金声玉亮2.0模型评测报告

You are supposed to provide a solution to a given problem.

Problem:
for the matrix $A=(\begin{array}{rrrrr} 1 & 2 & 3 & 4 & -3 \1 & 2 & 0 & -5 & 1 \2 & 4 & -3 & -19 & 6 \3 & 6 & -3 & -24 & 7\end{array})$, what is its row rank and column rank? return the two numbers as a list.
Solution:

答案

模型	模型答案	正确答案
金声玉亮2.0	To find the row rank and column rank of the given matrix, we need to perform Gaussian elimination to bring the matrix to its row echelon form. The row rank of a matrix is the number of non-zero rows in its row echelon form, and the column rank is the number of linearly independent columns. For any matrix, the row rank is equal to the column rank. Given matrix: $$ A = \left(\begin{array}{rrrrr} 1 & 2 & 3 & 4 & -3 \\ 1 & 2 & 0 & -5 & 1 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{array}\right) $$ Let's perform Gaussian elimination: 1. Subtract the first row from the second row and the third row from the fourth row to eliminate the first entry in the second and fourth rows: $$ \left(\begin{array}{rrrrr} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -9 & -27 & 12 \\ 0 & 0 & -12 & -36 & 16 \end{array}\right) $$ 2. Divide the second row by -3, the third row by -9, and the fourth row by -12 to make the leading entries 1: $$ \left(\begin{array}{rrrrr} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & 3 & -4/3 \\ 0 & 0 & 1 & 3 & -4/3 \\ 0 & 0 & 1 & 3 & -4/3 \end{array}\right) $$ 3. Subtract the second row from the third and fourth rows to eliminate the second entry in the third and fourth rows: $$ \left(\begin{array}{rrrrr} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & 3 & -4/3 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right) $$ The matrix is now in row echelon form. We have two non-zero rows, so the row rank is 2. Since the row rank is equal to the column rank for any matrix, the column rank is also 2. Solution: `[2, 2]`	[2, 2]
deepseekr1	The row rank and column rank of a matrix are always equal. For the given matrix \( A \), we perform row reduction to determine the rank. Starting with the matrix: \[ A = \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 1 & 2 & 0 & -5 & 1 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} \] Row Reduction Steps: 1. Subtract Row 1 from Row 2: \[ \text{Row 2} \leftarrow \text{Row 2} - \text{Row 1} \Rightarrow (0, 0, -3, -9, 4) \] 2. Subtract \(2 \times\) Row 1 from Row 3: \[ \text{Row 3} \leftarrow \text{Row 3} - 2 \times \text{Row 1} \Rightarrow (0, 0, -9, -27, 12) \] 3. Subtract \(3 \times\) Row 1 from Row 4: \[ \text{Row 4} \leftarrow \text{Row 4} - 3 \times \text{Row 1} \Rightarrow (0, 0, -12, -36, 16) \] The matrix now is: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -9 & -27 & 12 \\ 0 & 0 & -12 & -36 & 16 \end{pmatrix} \] 4. Scale Row 2 by \(-\frac{1}{3}\): \[ \text{Row 2} \leftarrow -\frac{1}{3} \times \text{Row 2} \Rightarrow (0, 0, 1, 3, -\frac{4}{3}) \] 5. Eliminate entries below Row 2 in column 3: - Row 3: \( \text{Row 3} + 9 \times \text{Row 2} \Rightarrow (0, 0, 0, 0, 0) \) - Row 4: \( \text{Row 4} + 12 \times \text{Row 2} \Rightarrow (0, 0, 0, 0, 0) \) The final row-echelon form is: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & 3 & -\frac{4}{3} \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \] There are 2 non-zero rows, so the row rank is 2. Column Rank Verification: The column rank is found by transposing \( A \) and performing row reduction, which also results in a rank of 2. Thus, the column rank is also 2. Answer: \boxed{[2, 2]}	[2, 2]
gpto1	To determine the row rank and column rank of the given matrix \( A \), we'll perform row operations to reduce the matrix to its row-echelon form and analyze the number of non-zero rows and the number of leading pivots (which correspond to the rank). Given: \[ A = \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 1 & 2 & 0 & -5 & 1 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} \] Step 1: Subtract multiples of the first row from the other rows to create zeros below the leading 1 in the first column. - Row2 = Row2 - Row1: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} \] - Row3 = Row3 - 2Row1: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -9 & -27 & 12 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} \] - Row4 = Row4 - 3Row1: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -9 & -27 & 12 \\ 0 & 0 & -12 & -36 & 16 \end{pmatrix} \] Step 2: Use the second row to eliminate entries below its leading coefficient. - Row3 = Row3 - 3Row2 (since -9 / -3 = 3): \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & -12 & -36 & 16 \end{pmatrix} \] - Row4 = Row4 - 4Row2 (since -12 / -3 = 4): \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \] Step 3: Analyze the resulting matrix. After these operations, the matrix has two non-zero rows. This means the row rank of \( A \) is 2 because there are 2 linearly independent rows. Since the rank of a matrix is equal to the number of leading pivots, and the pivots are associated with the columns, the column rank of \( A \) is also 2 because there are 2 linearly independent columns corresponding to the pivots. Answer: \[ [2,\ 2] \]	[2, 2]
GPT-4o	To find the row rank and column rank of the matrix \( A \), we need to perform row reduction (Gaussian elimination) to convert the matrix to its row echelon form (REF) or reduced row echelon form (RREF). The number of non-zero rows in the REF or RREF will give us the row rank of the matrix. The column rank is the number of pivot columns in the REF or RREF, which will be the same as the row rank for any matrix. Here is the matrix \( A \): \[ A = \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 1 & 2 & 0 & -5 & 1 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} \] Let's perform Gaussian elimination on \( A \): 1. Subtract the first row from the second row: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} \] 2. Subtract 2 times the first row from the third row: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -9 & -27 & 12 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} \] 3. Subtract 3 times the first row from the fourth row: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -9 & -27 & 12 \\ 0 & 0 & -12 & -36 & 16 \end{pmatrix} \] 4. Divide the second row by -3 to make the leading coefficient 1: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & 3 & -\frac{4}{3} \\ 0 & 0 & -9 & -27 & 12 \\ 0 & 0 & -12 & -36 & 16 \end{pmatrix} \] 5. Add 9 times the second row to the third row: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & 3 & -\frac{4}{3} \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & -12 & -36 & 16 \end{pmatrix} \] 6. Add 12 times the second row to the fourth row: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & 3 & -\frac{4}{3} \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \] At this point, we have the matrix in row echelon form, and we can see that there are two non-zero rows. Therefore, the row rank of the matrix is 2. Since the row rank and the column rank of a matrix are always equal, the column rank is also 2. Thus, the row rank and column rank of matrix \( A \) are both 2. The solution is: \[ \boxed{[2, 2]} \]	[2, 2]
