A is a 4 x 4 matrix. By applying a few elementary transformations on A, if we get the matrix [2 1 0 0; 0 0 0 0; 0 0 1 0; 3 3/2 0 0], which is equivalent to A, then rank of A is

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q5 – AP-EAMCET 2026

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Home Maths MatricesQuestion #281

A is a

4 \times 4

matrix. By applying a few elementary transformations on A, if we get the matrix

(\begin{matrix} 2 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 3 & \frac{3}{2} & 0 & 0 \end{matrix})

, which is equivalent to A, then rank of A is

Clue 1 of 3: Rank Definition

The rank of a matrix is the maximum number of linearly independent row or column vectors. Elementary row/column operations do not change the rank of a matrix.

🧠 Full Solution Path

Step 1: Understand the Problem and Matrix Properties✦ Active

The problem states that matrix A is a $4 \times 4$ matrix. After applying elementary transformations on A, we obtain an equivalent matrix. A fundamental property of elementary row/column operations is that they do not change the rank of a matrix. Therefore, the rank of A is equal to the rank of the given transformed matrix.

M = (\begin{matrix} 2 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 3 & \frac{3}{2} & 0 & 0 \end{matrix})

💡 Teacher's Secret Hint

Remember that rank is invariant under elementary row and column operations. This simplifies the problem as we only need to find the rank of the given matrix.

Step 2: Identify Linearly Dependent Rows○ Expand

We examine the rows of matrix M to find any linear dependencies. Let $R_{1}, R_{2}, R_{3}, R_{4}$ be the first, second, third, and fourth rows, respectively. We observe that $R_{2}$ is already a zero row: $R_{2} = (0, 0, 0, 0)$ . Now let's compare $R_{1}$ and $R_{4}$ : $R_{1} = (2, 1, 0, 0)$ and $R_{4} = (3, \frac{3}{2}, 0, 0)$ . We can see that $R_{4} = \frac{3}{2} R_{1}$ , because $3 = \frac{3}{2} \times 2$ and $\frac{3}{2} = \frac{3}{2} \times 1$ .

💡 Teacher's Secret Hint

Always look for obvious zero rows or rows that are scalar multiples of other rows first, as these are the easiest to identify as linearly dependent.

Step 3: Perform Row Operations to Achieve Row Echelon Form○ Expand

Since $R_{4} = \frac{3}{2} R_{1}$ , we can eliminate $R_{4}$ by performing the elementary row operation $R_{4} \to R_{4} - \frac{3}{2} R_{1}$ . This will make $R_{4}$ a zero row.

M^{'} = (\begin{matrix} 2 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 3 - \frac{3}{2} (2) & \frac{3}{2} - \frac{3}{2} (1) & 0 - \frac{3}{2} (0) & 0 - \frac{3}{2} (0) \end{matrix}) = (\begin{matrix} 2 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{matrix})

💡 Teacher's Secret Hint

The goal of row operations is to simplify the matrix to a form where the number of non-zero rows is easily countable. Row echelon form is ideal for this.

Step 4: Count Non-Zero Rows to Determine Rank○ Expand

In the transformed matrix $M^{'}$ , we have two non-zero rows: $R_{1}^{'} = (2, 1, 0, 0)$ and $R_{3}^{'} = (0, 0, 1, 0)$ . These two rows are linearly independent (one cannot be obtained by scaling the other). The other two rows ( $R_{2}^{'}$ and $R_{4}^{'}$ ) are zero rows. The rank of a matrix is defined as the number of non-zero rows in its row echelon form.

Therefore, the rank of matrix M (and consequently, matrix A) is 2.

💡 Teacher's Secret Hint

A non-zero row is any row that contains at least one non-zero element. Make sure you don't confuse the number of rows with the rank itself.

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Maths Question 5 – AP-EAMCET 2026

Clue 1 of 3: Rank Definition

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