Question

Let B = \(\begin{bmatrix} 5 & -9 & 6 0 & 2 & 3 0 & 0 & 7 \end{bmatrix}\) , then its reduced echelon form is :

• A. \(\begin{bmatrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{bmatrix}\)
• B. \(\begin{bmatrix} 1 & 0 & 0 0 & 0 & 1 0 & 1 & 0 \end{bmatrix}\)
• C. \(\begin{bmatrix} 5 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{bmatrix}\)
• D. \(\begin{bmatrix} 1 & -9 & 0 0 & 1 & 0 0 & 0 & 1 \end{bmatrix}\)

Hint:

To solve this instantly without calculations:
For any square matrix, if its determinant is non-zero (which we know here because the diagonal product of the triangular matrix is \(5 \times 2 \times 7 = 70 \ne 0\)), its reduced row echelon form (RREF) is always the identity matrix \(I\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The reduced row echelon form (RREF) of a matrix is obtained by performing a sequence of elementary row operations.
A matrix is in RREF if:
1. The first non-zero entry in any row (called the leading entry) is 1.
2. The leading entry of any row is to the right of the leading entry of the row above it.
3. All entries in a column above and below a leading 1 are zero.

Step 2: Detailed Explanation:
Let us analyze the given upper triangular matrix \(B\):
\[ B = \begin{bmatrix} 5 & -9 & 6 0 & 2 & 3 0 & 0 & 7 \end{bmatrix} \]
Notice that the diagonal entries are \(5\), \(2\), and \(7\).
Since all diagonal entries of this upper triangular matrix are non-zero, the determinant of the matrix is:
\[ \det(B) = 5 \times 2 \times 7 = 70 \ne 0 \]
Since the determinant is non-zero, the matrix is invertible (non-singular).
By a fundamental theorem of linear algebra, any invertible \(n \times n\) matrix can be reduced via elementary row operations to the \(n \times n\) identity matrix \(I_n\).
Let us show the row operations explicitly to confirm this:
-

Step 1: Divide Row 3 by 7 to make its leading entry 1:
\[ R_3 \to \frac{1}{7} R_3 \implies \begin{bmatrix} 5 & -9 & 6 0 & 2 & 3 0 & 0 & 1 \end{bmatrix} \]
- Use the leading 1 in Row 3 to eliminate the entries above it in Column 3:
\[ R_2 \to R_2 - 3 R_3 \implies \begin{bmatrix} 5 & -9 & 6 0 & 2 & 0 0 & 0 & 1 \end{bmatrix} \]
\[ R_1 \to R_1 - 6 R_3 \implies \begin{bmatrix} 5 & -9 & 0 0 & 2 & 0 0 & 0 & 1 \end{bmatrix} \]
-

Step 2: Divide Row 2 by 2 to make its leading entry 1:
\[ R_2 \to \frac{1}{2} R_2 \implies \begin{bmatrix} 5 & -9 & 0 0 & 1 & 0 0 & 0 & 1 \end{bmatrix} \]
-

Step 3: Use the leading 1 in Row 2 to eliminate the entry above it in Column 2:
\[ R_1 \to R_1 + 9 R_2 \implies \begin{bmatrix} 5 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{bmatrix} \]
-

Step 5: Divide Row 1 by 5 to make its leading entry 1:
\[ R_1 \to \frac{1}{5} R_1 \implies \begin{bmatrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{bmatrix} \]
The matrix is now in reduced row-echelon form, which is the identity matrix \(I_3\).

Step 3: Final Answer:
The correct option is (A).