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of I MON re now in a position to prove that if A is an m × n mat ces H and J from A using a finite sequence of elementary 1 d J are in reduced row echelon form, then H = J. Thus, the ise the Principle of Mathematical Induction.

of I MON re now in a position to prove that if A is an m × n mat ces H and J from A using a finite sequence of elementary 1 d J are in reduced row echelon form, then H = J. Thus, the ise the Principle of Mathematical Induction.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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2.3 #9

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Please answer d and e

9. Proof of Theorem 1.5.3: The Uniqueness of the Reduced Row Echelon Form: We are now in a position to prove that if A is an m × n matrix, and we obtain two matrices H and J from A using a finite sequence of elementary row operations, and both H and J are in reduced row echelon form, then H = J. Thus, the rref of A is unique. We will use the Principle of Mathematical Induction.
Transcribed Image Text:9. Proof of Theorem 1.5.3: The Uniqueness of the Reduced Row Echelon Form: We are now in a position to prove that if A is an m × n matrix, and we obtain two matrices H and J from A using a finite sequence of elementary row operations, and both H and J are in reduced row echelon form, then H = J. Thus, the rref of A is unique. We will use the Principle of Mathematical Induction.
d. Both H and J below are in rref and have rank 3: 1 0 0 6. 7 1 0 -2 0 7 H = 0 1 0 -2 4 J = 0 1 4 -0 4 0 0 1 5 -3 0 0 0 1 -3 Explain why the 3rd row of J cannot be expressed as a linear combination of the three rows of H. Hint: use the fact that the leading 1 is in the 4th column and every entry to its left is zero. Now, explain in general that the leading 1 in row k of H must be in the same column as the leading 1 in row k of J. Hint: pick the matrix whose leading one in row k is further to the right. е.
Transcribed Image Text:d. Both H and J below are in rref and have rank 3: 1 0 0 6. 7 1 0 -2 0 7 H = 0 1 0 -2 4 J = 0 1 4 -0 4 0 0 1 5 -3 0 0 0 1 -3 Explain why the 3rd row of J cannot be expressed as a linear combination of the three rows of H. Hint: use the fact that the leading 1 is in the 4th column and every entry to its left is zero. Now, explain in general that the leading 1 in row k of H must be in the same column as the leading 1 in row k of J. Hint: pick the matrix whose leading one in row k is further to the right. е.
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