Question

The rank of the matrix \( A = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 0 & 1 \\ 4 & 1 & 4 \end{bmatrix} \) is

Hint:

The rank of a matrix is the maximum number of linearly independent rows (or columns).
It can be found by reducing the matrix to row echelon form and counting the number of non-zero rows.
For a square matrix, if its determinant is non-zero, its rank is equal to its order. If the determinant is zero, the rank is less than its order.
Rank is also the order of the largest non-singular (non-zero determinant) square submatrix.

Answer 1

The matrix is \[ A = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 0 & 1 \\ 4 & 1 & 4 \end{bmatrix}. \]

To find the rank, we can use row operations to reduce it to row echelon form or calculate its determinant. If \( \det(A) \neq 0 \), the rank is 3. If \( \det(A) = 0 \), then rank is less than 3.

Step 1: Calculate Determinant

\[ \det(A) = 2 \begin{vmatrix} 0 & 1 \\ 1 & 4 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 4 & 4 \end{vmatrix} + 2 \begin{vmatrix} 1 & 0 \\ 4 & 1 \end{vmatrix} \] \[ = 2(0 \cdot 4 - 1 \cdot 1) - 1(1 \cdot 4 - 1 \cdot 4) + 2(1 \cdot 1 - 0 \cdot 4) \] \[ = 2(-1) - 1(0) + 2(1) = -2 + 0 + 2 = 0 \]

Since \( \det(A) = 0 \), the rank is less than 3.

Step 2: Check for Rank 2 (2x2 Submatrix)

Consider the top-left 2x2 submatrix: \[ \begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix} \] Its determinant is \( (2)(0) - (1)(1) = -1 \neq 0 \).

So, a 2x2 submatrix has a non-zero determinant → rank is at least 2.

Step 3: Use Row Operations

\[ A = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 0 & 1 \\ 4 & 1 & 4 \end{bmatrix} \Rightarrow R_1 \leftrightarrow R_2 \Rightarrow \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 2 \\ 4 & 1 & 4 \end{bmatrix} \] \[ R_2 \rightarrow R_2 - 2R_1, \quad R_3 \rightarrow R_3 - 4R_1 \Rightarrow \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \end{bmatrix} \] \[ R_3 \rightarrow R_3 - R_2 \Rightarrow \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]

This is row echelon form. There are two non-zero rows → rank is:

\(\boxed{2}\)