1.

3x	+	2y	+	z	=	0
x	-	y	+	z	=	0
0x	+	2y	+	3z	=	0

...has augmented matrix

3	2	1	|	0
1	-1	1	|	0
0	2	3	|	0

...which has reduced row-echelon form

1	0	0	|	0
0	1	0	|	0
0	0	1	|	0

...and hence has only the trivial solution x₁=x₂=x₃=0.

x	+	2y	+	z	=	2
x	-	y	+	z	=	1
-x	+	4y	-	z	=	3

...has augmented matrix

1	2	1	|	2
1	-1	1	|	1
-1	4	-1	|	3

...which can be row-reduced to

1	2	1	|	2
0	6	0	|	2
0	0	0	|	3

...and is therefore inconsistent: there are no solutions, since it is a well-known fact that 0 does not equal 3.

2. Suppose p(t) is a degree two polynomial:

p(t)=at²+bt+c

We want to have p(-1)=2,p(1)=5,p(2)=6:

p(-1)=-a-b+c=2
p(1)=a+b+c=5
p(2)=4a+2b+c=6

This gives us three equations, which we can solve via row-reduction for the coefficients a,b,c.

3. Are the following sets of vectors linearly independent?

• The first set must be dependent; if we set up a system of equations to test for dependence, we have a homogeneous system of two equations in three unknowns, which means we're guaranteed at least one independent variable: there are infinitely many linear combinations of these vectors that sum to zero.
• The second set is independent: any set of two vectors is linearly dependent if and only if one vector is a scalar multiple of the other.
• The third set is dependent: it contains the zero vector, and any set of vectors containing the zero vector is dependent.

4.

[	1	2	]	[	2	0	]	=	[	0	6	]
[	3	4	]	[	-1	3	]	=	[	2	12	]

[	1	2	]	[	1	4	]	=	[	-3	-12	]
[	3	4	]	[	-2	-8	]	=	[	-5	-20	]

To find matrix inverses, we set up big honking augmented matrices, with the matrix we're inverting on one side, and the identity matrix on the other. We reduce to RREF; if we get a row of zeroes, our matrix is not invertible. If we get the identity on the left, whatever's left on the right-hand side is our inverse.

1	2	|	1	0
3	4	|	0	1

...reduces to

1	0	|	-2	1
0	1	|	1.5	-0.5

...So A is invertible, with inverse as above. On the other hand, if we try this on C...

1	4	|	1	0
-2	-8	|	0	1

...reduces to

1	4	|	1	0
0	0	|	2	1

...which is crap. So C is not invertible.

5. Suppose A is invertible (or nonsingular; same thing). Then...

(A+A^-1)(A-A^-1)=A²-AA^-1+A^-1A -(A^-1)²
=A²-(A^-1)²

...since A and A^-1 commute. If we replace A^-1 with a general matrix B such that AB is not equal to BA, then this is no longer true, as you are invited to confirm in the privacy of your own homes.

Had enough? Return to the 308B page.