Problem 268 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Find the rank of the matrix \(A=\left[\begin{array}{lll}1 & 2 & 3 \\ 3 & 6 & 9 \\\ 1 & 2 & 3\end{array}\right]\).

Step-by-Step Solution

Answer

The rank of the matrix A is 1.

Step 1: Identify Linearly Independent Rows

In the matrix \( A=\left[\begin{array}{lll}1 & 2 & 3 \ 3 & 6 & 9 \ 1 & 2 & 3\end{array}\right] \), it is observed that Rows 1 and 3 are identical, and Row 2 is a multiple of Row 1, specifically it's 3 times Row 1.

Step 2: Reduce to Row Echelon Form

If multiple rows are the same or multiples of each other, they are linearly dependent and can be reduced to a simpler row through the Gaussian elimination process. After reducing, the Matrix A becomes \( A=\left[\begin{array}{lll}1 & 2 & 3 \ 0 & 0 & 0 \ 0 & 0 & 0\end{array}\right] \).

Step 3: Determine the Rank

In the simplified matrix, the rank is given by the number of non-zero rows. Here, only one row is non-zero.

Key Concepts

Linearly Independent Rows

Understanding linearly independent rows is crucial for determining the rank of a matrix. In simple terms, rows in a matrix are linearly independent if no row can be written as a combination of other rows. For example, consider a matrix where two rows are identical or one row is a scaled version of another. These rows are linearly dependent.

In the given matrix:

• Row 1 is \([1, 2, 3]\)
• Row 2 is \([3, 6, 9]\), which is precisely 3 times Row 1.
• Row 3 is also \([1, 2, 3]\), identical to Row 1.

As seen here, Rows 1 and 3 are identical, and Row 2 is a multiple of Row 1. This confirms that all rows in this matrix are linearly dependent on Row 1. To find the rank, we must first identify and keep only the linearly independent rows.

Gaussian Elimination

Gaussian elimination is a method used to simplify matrices, making it easier to determine their rank and solve systems of linear equations. This process involves performing operations on the rows of a matrix to transform it into a simpler form, typically aiming for what's known as "Row Echelon Form."

In our scenario, since Rows 2 and 3 are linearly dependent on Row 1, we can use Gaussian elimination to zero them out. This is achieved by:

• Subtracting multiples of Row 1 from Row 2 and Row 3.
• The result is a new matrix:\[\begin{array}{c c c}1 & 2 & 3 \0 & 0 & 0 \0 & 0 & 0\end{array}\]

This approach effectively makes the dependency explicit, leaving only the non-zero row. Remember, this method allows us to systematically simplify matrices and uncover their essential structure without changing their rank.

Row Echelon Form

The row echelon form (REF) of a matrix is a type of matrix form that simplifies various matrix operations, including rank determination. In REF, all non-zero rows appear above any rows of all zeros, and the leading entry (or "pivot") of each non-zero row is to the right of the leading entry of the row above it.

For our matrix:

• The transformation through Gaussian elimination provided us with the row echelon form:\[\begin{array}{c c c}1 & 2 & 3 \0 & 0 & 0 \0 & 0 & 0\end{array}\]
• This arrangement is simple, as it consists of only one non-zero row, with zeros following immediately, making it easy to determine the rank.

The rank of the matrix is given by the number of non-zero rows in its row echelon form. Hence, in this case, the rank of the matrix is 1. This representation helps streamline matrix-related problems and gives a clear count of independent rows.