# 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 R_{1} and R_{2}
.

Eliminate from 2nd and 3rd row (R_{1} + R_{3})

. Switch R_{2} and R_{3}
.

Eliminate from third row (R_{3} + 2R_{2}):

. Make leading entry in
third row 1 (R_{3}/3)

Eliminate from first row: (R_{1} − R_{2}):

Eliminate from first and second row (R_{1} + 3R_{3} and R_{2} −
R_{3}):

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

and

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).

AB_{34} 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} = I_{4}
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: (R_{2} + 2R_{1} and R_{3} −
3R_{1})

Now we eliminate
from the third row: (R_{3} + 2R_{2})

We make the leading entry in second row 1 (−R_{2}) 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: (R_{2} + 3R_{1}
and R_{3} − 2R_{2}

.

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 A^{2} = A.

I_{4} the 4x4 identity matrix is nonsingular and I^{2} = I.

b) Find all 4x4 nonsingular matrices that satisfy A^{2} = 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^{-1}A^{2} = A^{-1}A which means that IA = I and therefore A = I. Thus the identity
matrix

is the only 4x4 nonsingular matrix that satisfies A^{2} = A.

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