Linear Algebra

1. Let

a) Row reduce to put A in echelon form.
b) Continue row reducing to put A in reduced echelon form.
c) Identify the basic columns of A, the free columns of A and the rank of A.


After row reducing in any one of a variety of ways, you get the reduced echelon form of A:

Thus there are pivots in columns 1 and 3, so those are the basic columns. The free columns are 2
and 4, and rank(A) = 2.

2. Consider the inhomogeneous linear system

3x + 2y + z = 13
−x + 4y + 2z = −2
2x + 6y + 3z = 11

a) Write the augmented matrix for the system.
b) Row reduce to echelon form.
c) Find a particular solution to the inhomogeneous system.
d) Determine and describe parametrically all solutions to the corresponding homogeneous system.
e) Describe all solutions to the original inhomogeneous system.


The augmented matrix for the system is

You can row reduce to echelon form, or go all the way to reduced echelon form, which is

So z is a free variable , and you can find a particular solution by taking z = anything, say z=0.

Then the second line says

y + 1/2z = 1/2, so for z = 0, y = 1/2

The the first line says

x = 4

so our particular solution is (x,y,z) = (4, 1/2, 0)

For the homogeneous system ignore column 4, consider the constants to be 0. Then the reduced matrix says:

y + 1/2 z = 0 , which implies y = -1/2 z, and x = 0
So the general homogeneous solution set is {(0,-1/2 z, z)}.
Scaling by a factor of 2, writing z = 2t, we can say that the general homogeneous solution is parametrically
described by (x, y, z) = t (0, -1, 2)

So the general solution to the inhomogeneous system is
{(x, y, z) = (4, 1/2, 0 ) + t (0, -1, 2)}

3. Consider the linear system

x + (a + 2)y = a + 2
(a + 1)x + 2y = 3a + 11

For which values of the parameter a does the system have
a) a unique solution
b) no solutions
c) infinitely many solutions


The associated augmented matrix is

Row reducing to echelon form (without dividing by something that might be zero) we get

Case 1: If a ≠ 0 or -3, then −a^2 − 3a ≠ 0, so there is a pivot in the second row. Thus the rank of
the coefficient matrix is 2, and there is a unique solution.

Case 2: If a = -3, then the second line is identically zero, and then y is a free variable. Thus there
are infinitely many solutions.

Case 3: If a = 0, then the second row is (0 0 9), so there is a pivot in the augmentation column. The
equations are inconsistent , so no solutions.

4. Let

Write the vector

as a linear combination of v 1, v2, v3.


Set this up as a system of equations x1v1 + x2v2 + x3v3 = w. The augmented matrix is

The reduced echelon form of A is

Thus there is a unique solution

5. List seven 3 × 3 matrices, all with rank ≥ 2, that are not row equivalent . All matrices listed must
be in reduced echelon form.

6. Consider the vectors the diagonal of a cube given by the vector v = (1, 1, 1) and the diagonal of one
of its faces given by w = (1, 1, 0). Determine the lengths of v and w, and the angle between them.

The vectors are given, so just calculate :

so the angle between v and w is given by

7. Give examples of

(a) A system of four linear equations with 3 unknowns that has more than one solution .
(b) A system of three equations with 4 unknowns that has no solution .

There is a lot of free play here. Some really simple examples are possible, for example

8. Find all vectors in R^4 that are perpendicular (orthogonal) to

Let w = (a,b,c,d), then we have two equations in 4 unknowns:

w · v1 = 0,w · v2 = 0

This translates to
a - c = 0
a - b = 0

So a = b and a = c, where c and d are free variables. Thus the complete solutions set is
{(a, b, c, d) = (c, c, c, d)}
Parametrically, that is {c ( 1, 1, 1, 0) + d (0, 0, 0, 1)}

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