Question

Let
\(A=\begin{pmatrix} 1 & 1 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 3 \\ 1 & 1 & 4 & 4 &4 \\ \end{pmatrix}\)
and B be a 5 × 5 real matrix such that AB is the zero matrix. Then the maximum possible rank of B is equal to ____________.

Answer 1

Correct Answer: 2

To solve the problem of determining the maximum rank of matrix \(B\) given that \(AB\) is the zero matrix, we need to apply the rank-nullity theorem and properties of matrix multiplication.

1. **Identify the Rank of \(A\):**
Matrix \(A\) is a \(3 \times 5\) matrix. To determine its rank, we perform row operations to obtain its row echelon form:
\[A = \begin{pmatrix} 1 & 1 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 3 \\ 1 & 1 & 4 & 4 & 4 \\ \end{pmatrix}\]
- Subtract the first row from the second and third rows:
\(\begin{pmatrix} 1 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & 2 \\ 0 & 0 & 4 & 4 & 3 \end{pmatrix}\)
- Subtract quadruple of the second row from the third row:
\(\begin{pmatrix} 1 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 & -5 \end{pmatrix}\)
The rank of \(A\) is the number of non-zero rows, which is 3.

2. **Apply Matrix Rank Properties:**
The product \(AB\) being a zero matrix implies that the column space of \(B\) is in the null space of \(A\).
According to the rank-nullity theorem, for \(A\) (a \(3 \times 5\) matrix), we have:
\(\text{rank}(A) + \text{nullity}(A) = 5\)
Since \(\text{rank}(A) = 3\), then \(\text{nullity}(A) = 2\).
The nullity of \(A\) provides the dimensionality of the solution space in which the columns of \(B\) must lie.

3. **Determine the Maximum Rank of \(B\):**
The maximum number of linearly independent columns that \(B\) can have corresponds to the dimensionality of the null space of \(A\), which is its nullity.
Therefore, the maximum rank of \(B\) is equal to the nullity of \(A\):
\(\text{max rank}(B) = 2\).

This value, 2, falls within the expected range (2,2) as provided. Thus, the maximum possible rank of \(B\) is confirmed to be 2.