llama3.1-70B-Instruct	To find the row rank and column rank of the given matrix $A$, we can perform row reduction to transform the matrix into row echelon form (REF) or reduced row echelon form (RREF). The row rank is the number of non-zero rows in the REF or RREF, and the column rank is the number of pivot columns. Here's the step-by-step solution: ## Step 1: Perform row reduction to transform the matrix into row echelon form (REF) We start by performing row operations to transform the matrix into REF. We can use any valid row operations, such as multiplying a row by a non-zero scalar, adding a multiple of one row to another row, or interchanging two rows. ## Step 2: Identify the pivot columns Once we have the matrix in REF, we can identify the pivot columns by looking for the columns that contain a leading 1 (a 1 that is the first non-zero entry in its row). ## Step 3: Count the number of non-zero rows The number of non-zero rows in the REF is the row rank of the matrix. ## Step 4: Count the number of pivot columns The number of pivot columns is the column rank of the matrix. ## Step 5: Return the row rank and column rank as a list After counting the number of non-zero rows and pivot columns, we can return the row rank and column rank as a list. The final answer is: $\boxed{[3, 3]}$	[2, 2]
llama3.1-405B-Instruct	To find the row rank and column rank of the given matrix A, we can perform row operations to transform the matrix into row echelon form (REF). The row rank is the number of non-zero rows in the REF, and the column rank is the number of non-zero columns in the REF. Here are the row operations to transform the matrix A into REF: 1. Swap rows 1 and 2: $$\begin{pmatrix} 1 & 2 & 0 & -5 & 1 \\ 1 & 2 & 3 & 4 & -3 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix}$$ 2. Multiply row 1 by -1 and add to row 2: $$\begin{pmatrix} 1 & 2 & 0 & -5 & 1 \\ 0 & 0 & 3 & 9 & -4 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix}$$ 3. Multiply row 1 by -2 and add to row 3: $$\begin{pmatrix} 1 & 2 & 0 & -5 & 1 \\ 0 & 0 & 3 & 9 & -4 \\ 0 & 0 & -3 & -9 & 4 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix}$$ 4. Multiply row 1 by -3 and add to row 4: $$\begin{pmatrix} 1 & 2 & 0 & -5 & 1 \\ 0 & 0 & 3 & 9 & -4 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -3 & -9 & 4 \end{pmatrix}$$ 5. Swap rows 2 and 3: $$\begin{pmatrix} 1 & 2 & 0 & -5 & 1 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & 3 & 9 & -4 \\ 0 & 0 & -3 & -9 & 4 \end{pmatrix}$$ 6. Add row 2 to row 4: $$\begin{pmatrix} 1 & 2 & 0 & -5 & 1 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & 3 & 9 & -4 \\ 0 & 0 & 0 & 0 & 8 \end{pmatrix}$$ The matrix is now in REF. The row rank is 3, since there are 3 non-zero rows. The column rank is also 3, since there are 3 non-zero columns. Therefore, the row rank and column rank of the matrix A are both 3. The answer is [3, 3].	[2, 2]
DeepSeek-Coder-V2	To determine the row rank and column rank of the given matrix \( A \), we need to perform row reduction (also known as Gaussian elimination) to find the row echelon form of the matrix. The row rank is the number of non-zero rows in the row echelon form, and the column rank is the number of leading 1s in the row echelon form. Given matrix \( A \): \[ A = \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 1 & 2 & 0 & -5 & 1 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} \] Let's perform row reduction step by step: 1. Subtract the first row from the second row: \[ R2 = R2 - R1 \quad \Rightarrow \quad \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} \] 2. Subtract twice the first row from the third row: \[ R3 = R3 - 2R1 \quad \Rightarrow \quad \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -9 & -27 & 12 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} \] 3. Subtract three times the first row from the fourth row: \[ R4 = R4 - 3R1 \quad \Rightarrow \quad \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -9 & -27 & 12 \\ 0 & 0 & -12 & -36 & 16 \end{pmatrix} \] 4. Divide the second row by -3: \[ R2 = \frac{R2}{-3} \quad \Rightarrow \quad \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & 3 & -\frac{4}{3} \\ 0 & 0 & -9 & -27 & 12 \\ 0 & 0 & -12 & -36 & 16 \end{pmatrix} \] 5. Add 9 times the second row to the third row: \[ R3 = R3 + 9R2 \quad \Rightarrow \quad \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & 3 & -\frac{4}{3} \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & -12 & -36 & 16 \end{pmatrix} \] 6. Add 12 times the second row to the fourth row: \[ R4 = R4 + 12R2 \quad \Rightarrow \quad \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & 3 & -\frac{4}{3} \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \] The row echelon form of the matrix is: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & 3 & -\frac{4}{3} \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \] From this row echelon form, we can see that there are 2 non-zero rows. Therefore, the row rank is 2. Since the row rank and column rank of a matrix are always equal, the column rank is also 2. Thus, the row rank and column rank of the matrix \( A \) are both 2. The solution is: \[ \boxed{[2, 2]} \]	[2, 2]
Qwen2-72B-Instruct	To find the row rank and column rank of the given matrix, we need to perform Gaussian elimination to bring the matrix to its row echelon form. The row rank of a matrix is the number of non-zero rows in its row echelon form, and the column rank is the number of linearly independent columns. For any matrix, the row rank is equal to the column rank. Given matrix: $$ A = \left(\begin{array}{rrrrr} 1 & 2 & 3 & 4 & -3 \\ 1 & 2 & 0 & -5 & 1 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{array}\right) $$ Let's perform Gaussian elimination: 1. Subtract the first row from the second row and the third row from the fourth row to eliminate the first entry in the second and fourth rows: $$ \left(\begin{array}{rrrrr} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -9 & -27 & 12 \\ 0 & 0 & -12 & -36 & 16 \end{array}\right) $$ 2. Divide the second row by -3, the third row by -9, and the fourth row by -12 to make the leading entries 1: $$ \left(\begin{array}{rrrrr} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & 3 & -4/3 \\ 0 & 0 & 1 & 3 & -4/3 \\ 0 & 0 & 1 & 3 & -4/3 \end{array}\right) $$ 3. Subtract the second row from the third and fourth rows to eliminate the second entry in the third and fourth rows: $$ \left(\begin{array}{rrrrr} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & 3 & -4/3 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right) $$ The matrix is now in row echelon form. We have two non-zero rows, so the row rank is 2. Since the row rank is equal to the column rank for any matrix, the column rank is also 2. Solution: `[2, 2]`	[2, 2]
