GATE CS 1994 Q1.9 — Linear Algebra: Matrices, Determinants, System of Linear Equations

We can find the rank of the matrix $A = \begin{bmatrix} 0 & 0 & -3 \\ 9 & 3 & 5 \\ 3 & 1 & 1 \end{bmatrix}$ using two standard methods:

Method 1: Determinant and Minors

First, let's calculate the determinant of the $3 \times 3$ matrix $A$ by expanding along the first row:

Since $\det(A) = 0$, the rank of the matrix must be strictly less than $3$.

Now, let's check if there exists any $2 \times 2$ submatrix (minor) with a non-zero determinant. Consider the submatrix formed by the last two rows and the last two columns:

Since there is at least one non-zero minor of order $2$, the rank of the matrix is $2$.

Method 2: Row Reduction (Echelon Form)

Let's perform elementary row operations to reduce the matrix to row echelon form:

1. Swap row 1 ($R_1$) and row 3 ($R_3$): $$\begin{bmatrix} 3 & 1 & 1 \\ 9 & 3 & 5 \\ 0 & 0 & -3 \end{bmatrix}$$
2. Perform $R_2 \to R_2 - 3R_1$: $$\begin{bmatrix} 3 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 0 & -3 \end{bmatrix}$$
3. Perform $R_3 \to R_3 + \frac{3}{2}R_2$: $$\begin{bmatrix} 3 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{bmatrix}$$

The number of non-zero rows in the row echelon form is $2$. Therefore, the rank of the matrix is $2$.