Step 1: Understanding the Question:
The rank of a matrix is the maximum number of linearly independent row vectors (or column vectors) in the matrix.
To find the rank of a matrix, we can reduce it to row echelon form using elementary row operations.
Key Formula or Approach:
Perform Gaussian elimination (elementary row operations) on matrix \(A\):
- Row replacement: \(R_i \to R_i - c R_j\)
The number of non-zero rows in the resulting echelon form will be the rank of the matrix.
Step 2: Detailed Explanation:
• Write down the original matrix \(A\):
\[ A = \begin{bmatrix} 1 & 2 & 0 & -1 2 & 6 & -3 & -3 3 & 10 & -6 & -5 \end{bmatrix} \]
• We want to create zeros below the leading 1 in the first column.
Perform the row operations:
\[ R_2 \to R_2 - 2R_1 \]
\[ R_3 \to R_3 - 3R_1 \]
• Calculating the new rows:
For \(R_2\):
\[ R_2 - 2R_1 = [2, 6, -3, -3] - [2, 4, 0, -2] = [0, 2, -3, -1] \]
For \(R_3\):
\[ R_3 - 3R_1 = [3, 10, -6, -5] - [3, 6, 0, -3] = [0, 4, -6, -2] \]
Thus, the matrix becomes:
\[ \begin{bmatrix} 1 & 2 & 0 & -1 0 & 2 & -3 & -1 0 & 4 & -6 & -2 \end{bmatrix} \]
• Now, we want to create a zero below the leading entry in the second column.
Perform the row operation:
\[ R_3 \to R_3 - 2R_2 \]
• Calculating the new \(R_3\):
\[ R_3 - 2R_2 = [0, 4, -6, -2] - [0, 4, -6, -2] = [0, 0, 0, 0] \]
Thus, the row echelon form of the matrix is:
\[ \begin{bmatrix} 1 & 2 & 0 & -1 0 & 2 & -3 & -1 0 & 0 & 0 & 0 \end{bmatrix} \]
• Count the number of non-zero rows in the echelon form.
There are 2 non-zero rows (Row 1 and Row 2).
Therefore, the rank of the matrix is 2.
Step 3: Final Answer:
The rank of matrix \(A\) is 2.