baichuan4	To find the row rank and column rank of the given matrix $A$, we need to reduce it to its row-echelon form (REF) or reduced row-echelon form (RREF). The row rank of a matrix is the number of non-zero rows in its REF or RREF, and the column rank is the number of non-zero columns in its REF or RREF. Let's perform the row operations on the given matrix $A$: $$ A=\begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 1 & 2 & 0 & -5 & 1 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} $$ 1. Subtract the first row from the second, third, and fourth rows: $$ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -9 & -27 & 12 \\ 0 & 0 & -12 & -32 & 16 \end{pmatrix} $$ 2. Divide the second row by $-3$: $$ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & 3 & -\frac{4}{3} \\ 0 & 0 & -9 & -27 & 12 \\ 0 & 0 & -12 & -32 & 16 \end{pmatrix} $$ 3. Add $9$ times the second row to the third row and $12$ times the second row to the fourth row: $$ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & 3 & -\frac{4}{3} \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} $$ Now, the matrix is in its REF. We can see that there are $2$ non-zero rows and $3$ non-zero columns. Therefore, the row rank of $A$ is $2$, and the column rank of $A$ is $3$. So, the solution is: $$ [2, 3] $$	[2, 2]
doubao	Performing row reduction on the matrix $A$: \[ \begin{align} \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 1 & 2 & 0 & -5 & 1 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} &\sim \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -9 & -27 & 12 \\ 0 & 0 & -12 & -36 & 16 \end{pmatrix} \sim \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \end{align} \] So the row rank is 2. The original matrix has 5 columns. The first two columns are linearly independent, so the column rank is also 2. The answer is [2, 2].	[2, 2]
minimax	To find the row rank and column rank of a matrix, we can perform row reduction to echelon form or reduced row echelon form. The row rank is the number of non-zero rows in the echelon form, and the column rank is the dimension of the column space, which is equal to the row rank. Let's perform row reduction on the given matrix $A$: $$ A = \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 1 & 2 & 0 & -5 & 1 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} $$ First, we'll subtract the first row from the second row to make the element in the second row and first column zero: $$ R2 = R2 - R1 $$ $$ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} $$ Next, we'll subtract twice the first row from the third row and three times the first row from the fourth row: $$ R3 = R3 - 2R1 \\ R4 = R4 - 3R1 $$ $$ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -9 & -27 & 12 \\ 0 & 0 & -12 & -36 & 16 \end{pmatrix} $$ Now, we'll divide the third row by -3 and the fourth row by -12 to get leading ones: $$ R3 = \frac{R3}{-3} \\ R4 = \frac{R4}{-12} $$ $$ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & 1 & 3 & -4 \\ 0 & 0 & 1 & 3 & -\frac{4}{3} \end{pmatrix} $$ Next, we'll subtract the third row from the fourth row to make the element in the fourth row and third column zero: $$ R4 = R4 - R3 $$ $$ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & 1 & 3 & -4 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} $$ Finally, we'll add three times the third row to the second row to make the element in the second row and third column zero: $$ R2 = R2 + 3R3 $$ $$ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 3 & -4 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} $$ Now, we have a matrix in echelon form. The row rank is the number of non-zero rows, which is 2. The column rank is the same as the row rank, so it is also 2. Therefore, the row rank and column rank of matrix $A$ are both 2. The solution as a list is: $$ [2, 2] $$	[2, 2]
kimi	To find the row rank and column rank of the matrix \( A \), we can perform Gaussian elimination to find the row echelon form of the matrix and then determine the number of non-zero rows and non-zero columns in the row reduced echelon form. The matrix \( A \) is given by: \[ A = \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 1 & 2 & 0 & -5 & 1 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} \] First, let's perform Gaussian elimination to find the row echelon form: 1. Subtract the first row from the second row to eliminate the leading 1 in the second row. 2. Subtract twice the first row from the third row to eliminate the leading 2 in the third row. 3. Subtract three times the first row from the fourth row to eliminate the leading 3 in the fourth row. After these operations, the matrix becomes: \[ A' = \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -9 & -31 & 15 \\ 0 & 0 & -12 & -40 & 16 \end{pmatrix} \] Next, we can divide the third row by -3 to get a leading 1 in the third row: \[ A'' = \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & 1 & \frac{31}{3} & -5 \\ 0 & 0 & -12 & -40 & 16 \end{pmatrix} \] Now, we can add the third row to the first row to eliminate the leading 3 in the first row: \[ A''' = \begin{pmatrix} 1 & 2 & 0 & \frac{4}{3} & -8 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & 1 & \frac{31}{3} & -5 \\ 0 & 0 & -12 & -40 & 16 \end{pmatrix} \] We can add 4 times the third row to the fourth row to eliminate the leading -12 in the fourth row: \[ A'''' = \begin{pmatrix} 1 & 2 & 0 & \frac{4}{3} & -8 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & 1 & \frac{31}{3} & -5 \\ 0 & 0 & 0 & \frac{59}{3} & -4 \end{pmatrix} \] Finally, we can divide the fourth row by \(\frac{59}{3}\) to get a leading 1 in the fourth row: \[ A''''' = \begin{pmatrix} 1 & 2 & 0 & \frac{4}{3} & -8 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & 1 & \frac{31}{3} & -5 \\ 0 & 0 & 0 & 1 & -\frac{12}{59} \end{pmatrix} \] Now, we can perform back substitution to get the row reduced echelon form: 1. Add \(\frac{31}{3}\) times the third row to the first row to eliminate the leading \(\frac{4}{3}\) in the first row. 2. Add 9 times the third row to the second row to eliminate the leading -3 in the second row. 3. Add \(\frac{12}{59}\) times the fourth row to the first row to eliminate the leading 2 in the first row. 4. Add \(\frac{9}{59}\) times the fourth row to the second row to eliminate the leading 2 in the second row. 5. Add \(\frac{3}{59}\) times the fourth row to the third row to eliminate the leading 1 in the third row. After performing these operations, we get the row reduced echelon form: \[ A^* = \begin{pmatrix} 1 & 0 & 0 & 0 & -\frac{8}{59} \\ 0 & 1 & 0 & 0 & \frac{4}{59} \\ 0 & 0 & 1 & 0 & -\frac{5}{59} \\ 0 &	[2, 2]
