Question

The rank of the matrix A = $ \begin{bmatrix} 1 & 3 & 4 & 3 \\ 3 & 9 & 12 & 9 \\ -1 & -3 & -4 & -3 \end{bmatrix} $ is

• A. \( 3 \)
• B. \( 2 \)
• C. \( 1 \)
• D. \( 0 \)

Hint:

The rank of a matrix is the maximum number of linearly independent row vectors or column vectors. In this specific case, notice that the second row ($[3 \ 9 \ 12 \ 9]$) is exactly 3 times the first row ($[1 \ 3 \ 4 \ 3]$), and the third row ($[-1 \ -3 \ -4 \ -3]$) is exactly -1 times the first row. This indicates that all rows are scalar multiples of the first row, meaning there is only one linearly independent row. Hence, the rank is 1. This can often be a faster way to determine rank for matrices where rows/columns are clearly dependent.

Answer 1

The Correct Option is C

To find the rank of a matrix, we can reduce it to row echelon form using elementary row operations. The number of non-zero rows in the row echelon form will be the rank of the matrix. Given matrix A: $$ A = \begin{bmatrix} 1 & 3 & 4 & 3 \\ 3 & 9 & 12 & 9 \\ -1 & -3 & -4 & -3 \end{bmatrix} $$
Step 1: Perform row operations to make elements below the leading 1 in the first column zero. Apply $R_2 \to R_2 - 3R_1$: $$ \begin{bmatrix} 1 & 3 & 4 & 3 \\ 3 - 3(1) & 9 - 3(3) & 12 - 3(4) & 9 - 3(3) \\ -1 & -3 & -4 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 3 & 4 & 3 \\ 0 & 0 & 0 & 0 \\ -1 & -3 & -4 & -3 \end{bmatrix} $$
Step 2: Apply $R_3 \to R_3 + R_1$: $$ \begin{bmatrix} 1 & 3 & 4 & 3 \\ 0 & 0 & 0 & 0 \\ -1 + 1 & -3 + 3 & -4 + 4 & -3 + 3 \end{bmatrix} = \begin{bmatrix} 1 & 3 & 4 & 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$
Step 3: Determine the rank. The matrix is now in row echelon form. We can see that there is only one non-zero row. Therefore, the rank of the matrix A is 1.