Math Exam 1 Solutions

1) (20 points) Consider the following system of linear equations :

a) Write the augmented matrix for this system.

b) Reduce the system to Reduced Echelon Form.

.Switch R1 and R2 .

Eliminate from 2nd and 3rd row (R1 + R3)

. Switch R2 and R3 .

Eliminate from third row (R3 + 2R2):

. Make leading entry in third row 1 (R3/3)

Eliminate from first row: (R1 − R2):

Eliminate from first and second row (R1 + 3R3 and R2 − R3):

c) Find the general solution to this system of equations.

From the reduced echelon form we see that the system reduced to the following equations:

Therefore the general solution is :

where is an independent variable and can take any value.

d) Write the general solution in vector form.

2) (20 points) Let


a) Find the third column of AB.

The third column of AB comes from multiplying A by the third column of B:

b) Find (AB)34 (the entry of AB in third row and fourth column).

AB34 comes from multiplying third row of A by fourth column of B:

c) Let C be a non-singular 4x4 matrix with rows . Let the columns of C-1 be
. Calculate   for 1≤ i, j ≤ 4.

is the entry in i-th row and j-th column of CC-1. We know that CC-1 = I4 the 4x4
identity matrix. Therefore = 1 when i = j and = 0 when i ≠ j.

3)(20 points) Determine whether the vectors and
are linearly dependent or linearly independent. If the vectors are linearly dependent,
express 0 as a nontrivial linear combination of the vectors .

In order to find out whether vectors are linearly dependent we need to find all of their
linear combinations that equal 0:

This is the same as finding all of the solutions of:

The augmented matrix of this system is:

We reduce it to reduced echelon form: (R2 + 2R1 and R3 − 3R1)

Now we eliminate from the third row: (R3 + 2R2)

We make the leading entry in second row 1 (−R2) and use second row to eliminate from
the first row:

The general solution is: and with an independent variable. Since
there are infinitely many solutions the vectors are linearly dependent . To find a particular
nontrivial combination we need to pick a value of . Lets choose = 1. Then x=11 and
= −4 and the linear combination is:

4) (20 points) Find all of the values of a such that the matrix

is singular. Briefly explain your reasoning.
In order to find out of A is singular we need to find all of the solutions to Ax = 0. We begin
by reducing the augmented matrix to echelon form:

We eliminate from the second and third row: (R2 + 3R1 and R3 − 2R2

When a ≠ 6 it follows that = 0 and then from row 2 = 0 and then = 0. Thus when
a ≠ 6 the only solution to Ax = 0 is x = 0 and therefore A is nonsingular.

When a = 6 we have the system:

Here is an independent variable and thus there are infinitely many solutions to Ax = 0.
Therefore when a = 6 the matrix A is singular and A is nonsingular for any other value of a.

5) (20 points) a) Find one example of a nonsingular 4x4 matrix A that satisfies A2 = A.

I4 the 4x4 identity matrix is nonsingular and I2 = I.

b) Find all 4x4 nonsingular matrices that satisfy A2 = A. Explain your reasoning.

Since A is nonsingular A has an inverse matrix A-1. We can multiply both sides of the
equation by A -1:

A-1A2 = A-1A which means that IA = I and therefore A = I. Thus the identity matrix
is the only 4x4 nonsingular matrix that satisfies A2 = A.

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