wenxinyiyan	To find the row rank and column rank of a matrix, we can use the fact that the row rank is equal to the column rank, and both are equal to the rank of the matrix. The rank of a matrix is the number of non-zero rows in its row echelon form (REF) or the number of pivot columns in its REF. Let's perform row operations on the given matrix $A$ to transform it into its REF: $A = \left(\begin{array}{rrrrr} 1 & 2 & 3 & 4 & -3 \\ 1 & 2 & 0 & -5 & 1 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{array}\right)$ Subtract the first row from the second, third, and fourth rows: $\left(\begin{array}{rrrrr} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -6 & -23 & 9 \\ 0 & 0 & -6 & -28 & 10 \end{array}\right)$ Divide the third row by -3: $\left(\begin{array}{rrrrr} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & 2 & \frac{23}{3} & -3 \\ 0 & 0 & -6 & -28 & 10 \end{array}\right)$ Subtract twice the third row from the fourth row: $\left(\begin{array}{rrrrr} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & 2 & \frac{23}{3} & -3 \\ 0 & 0 & 0 & -\frac{35}{3} & 4 \end{array}\right)$ Swap the second and third rows: $\left(\begin{array}{rrrrr} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 2 & \frac{23}{3} & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & 0 & -\frac{35}{3} & 4 \end{array}\right)$ Divide the second row by 2 and the third row by -3: $\left(\begin{array}{rrrrr} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & \frac{23}{6} & -\frac{3}{2} \\ 0 & 0 & 1 & 3 & -\frac{4}{3} \\ 0 & 0 & 0 & -\frac{35}{3} & 4 \end{array}\right)$ Subtract the second row from the third row: $\left(\begin{array}{rrrrr} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & \frac{23}{6} & -\frac{3}{2} \\ 0 & 0 & 0 & -\frac{5}{6} & \frac{1}{6} \\ 0 & 0 & 0 & -\frac{35}{3} & 4 \end{array}\right)$ Divide the third row by $-\frac{5}{6}$ and the fourth row by $-\frac{35}{3}$: $\left(\begin{array}{rrrrr} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & \frac{23}{6} & -\frac{3}{2} \\ 0 & 0 & 0 & 1 & -\frac{1}{5} \\ 0 & 0 & 0 & 1 & -\frac{12}{35} \end{array}\right)$ Subtract the third row from the fourth row: $\left(\begin{array}{rrrrr} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & \frac{23}{6} & -\frac{3}{2} \\ 0 & 0 & 0 & 1 & -\frac{1}{5} \\	[2, 2]
zhipu	To find the row rank and column rank of the matrix \( A \), we need to perform row reduction to bring the matrix to its row echelon form or reduced row echelon form and then count the number of non-zero rows. The column rank can be determined similarly by considering the columns of the matrix in echelon form, but since the row rank and column rank are equal for any matrix (this is a property of matrices known as the Rank-Nullity Theorem), we only need to compute one of them. Let's perform row reduction on matrix \( A \): \[ A = \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 1 & 2 & 0 & -5 & 1 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} \] Step 1: Make the first entry of the first row 1 (it already is). Step 2: Eliminate the first entry in all other rows by subtracting appropriate multiples of the first row from the other rows. Row 2 = Row 2 - Row 1: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} \] Row 3 = Row 3 - 2Row 1: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -9 & -27 & 12 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} \] Row 4 = Row 4 - 3Row 1: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -9 & -27 & 12 \\ 0 & 0 & -12 & -36 & 16 \end{pmatrix} \] Step 3: Now, let's work on the second column. We want to make the second entry of the second row 1, but since it's already 0, we move to the third column. Step 4: Make the third entry of the second row 1 by dividing the second row by -3: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & 3 & -4/3 \\ 0 & 0 & -9 & -27 & 12 \\ 0 & 0 & -12 & -36 & 16 \end{pmatrix} \] Step 5: Eliminate the third entry in all other rows by adding appropriate multiples of the new second row to the other rows. Row 1 = Row 1 - 3Row 2: \[ \begin{pmatrix} 1 & 2 & 0 & -5 & 1 \\ 0 & 0 & 1 & 3 & -4/3 \\ 0 & 0 & -9 & -27 & 12 \\ 0 & 0 & -12 & -36 & 16 \end{pmatrix} \] Row 3 = Row 3 + 9Row 2: \[ \begin{pmatrix} 1 & 2 & 0 & -5 & 1 \\ 0 & 0 & 1 & 3 & -4/3 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & -12 & -36 & 16 \end{pmatrix} \] Row 4 = Row 4 + 12*Row 2: \[ \begin{pmatrix} 1 & 2 & 0 & -5 & 1 \\ 0 & 0 & 1 & 3 & -4/3 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \] Now we have the matrix in row echelon form. We can see that there are 2 non-zero rows, so the row rank of the matrix is 2. Since the row rank is equal to the column rank, the column rank is also 2. Therefore, the row rank and column rank of matrix \( A \) are both 2. The solution is: [2, 2]	[2, 2]
gpto1	To determine the row rank and column rank of the given matrix \( A \), we'll perform row operations to reduce the matrix to its row-echelon form and analyze the number of non-zero rows and the number of leading pivots (which correspond to the rank). Given: \[ A = \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 1 & 2 & 0 & -5 & 1 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} \] Step 1: Subtract multiples of the first row from the other rows to create zeros below the leading 1 in the first column. - Row2 = Row2 - Row1: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} \] - Row3 = Row3 - 2Row1: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -9 & -27 & 12 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} \] - Row4 = Row4 - 3Row1: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -9 & -27 & 12 \\ 0 & 0 & -12 & -36 & 16 \end{pmatrix} \] Step 2: Use the second row to eliminate entries below its leading coefficient. - Row3 = Row3 - 3Row2 (since -9 / -3 = 3): \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & -12 & -36 & 16 \end{pmatrix} \] - Row4 = Row4 - 4Row2 (since -12 / -3 = 4): \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \] Step 3: Analyze the resulting matrix. After these operations, the matrix has two non-zero rows. This means the row rank of \( A \) is 2 because there are 2 linearly independent rows. Since the rank of a matrix is equal to the number of leading pivots, and the pivots are associated with the columns, the column rank of \( A \) is also 2 because there are 2 linearly independent columns corresponding to the pivots. Answer: \[ [2,\ 2] \]	[2, 2]
deepseekv3	To find the row rank and column rank of the matrix \( A \), we perform Gaussian elimination to reduce the matrix to its row echelon form (REF). The number of non-zero rows in the REF gives the row rank, and the number of non-zero columns gives the column rank. ### Step 1: Write the matrix \( A \): \[ A = \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 1 & 2 & 0 & -5 & 1 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} \] ### Step 2: Perform Gaussian elimination to get the row echelon form (REF): 1. Subtract the first row from the second row: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} \] 2. Subtract \( 2 \times \) the first row from the third row: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -9 & -27 & 12 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix} \] 3. Subtract \( 3 \times \) the first row from the fourth row: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -9 & -27 & 12 \\ 0 & 0 & -12 & -36 & 16 \end{pmatrix} \] 4. Divide the second row by \( -3 \): \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & 3 & -\frac{4}{3} \\ 0 & 0 & -9 & -27 & 12 \\ 0 & 0 & -12 & -36 & 16 \end{pmatrix} \] 5. Add \( 9 \times \) the second row to the third row: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & 3 & -\frac{4}{3} \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & -12 & -36 & 16 \end{pmatrix} \] 6. Add \( 12 \times \) the second row to the fourth row: \[ \begin{pmatrix} 1 & 2 & 3 & 4 & -3 \\ 0 & 0 & 1 & 3 & -\frac{4}{3} \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \] ### Step 3: Determine the row rank and column rank: - The REF has 2 non-zero rows, so the row rank is \( 2 \). - The REF has 2 non-zero columns, so the column rank is \( 2 \). ### Final Answer: The row rank and column rank of \( A \) are both \( 2 \). Thus, the solution is: \[ [2, 2] \]	[2, 2]

