Question

The rank of the matrix \( \begin{bmatrix} 2 & -3 & 4 & 0 \\ 5 & -4 & 2 & 1 \\ 1 & -3 & 5 & -4 \end{bmatrix} \) is

• A. \(0 \)
• B. \(3 \)
• C. \(2 \)
• D. \(1 \)

Hint:

The rank of a matrix can be found by reducing it to row echelon form using elementary row operations. The number of non-zero rows in the row echelon form is the rank. Alternatively, find the largest square submatrix whose determinant is non-zero; the order of this submatrix is the rank.

Answer 1

The Correct Option is B

Step 1: Understand the definition of the rank of a matrix.
The rank of a matrix is the maximum number of linearly independent row vectors or column vectors. It is also equal to the order of the largest non-zero minor (determinant of a submatrix).
The given matrix \(A\) is a \(3 \times 4\) matrix. The maximum possible rank of this matrix is \(\min(3, 4) = 3\).
Step 2: Reduce the matrix to row echelon form using elementary row operations.
Let the given matrix be \(A\):
\[ A = \begin{bmatrix} 2 & -3 & 4 & 0 \\ 5 & -4 & 2 & 1 \\ 1 & -3 & 5 & -4 \end{bmatrix} \] Swap \(R_1\) and \(R_3\) to get a '1' in the top-left corner for easier calculations: \[ A \sim \begin{bmatrix} 1 & -3 & 5 & -4 \\ 5 & -4 & 2 & 1 \\ 2 & -3 & 4 & 0 \end{bmatrix} \] Perform row operations to make elements below the leading '1' in the first column zero:
\(R_2 \to R_2 - 5R_1\)
\(R_3 \to R_3 - 2R_1\)
\[ A \sim \begin{bmatrix} 1 & -3 & 5 & -4 \\ 5 - 5(1) & -4 - 5(-3) & 2 - 5(5) & 1 - 5(-4) \\ 2 - 2(1) & -3 - 2(-3) & 4 - 2(5) & 0 - 2(-4) \end{bmatrix} \] \[ A \sim \begin{bmatrix} 1 & -3 & 5 & -4 \\ 0 & 11 & -23 & 21 \\ 0 & 3 & -6 & 8 \end{bmatrix} \] Now, make the element in the second column of the third row zero: \(R_3 \to 11R_3 - 3R_2\) (to avoid fractions) \[ A \sim \begin{bmatrix} 1 & -3 & 5 & -4 \\ 0 & 11 & -23 & 21 \\ 0 & 11(3) - 3(11) & 11(-6) - 3(-23) & 11(8) - 3(21) \end{bmatrix} \] \[ A \sim \begin{bmatrix} 1 & -3 & 5 & -4 \\ 0 & 11 & -23 & 21 \\ 0 & 0 & -66 + 69 & 88 - 63 \end{bmatrix} \] \[ A \sim \begin{bmatrix} 1 & -3 & 5 & -4 \\ 0 & 11 & -23 & 21 \\ 0 & 0 & 3 & 25 \end{bmatrix} \] Step 3: Determine the rank from the row echelon form.
The matrix is now in row echelon form. The number of non-zero rows is 3.
Therefore, the rank of the matrix is 3.
Alternative Method (using determinants of minors):
The maximum possible rank is 3. We can check if there exists a \(3 \times 3\) submatrix whose determinant is non-zero.
Consider the submatrix formed by the first three columns: \[ M = \begin{vmatrix} 2 & -3 & 4 \\ 5 & -4 & 2 \\ 1 & -3 & 5 \end{vmatrix} \] Calculate its determinant: \[ \det(M) = 2 \begin{vmatrix} -4 & 2 \\ -3 & 5 \end{vmatrix} - (-3) \begin{vmatrix} 5 & 2 \\ 1 & 5 \end{vmatrix} + 4 \begin{vmatrix} 5 & -4 \\ 1 & -3 \end{vmatrix} \] \[ = 2((-4)(5) - (2)(-3)) + 3((5)(5) - (2)(1)) + 4((5)(-3) - (-4)(1)) \] \[ = 2(-20 + 6) + 3(25 - 2) + 4(-15 + 4) \] \[ = 2(-14) + 3(23) + 4(-11) \] \[ = -28 + 69 - 44 \] \[ = 41 - 44 = -3. \] Since the determinant of this \(3 \times 3\) minor is \(-3\), which is non-zero, the rank of the matrix is 3.
The final answer is \( \boxed{3} \).