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:

An extremely useful shortcut:
Any square matrix of size \(n \times n\) with full rank (determinant \(\neq 0\)) always reduces to the identity matrix \(I_n\) in RREF.
Since \(B\) is upper triangular with all non-zero diagonal entries, it has full rank. Its RREF is immediately the identity matrix.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given an upper triangular \(3 \times 3\) matrix \(B\).
We need to find its Reduced Row Echelon Form (RREF).
Key Formula or Approach:
The Reduced Row Echelon Form of a matrix has the following properties:
1. If a row does not consist entirely of zeros, then the first non-zero number in the row is a 1 (called a leading 1).
2. If there are any rows that consist entirely of zeros, they are grouped at the bottom.
3. In any two successive rows that do not consist entirely of zeros, the leading 1 in the lower row occurs farther to the right than the leading 1 in the higher row.
4. Each column that contains a leading 1 has zeros everywhere else in that column.

Step 2: Detailed Explanation:

• Let us analyze the diagonal elements of the upper triangular matrix \(B\):
The diagonal entries are 5, 2, and 7.
Since all diagonal entries of this upper triangular matrix are non-zero, the determinant of \(B\) is the product of the diagonal entries:
\[ \det(B) = 5 \times 2 \times 7 = 70 \neq 0 \]

• Since the determinant is non-zero, the matrix is invertible (has full rank of 3).

• For any invertible \(n \times n\) matrix, its unique reduced row echelon form is always the \(n \times n\) identity matrix \(I_n\).

• Let us perform the row operations explicitly to confirm:
- Normalizing rows:
\(R_1 \to \frac{1}{5} R_1 \implies [1, -1.8, 1.2]\)
\(R_2 \to \frac{1}{2} R_2 \implies [0, 1, 1.5]\)
\(R_3 \to \frac{1}{7} R_3 \implies [0, 0, 1]\)
Matrix becomes:
\[ \begin{bmatrix} 1 & -1.8 & 1.2 0 & 1 & 1.5 0 & 0 & 1 \end{bmatrix} \]
- Eliminating entries above the leading 1 of \(R_3\) in column 3:
\(R_2 \to R_2 - 1.5 R_3 \implies [0, 1, 0]\)
\(R_1 \to R_1 - 1.2 R_3 \implies [1, -1.8, 0]\)
Matrix becomes:
\[ \begin{bmatrix} 1 & -1.8 & 0 0 & 1 & 0 0 & 0 & 1 \end{bmatrix} \]
- Eliminating the entry above the leading 1 of \(R_2\) in column 2:
\(R_1 \to R_1 + 1.8 R_2 \implies [1, 0, 0]\)
Final RREF:
\[ \begin{bmatrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{bmatrix} \]

Step 3: Final Answer:
The reduced echelon form of matrix \(B\) is the identity matrix \(\begin{bmatrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{bmatrix}\).