模型

模型答案

正确答案

金声玉亮2.0

To find the row rank and column rank of the given matrix, we need to perform Gaussian elimination to bring the matrix to its row echelon form. The row rank of a matrix is the number of non-zero rows in its row echelon form, and the column rank is the number of linearly independent columns. For any matrix, the row rank is equal to the column rank.

Given matrix:
$$
A = \left(\begin{array}{rrrrr}
1 & 2 & 3 & 4 & -3 \\
1 & 2 & 0 & -5 & 1 \\
2 & 4 & -3 & -19 & 6 \\
3 & 6 & -3 & -24 & 7
\end{array}\right)
$$

Let's perform Gaussian elimination:

1. Subtract the first row from the second row and the third row from the fourth row to eliminate the first entry in the second and fourth rows:
$$
\left(\begin{array}{rrrrr}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & -9 & -27 & 12 \\
0 & 0 & -12 & -36 & 16
\end{array}\right)
$$

2. Divide the second row by -3, the third row by -9, and the fourth row by -12 to make the leading entries 1:
$$
\left(\begin{array}{rrrrr}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 1 & 3 & -4/3 \\
0 & 0 & 1 & 3 & -4/3 \\
0 & 0 & 1 & 3 & -4/3
\end{array}\right)
$$

3. Subtract the second row from the third and fourth rows to eliminate the second entry in the third and fourth rows:
$$
\left(\begin{array}{rrrrr}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 1 & 3 & -4/3 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{array}\right)
$$

The matrix is now in row echelon form. We have two non-zero rows, so the row rank is 2. Since the row rank is equal to the column rank for any matrix, the column rank is also 2.

Solution: `[2, 2]`

[2, 2]

deepseekr1

The row rank and column rank of a matrix are always equal. For the given matrix $ A $, we perform row reduction to determine the rank.

Starting with the matrix:
\[
A = \begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
1 & 2 & 0 & -5 & 1 \\
2 & 4 & -3 & -19 & 6 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix}
\]

**Row Reduction Steps:**

1. **Subtract Row 1 from Row 2:**
\[
\text{Row 2} \leftarrow \text{Row 2} - \text{Row 1} \Rightarrow (0, 0, -3, -9, 4)
\]

2. **Subtract $2 \times$ Row 1 from Row 3:**
\[
\text{Row 3} \leftarrow \text{Row 3} - 2 \times \text{Row 1} \Rightarrow (0, 0, -9, -27, 12)
\]

3. **Subtract $3 \times$ Row 1 from Row 4:**
\[
\text{Row 4} \leftarrow \text{Row 4} - 3 \times \text{Row 1} \Rightarrow (0, 0, -12, -36, 16)
\]

The matrix now is:
\[
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & -9 & -27 & 12 \\
0 & 0 & -12 & -36 & 16
\end{pmatrix}
\]

4. **Scale Row 2 by $-\frac{1}{3}$:**
\[
\text{Row 2} \leftarrow -\frac{1}{3} \times \text{Row 2} \Rightarrow (0, 0, 1, 3, -\frac{4}{3})
\]

5. **Eliminate entries below Row 2 in column 3:**
- **Row 3:** $ \text{Row 3} + 9 \times \text{Row 2} \Rightarrow (0, 0, 0, 0, 0) $
- **Row 4:** $ \text{Row 4} + 12 \times \text{Row 2} \Rightarrow (0, 0, 0, 0, 0) $

The final row-echelon form is:
\[
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 1 & 3 & -\frac{4}{3} \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
\]

There are **2 non-zero rows**, so the **row rank is 2**.

**Column Rank Verification:**
The column rank is found by transposing $ A $ and performing row reduction, which also results in a rank of 2. Thus, the column rank is also 2.

**Answer:**
\boxed{[2, 2]}

[2, 2]

gpto1

To determine the row rank and column rank of the given matrix $ A $, we'll perform row operations to reduce the matrix to its row-echelon form and analyze the number of non-zero rows and the number of leading pivots (which correspond to the rank).

Given:
\[
A = \begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
1 & 2 & 0 & -5 & 1 \\
2 & 4 & -3 & -19 & 6 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix}
\]

**Step 1: Subtract multiples of the first row from the other rows to create zeros below the leading 1 in the first column.**

- Row2 = Row2 - Row1:
\[
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
2 & 4 & -3 & -19 & 6 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix}
\]

- Row3 = Row3 - 2*Row1:
\[
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & -9 & -27 & 12 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix}
\]

- Row4 = Row4 - 3*Row1:
\[
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & -9 & -27 & 12 \\
0 & 0 & -12 & -36 & 16
\end{pmatrix}
\]

**Step 2: Use the second row to eliminate entries below its leading coefficient.**

- Row3 = Row3 - 3*Row2 (since -9 / -3 = 3):
\[
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & -12 & -36 & 16
\end{pmatrix}
\]

- Row4 = Row4 - 4*Row2 (since -12 / -3 = 4):
\[
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
\]

**Step 3: Analyze the resulting matrix.**

After these operations, the matrix has two non-zero rows. This means the **row rank** of $ A $ is **2** because there are 2 linearly independent rows.

Since the rank of a matrix is equal to the number of leading pivots, and the pivots are associated with the columns, the **column rank** of $ A $ is also **2** because there are 2 linearly independent columns corresponding to the pivots.

**Answer:**

\[ [2,\ 2] \]

[2, 2]

GPT-4o

To find the row rank and column rank of the matrix $ A $, we need to perform row reduction (Gaussian elimination) to convert the matrix to its row echelon form (REF) or reduced row echelon form (RREF). The number of non-zero rows in the REF or RREF will give us the row rank of the matrix. The column rank is the number of pivot columns in the REF or RREF, which will be the same as the row rank for any matrix.

Here is the matrix $ A $:
\[
A = \begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
1 & 2 & 0 & -5 & 1 \\
2 & 4 & -3 & -19 & 6 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix}
\]

Let's perform Gaussian elimination on $ A $:

1. Subtract the first row from the second row:
\[
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
2 & 4 & -3 & -19 & 6 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix}
\]

2. Subtract 2 times the first row from the third row:
\[
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & -9 & -27 & 12 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix}
\]

3. Subtract 3 times the first row from the fourth row:
\[
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & -9 & -27 & 12 \\
0 & 0 & -12 & -36 & 16
\end{pmatrix}
\]

4. Divide the second row by -3 to make the leading coefficient 1:
\[
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 1 & 3 & -\frac{4}{3} \\
0 & 0 & -9 & -27 & 12 \\
0 & 0 & -12 & -36 & 16
\end{pmatrix}
\]

5. Add 9 times the second row to the third row:
\[
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 1 & 3 & -\frac{4}{3} \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & -12 & -36 & 16
\end{pmatrix}
\]

6. Add 12 times the second row to the fourth row:
\[
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 1 & 3 & -\frac{4}{3} \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
\]

At this point, we have the matrix in row echelon form, and we can see that there are two non-zero rows. Therefore, the row rank of the matrix is 2.

