Question

Let \( T: \mathbb{R}^3 \to \mathbb{R}^3 \) be a linear map defined by \[ T(x_1, x_2, x_3) = (3x_1 + 5x_2 + x_3, x_3, 2x_1 + 2x_3). \] {Then the rank of \( T \) is equal to ________ (answer in integer).}

Hint:

To find the rank of a matrix, reduce it to row echelon form and count the number of non-zero rows. The rank is equal to the number of linearly independent rows.

Answer 1

To find the rank of the linear map \( T \), we need to determine the number of linearly independent rows in the matrix representation of \( T \). The map \( T \) is given by the following transformation: \[ T(x_1, x_2, x_3) = \begin{pmatrix} 3 & 5 & 1 \\ 0 & 0 & 1 \\ 2 & 0 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}. \] Thus, the matrix representation of \( T \) is: \[ A = \begin{pmatrix} 3 & 5 & 1 \\ 0 & 0 & 1 \\ 2 & 0 & 2 \end{pmatrix}. \] Now, we find the rank of this matrix by reducing it to row echelon form (REF). First, use row operations to simplify the matrix: Subtract \( \frac{2}{3} \) of the first row from the third row to make the element in the third row, first column, zero: \[ \begin{pmatrix} 3 & 5 & 1 \\ 0 & 0 & 1 \\ 0 & -\frac{10}{3} & \frac{4}{3} \end{pmatrix}. \] Multiply the third row by \( -\frac{3}{10} \) to make the second column entry of the third row equal to 1: \[ \begin{pmatrix} 3 & 5 & 1 \\ 0 & 0 & 1 \\ 0 & 1 & -\frac{2}{5} \end{pmatrix}. \] Finally, we can easily see that the first and second rows are linearly independent, and the third row is also linearly independent from the others. Thus, the matrix has 3 non-zero rows, so the rank of \( T \) is 3. 
Rank of \( T \) is: 3