \]
Step 2: Find the rank of matrix \(B\).
Given \[ B= \begin{bmatrix} 1 & 2 & 3 & 4\\ 2 & 4 & 6 & -8 \end{bmatrix} \] This is a \(2\times 4\) matrix, so its maximum possible rank is \(2\).
Now compare the two rows: \[ R_1=[1\ 2\ 3\ 4] \] and \[ R_2=[2\ 4\ 6\ -8] \] If \(R_2\) were a multiple of \(R_1\), then multiplying \(R_1\) by \(2\) would give \[ 2R_1=[2\ 4\ 6\ 8] \] But \[ R_2=[2\ 4\ 6\ -8] \] Since the last element is different, \[ R_2\neq 2R_1 \] Therefore, the two rows are linearly independent.
Hence, \[ r_2=2 \]
Step 3: Calculate \(r_1-r_2\).
We have \[ r_1=3 \] and \[ r_2=2 \] Therefore, \[ r_1-r_2=3-2 \] So, \[ r_1-r_2=1 \]
Step 4: Final conclusion.
Hence, \[ \boxed{1} \]
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