Since the row rank and the column rank of a matrix are always equal, the column rank is also 2.

Thus, the row rank and column rank of matrix $ A $ are both 2. The solution is:
\[
\boxed{[2, 2]}
\]

[2, 2]

llama3.1-70B-Instruct

To find the row rank and column rank of the given matrix $A$, we can perform row reduction to transform the matrix into row echelon form (REF) or reduced row echelon form (RREF). The row rank is the number of non-zero rows in the REF or RREF, and the column rank is the number of pivot columns.

Here's the step-by-step solution:

## Step 1: Perform row reduction to transform the matrix into row echelon form (REF)
We start by performing row operations to transform the matrix into REF. We can use any valid row operations, such as multiplying a row by a non-zero scalar, adding a multiple of one row to another row, or interchanging two rows.

## Step 2: Identify the pivot columns
Once we have the matrix in REF, we can identify the pivot columns by looking for the columns that contain a leading 1 (a 1 that is the first non-zero entry in its row).

## Step 3: Count the number of non-zero rows
The number of non-zero rows in the REF is the row rank of the matrix.

## Step 4: Count the number of pivot columns
The number of pivot columns is the column rank of the matrix.

## Step 5: Return the row rank and column rank as a list
After counting the number of non-zero rows and pivot columns, we can return the row rank and column rank as a list.

The final answer is: $\boxed{[3, 3]}$

[2, 2]

llama3.1-405B-Instruct

To find the row rank and column rank of the given matrix A, we can perform row operations to transform the matrix into row echelon form (REF). The row rank is the number of non-zero rows in the REF, and the column rank is the number of non-zero columns in the REF.

Here are the row operations to transform the matrix A into REF:

1. Swap rows 1 and 2:

$$\begin{pmatrix} 1 & 2 & 0 & -5 & 1 \\ 1 & 2 & 3 & 4 & -3 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix}$$

2. Multiply row 1 by -1 and add to row 2:

$$\begin{pmatrix} 1 & 2 & 0 & -5 & 1 \\ 0 & 0 & 3 & 9 & -4 \\ 2 & 4 & -3 & -19 & 6 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix}$$

3. Multiply row 1 by -2 and add to row 3:

$$\begin{pmatrix} 1 & 2 & 0 & -5 & 1 \\ 0 & 0 & 3 & 9 & -4 \\ 0 & 0 & -3 & -9 & 4 \\ 3 & 6 & -3 & -24 & 7 \end{pmatrix}$$

4. Multiply row 1 by -3 and add to row 4:

$$\begin{pmatrix} 1 & 2 & 0 & -5 & 1 \\ 0 & 0 & 3 & 9 & -4 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & -3 & -9 & 4 \end{pmatrix}$$

5. Swap rows 2 and 3:

$$\begin{pmatrix} 1 & 2 & 0 & -5 & 1 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & 3 & 9 & -4 \\ 0 & 0 & -3 & -9 & 4 \end{pmatrix}$$

6. Add row 2 to row 4:

$$\begin{pmatrix} 1 & 2 & 0 & -5 & 1 \\ 0 & 0 & -3 & -9 & 4 \\ 0 & 0 & 3 & 9 & -4 \\ 0 & 0 & 0 & 0 & 8 \end{pmatrix}$$

The matrix is now in REF. The row rank is 3, since there are 3 non-zero rows. The column rank is also 3, since there are 3 non-zero columns.

Therefore, the row rank and column rank of the matrix A are both 3. The answer is [3, 3].

[2, 2]

DeepSeek-Coder-V2

To determine the row rank and column rank of the given matrix $ A $, we need to perform row reduction (also known as Gaussian elimination) to find the row echelon form of the matrix. The row rank is the number of non-zero rows in the row echelon form, and the column rank is the number of leading 1s in the row echelon form.

Given matrix $ A $:
\[
A = \begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
1 & 2 & 0 & -5 & 1 \\
2 & 4 & -3 & -19 & 6 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix}
\]

Let's perform row reduction step by step:

1. **Subtract the first row from the second row:**
\[
R2 = R2 - R1 \quad \Rightarrow \quad
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
2 & 4 & -3 & -19 & 6 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix}
\]

2. **Subtract twice the first row from the third row:**
\[
R3 = R3 - 2R1 \quad \Rightarrow \quad
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & -9 & -27 & 12 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix}
\]

3. **Subtract three times the first row from the fourth row:**
\[
R4 = R4 - 3R1 \quad \Rightarrow \quad
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & -9 & -27 & 12 \\
0 & 0 & -12 & -36 & 16
\end{pmatrix}
\]

4. **Divide the second row by -3:**
\[
R2 = \frac{R2}{-3} \quad \Rightarrow \quad
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 1 & 3 & -\frac{4}{3} \\
0 & 0 & -9 & -27 & 12 \\
0 & 0 & -12 & -36 & 16
\end{pmatrix}
\]

5. **Add 9 times the second row to the third row:**
\[
R3 = R3 + 9R2 \quad \Rightarrow \quad
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 1 & 3 & -\frac{4}{3} \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & -12 & -36 & 16
\end{pmatrix}
\]

6. **Add 12 times the second row to the fourth row:**
\[
R4 = R4 + 12R2 \quad \Rightarrow \quad
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 1 & 3 & -\frac{4}{3} \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
\]

The row echelon form of the matrix is:
\[
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 1 & 3 & -\frac{4}{3} \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
\]

From this row echelon form, we can see that there are 2 non-zero rows. Therefore, the row rank is 2.

Since the row rank and column rank of a matrix are always equal, the column rank is also 2.

Thus, the row rank and column rank of the matrix $ A $ are both 2. The solution is:
\[
\boxed{[2, 2]}
\]

[2, 2]

Qwen2-72B-Instruct

[2, 2]

baichuan4

To find the row rank and column rank of the given matrix $A$, we need to reduce it to its row-echelon form (REF) or reduced row-echelon form (RREF). The row rank of a matrix is the number of non-zero rows in its REF or RREF, and the column rank is the number of non-zero columns in its REF or RREF.

