Assignment 1   –   Solutions

Identify which of the matrices below are in reduced row-echelon form and which are in row-echelon (but not fully reduced) form.   Where a matrix is in row-echelon (or reduced row-echelon) form, circle the leading ones.   Where a matrix is not in row-echelon form, state why not.

1. 

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   Row-echelon form (but not fully reduced, due to the non-zero entries above the two right-most leading ones).
   

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2. 

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   Reduced row-echelon form
   

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3. 

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   Not in row-echelon form.   There is a non-zero entry below the second leading one.

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4. 

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   Reduced row-echelon form
   

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5. 

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   Reduced row-echelon form
   

From equation 2 subtract equation 1:


Reading the reduced row-echelon form:
1 x + 0 y = 15   and
0 x + 1 y = 2
r = rank(A) = rank(A|b) = n = 2   Therefore there is a unique solution.
The output from the linear system reduction program is available here.
This system is so simple that other methods work well.
x and y are two numbers whose sum is 17 and whose difference is 13.
Obviously the larger number x is 15, halfway between 13 and 17, while y is the gap between 15 and each of 13 and 17.
Therefore this system has a

unique solution:     (x, y) = (15, 2)

r = rank(A) = rank(A|b) = 2   but   n = 4
p = n – r = 4 – 2 = 2
There is a two-parameter family of solutions.
w and x are leading variables and y and z are the free parameters (call them s and t).
Reading the reduced row-echelon form:
1 w + 0 x + 1 y – 1 z = 2
0 w + 1 x + 1 y + 1 z = 1
0 w + 0 x + 0 y + 0 z = 0
The third equation is redundant (it does not add any information not already provided by the first two equations).   Therefore there are

infinitely many solutions:

The output from the linear system reduction program is available here.
One can easily verify that this family of solutions satisfies all three equations of the linear system for all choices of the parameters s and t:
   
   
   

The output from the linear system reduction program is available here.
This row-echelon form has a leading one in the column for the right-side constants.
The system is inconsistent.
The bottom row is equivalent to the equation   0 x + 0 y + 0 z = 1, which has

no solution

The equation of the cubic curve is satisfied by all four points:

One could use row 2 to replace d by 1 in the other three equations, thereby reducing the system to three equations in three unknowns.   It would also make sense the rearrange the order of the equations before beginning the Gaussian elimination, placing rows 1, 2, 3 and 4 into rows 2, 4, 1 and 3 respectively, then removing new equation 4 upon noting d = 1:

This alternate solution is completed on another page.
Proceeding systematically with the original system instead,

The values of the coefficients are therefore

a = 3 , b = –4 , c = 2   and   d = 1

One can verify that the curve   y = 3x³ – 4x² + 2x + 1   does indeed pass through all four points:
(–1, –8) :     3×–1 – 4×1 + 2×–1 + 1 = –8    
(0, 1) :         3×0 – 4×0 + 2×0 + 1 = 1    
(1, 2) :         3×1 – 4×1 + 2×1 + 1 = 2    
(2, 13) :       3×8 – 4×4 + 2×2 + 1 = 13