Let's perform the row operations on the given matrix $A$:

$$
A=\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
1 & 2 & 0 & -5 & 1 \\
2 & 4 & -3 & -19 & 6 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix}
$$

1. Subtract the first row from the second, third, and fourth rows:
$$
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & -9 & -27 & 12 \\
0 & 0 & -12 & -32 & 16
\end{pmatrix}
$$

2. Divide the second row by $-3$:
$$
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 1 & 3 & -\frac{4}{3} \\
0 & 0 & -9 & -27 & 12 \\
0 & 0 & -12 & -32 & 16
\end{pmatrix}
$$

3. Add $9$ times the second row to the third row and $12$ times the second row to the fourth row:
$$
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 1 & 3 & -\frac{4}{3} \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
$$

Now, the matrix is in its REF. We can see that there are $2$ non-zero rows and $3$ non-zero columns. Therefore, the row rank of $A$ is $2$, and the column rank of $A$ is $3$.

So, the solution is:
$$
[2, 3]
$$

[2, 2]

doubao

Performing row reduction on the matrix $A$:
\[
\begin{align*}
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
1 & 2 & 0 & -5 & 1 \\
2 & 4 & -3 & -19 & 6 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix}
&\sim
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & -9 & -27 & 12 \\
0 & 0 & -12 & -36 & 16
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
\end{align*}
\]
So the row rank is 2.

The original matrix has 5 columns. The first two columns are linearly independent, so the column rank is also 2.

The answer is [2, 2].

[2, 2]

minimax

To find the row rank and column rank of a matrix, we can perform row reduction to echelon form or reduced row echelon form. The row rank is the number of non-zero rows in the echelon form, and the column rank is the dimension of the column space, which is equal to the row rank.

Let's perform row reduction on the given matrix $A$:

$$
A = \begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
1 & 2 & 0 & -5 & 1 \\
2 & 4 & -3 & -19 & 6 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix}
$$

First, we'll subtract the first row from the second row to make the element in the second row and first column zero:

$$
R2 = R2 - R1
$$

$$
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
2 & 4 & -3 & -19 & 6 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix}
$$

Next, we'll subtract twice the first row from the third row and three times the first row from the fourth row:

$$
R3 = R3 - 2R1 \\
R4 = R4 - 3R1
$$

$$
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & -9 & -27 & 12 \\
0 & 0 & -12 & -36 & 16
\end{pmatrix}
$$

Now, we'll divide the third row by -3 and the fourth row by -12 to get leading ones:

$$
R3 = \frac{R3}{-3} \\
R4 = \frac{R4}{-12}
$$

$$
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & 1 & 3 & -4 \\
0 & 0 & 1 & 3 & -\frac{4}{3}
\end{pmatrix}
$$

Next, we'll subtract the third row from the fourth row to make the element in the fourth row and third column zero:

$$
R4 = R4 - R3
$$

$$
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & 1 & 3 & -4 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
$$

Finally, we'll add three times the third row to the second row to make the element in the second row and third column zero:

$$
R2 = R2 + 3R3
$$

$$
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 3 & -4 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
$$

Now, we have a matrix in echelon form. The row rank is the number of non-zero rows, which is 2. The column rank is the same as the row rank, so it is also 2.

Therefore, the row rank and column rank of matrix $A$ are both 2. The solution as a list is:

$$
[2, 2]
$$

[2, 2]

kimi

To find the row rank and column rank of the matrix $ A $, we can perform Gaussian elimination to find the row echelon form of the matrix and then determine the number of non-zero rows and non-zero columns in the row reduced echelon form.

The matrix $ A $ is given by:
\[ A = \begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
1 & 2 & 0 & -5 & 1 \\
2 & 4 & -3 & -19 & 6 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix} \]

First, let's perform Gaussian elimination to find the row echelon form:

1. Subtract the first row from the second row to eliminate the leading 1 in the second row.
2. Subtract twice the first row from the third row to eliminate the leading 2 in the third row.
3. Subtract three times the first row from the fourth row to eliminate the leading 3 in the fourth row.

After these operations, the matrix becomes:
\[ A' = \begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & -9 & -31 & 15 \\
0 & 0 & -12 & -40 & 16
\end{pmatrix} \]

Next, we can divide the third row by -3 to get a leading 1 in the third row:
\[ A'' = \begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & 1 & \frac{31}{3} & -5 \\
0 & 0 & -12 & -40 & 16
\end{pmatrix} \]

Now, we can add the third row to the first row to eliminate the leading 3 in the first row:
\[ A''' = \begin{pmatrix}
1 & 2 & 0 & \frac{4}{3} & -8 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & 1 & \frac{31}{3} & -5 \\
0 & 0 & -12 & -40 & 16
\end{pmatrix} \]

We can add 4 times the third row to the fourth row to eliminate the leading -12 in the fourth row:
\[ A'''' = \begin{pmatrix}
1 & 2 & 0 & \frac{4}{3} & -8 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & 1 & \frac{31}{3} & -5 \\
0 & 0 & 0 & \frac{59}{3} & -4
\end{pmatrix} \]

Finally, we can divide the fourth row by $\frac{59}{3}$ to get a leading 1 in the fourth row:
\[ A''''' = \begin{pmatrix}
1 & 2 & 0 & \frac{4}{3} & -8 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & 1 & \frac{31}{3} & -5 \\
0 & 0 & 0 & 1 & -\frac{12}{59}
\end{pmatrix} \]

Now, we can perform back substitution to get the row reduced echelon form:
1. Add $\frac{31}{3}$ times the third row to the first row to eliminate the leading $\frac{4}{3}$ in the first row.
2. Add 9 times the third row to the second row to eliminate the leading -3 in the second row.
3. Add $\frac{12}{59}$ times the fourth row to the first row to eliminate the leading 2 in the first row.
4. Add $\frac{9}{59}$ times the fourth row to the second row to eliminate the leading 2 in the second row.
5. Add $\frac{3}{59}$ times the fourth row to the third row to eliminate the leading 1 in the third row.

After performing these operations, we get the row reduced echelon form:
\[ A^* = \begin{pmatrix}
1 & 0 & 0 & 0 & -\frac{8}{59} \\
0 & 1 & 0 & 0 & \frac{4}{59} \\
0 & 0 & 1 & 0 & -\frac{5}{59} \\
0 &

[2, 2]

wenxinyiyan

To find the row rank and column rank of a matrix, we can use the fact that the row rank is equal to the column rank, and both are equal to the rank of the matrix. The rank of a matrix is the number of non-zero rows in its row echelon form (REF) or the number of pivot columns in its REF.

Let's perform row operations on the given matrix $A$ to transform it into its REF:

$A = \left(\begin{array}{rrrrr}
1 & 2 & 3 & 4 & -3 \\
1 & 2 & 0 & -5 & 1 \\
2 & 4 & -3 & -19 & 6 \\
3 & 6 & -3 & -24 & 7
\end{array}\right)$

Subtract the first row from the second, third, and fourth rows:

$\left(\begin{array}{rrrrr}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & -6 & -23 & 9 \\
0 & 0 & -6 & -28 & 10
\end{array}\right)$

Divide the third row by -3:

$\left(\begin{array}{rrrrr}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & 2 & \frac{23}{3} & -3 \\
0 & 0 & -6 & -28 & 10
\end{array}\right)$

Subtract twice the third row from the fourth row:

$\left(\begin{array}{rrrrr}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & 2 & \frac{23}{3} & -3 \\
0 & 0 & 0 & -\frac{35}{3} & 4
\end{array}\right)$

Swap the second and third rows:

$\left(\begin{array}{rrrrr}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 2 & \frac{23}{3} & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & 0 & -\frac{35}{3} & 4
\end{array}\right)$

Divide the second row by 2 and the third row by -3:

$\left(\begin{array}{rrrrr}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 1 & \frac{23}{6} & -\frac{3}{2} \\
0 & 0 & 1 & 3 & -\frac{4}{3} \\
0 & 0 & 0 & -\frac{35}{3} & 4
\end{array}\right)$

Subtract the second row from the third row:

$\left(\begin{array}{rrrrr}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 1 & \frac{23}{6} & -\frac{3}{2} \\
0 & 0 & 0 & -\frac{5}{6} & \frac{1}{6} \\
0 & 0 & 0 & -\frac{35}{3} & 4
\end{array}\right)$

Divide the third row by $-\frac{5}{6}$ and the fourth row by $-\frac{35}{3}$:

$\left(\begin{array}{rrrrr}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 1 & \frac{23}{6} & -\frac{3}{2} \\
0 & 0 & 0 & 1 & -\frac{1}{5} \\
0 & 0 & 0 & 1 & -\frac{12}{35}
\end{array}\right)$

Subtract the third row from the fourth row:

$\left(\begin{array}{rrrrr}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 1 & \frac{23}{6} & -\frac{3}{2} \\
0 & 0 & 0 & 1 & -\frac{1}{5} \\

[2, 2]

zhipu

To find the row rank and column rank of the matrix $ A $, we need to perform row reduction to bring the matrix to its row echelon form or reduced row echelon form and then count the number of non-zero rows. The column rank can be determined similarly by considering the columns of the matrix in echelon form, but since the row rank and column rank are equal for any matrix (this is a property of matrices known as the Rank-Nullity Theorem), we only need to compute one of them.

Let's perform row reduction on matrix $ A $:

\[ A = \begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
1 & 2 & 0 & -5 & 1 \\
2 & 4 & -3 & -19 & 6 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix} \]

Step 1: Make the first entry of the first row 1 (it already is).

Step 2: Eliminate the first entry in all other rows by subtracting appropriate multiples of the first row from the other rows.

Row 2 = Row 2 - Row 1:
\[ \begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
2 & 4 & -3 & -19 & 6 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix} \]

Row 3 = Row 3 - 2*Row 1:
\[ \begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & -9 & -27 & 12 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix} \]

Row 4 = Row 4 - 3*Row 1:
\[ \begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & -9 & -27 & 12 \\
0 & 0 & -12 & -36 & 16
\end{pmatrix} \]

Step 3: Now, let's work on the second column. We want to make the second entry of the second row 1, but since it's already 0, we move to the third column.

Step 4: Make the third entry of the second row 1 by dividing the second row by -3:
\[ \begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 1 & 3 & -4/3 \\
0 & 0 & -9 & -27 & 12 \\
0 & 0 & -12 & -36 & 16
\end{pmatrix} \]

Step 5: Eliminate the third entry in all other rows by adding appropriate multiples of the new second row to the other rows.

Row 1 = Row 1 - 3*Row 2:
\[ \begin{pmatrix}
1 & 2 & 0 & -5 & 1 \\
0 & 0 & 1 & 3 & -4/3 \\
0 & 0 & -9 & -27 & 12 \\
0 & 0 & -12 & -36 & 16
\end{pmatrix} \]

Row 3 = Row 3 + 9*Row 2:
\[ \begin{pmatrix}
1 & 2 & 0 & -5 & 1 \\
0 & 0 & 1 & 3 & -4/3 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & -12 & -36 & 16
\end{pmatrix} \]

Row 4 = Row 4 + 12*Row 2:
\[ \begin{pmatrix}
1 & 2 & 0 & -5 & 1 \\
0 & 0 & 1 & 3 & -4/3 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix} \]

Now we have the matrix in row echelon form. We can see that there are 2 non-zero rows, so the row rank of the matrix is 2. Since the row rank is equal to the column rank, the column rank is also 2.

Therefore, the row rank and column rank of matrix $ A $ are both 2.

The solution is: [2, 2]

[2, 2]

gpto1

[2, 2]

deepseekv3

To find the **row rank** and **column rank** of the matrix $ A $, we perform **Gaussian elimination** to reduce the matrix to its **row echelon form** (REF). The number of non-zero rows in the REF gives the row rank, and the number of non-zero columns gives the column rank.

### Step 1: Write the matrix $ A $:
\[
A = \begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
1 & 2 & 0 & -5 & 1 \\
2 & 4 & -3 & -19 & 6 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix}
\]

### Step 2: Perform Gaussian elimination to get the row echelon form (REF):
1. Subtract the first row from the second row:
\[
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
2 & 4 & -3 & -19 & 6 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix}
\]

2. Subtract $ 2 \times $ the first row from the third row:
\[
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & -9 & -27 & 12 \\
3 & 6 & -3 & -24 & 7
\end{pmatrix}
\]

3. Subtract $ 3 \times $ the first row from the fourth row:
\[
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & -3 & -9 & 4 \\
0 & 0 & -9 & -27 & 12 \\
0 & 0 & -12 & -36 & 16
\end{pmatrix}
\]

4. Divide the second row by $ -3 $:
\[
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 1 & 3 & -\frac{4}{3} \\
0 & 0 & -9 & -27 & 12 \\
0 & 0 & -12 & -36 & 16
\end{pmatrix}
\]

5. Add $ 9 \times $ the second row to the third row:
\[
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 1 & 3 & -\frac{4}{3} \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & -12 & -36 & 16
\end{pmatrix}
\]

6. Add $ 12 \times $ the second row to the fourth row:
\[
\begin{pmatrix}
1 & 2 & 3 & 4 & -3 \\
0 & 0 & 1 & 3 & -\frac{4}{3} \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
\]

### Step 3: Determine the row rank and column rank:
- The REF has **2 non-zero rows**, so the **row rank** is $ 2 $.
- The REF has **2 non-zero columns**, so the **column rank** is $ 2 $.

### Final Answer:
The row rank and column rank of $ A $ are both $ 2 $. Thus, the solution is:
\[
[2, 2]
\]

[2, 2]

问题

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