Math 351
Solution Notes for the Second Exam

Name:K. F. Gauss

Signature:

1.

Answer all questions.

2.

Show ALL your work in the space provided. If work is continued on the back of any test page, clearly indicate its presence.

3.

All University rules and guidelines for student conduct are applicable.

4.

Calculators with algebraic, calculus or matrix functions may not be used.

5.

There are six problems on seven pages (this page and six others).

Results:

1.

out of 16 points you got 16

2.

out of 16 points you got 16

3.

out of 16 points you got 16

4.

out of 16 points you got 16

5.

out of 20 points you got 20

6.

out of 16 points you got 16

Your total: 100 out of 100

1.

Let

$$A=\left(\begin{array}{ccc}1&0&1\\ 3&3&4\\ 2&2&3\end{array}\right)$$

Find A^-1 and use it to solve

$$\left(\begin{array}{ccc}1&0&1\\ 3&3&4\\ 2&2&3\end{array}\righ... ...ray}\right)= \left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$$

To find the inverse, one sets up an augmented system with A on the left and I on the right:

$$\left( \begin{array}{ccc\vert ccc} 1&0&1&1&0&0\\ 3&3&4&0&1&0\\ 2&2&3&0&0&1 \end{array}\right).$$

If one uses row operations to convert A on the left into I, the entries on the right will be A^-1 so that

$$A^{-1}=\left(\begin{array}{ccc}1&2&-3\\ -1&1&-1\\ 0&-2&3\end{array}\right).$$

Once one finds the inverse, one can use it to solve for $\bf x$.

$$\begin{eqnarray*}\left(\begin{array}{c}x_1\\ x_2\\ x_3\end{array}\right) & = & \... ...ht) \\ & = & \left(\begin{array}{c}0\\ -1\\ 1\end{array}\right) \end{eqnarray*}$$

Now, you can move on to other things.

2.

Are

$${\rm span}\left\{\left(\begin{array}{c}1\\ 3\\ 1\end{array}\r... ...t), \left(\begin{array}{c}4\\ -2\\ 2\end{array}\right)\right\}$$

the same or not? Why?

There are several ways to solve this problem, all of which are mathematically equivalent.

(a)

One can determine the span for each set and compare them directly.

(b)

One can prove that each vector in the set on the right can be expressed a linear combination of the vectors on the left, and vice versa.

(c)

One can prove that any element in the left span is in the span on the right, and vice versa.

I will work the problem using the first method.

On the right, every element in the span can be described as

$$\left(\begin{array}{c}x_1\\ x_2\\ x_3\end{array}\right) = \al... ...+ \alpha_2 \left(\begin{array}{c}2\\ -1\\ 1\end{array}\right).$$

We seek a description of x₁, x₂, x₃ which will come from consistency conditions on the linear system above. The actually values of $\alpha_1, \alpha_2$ are irrelevant for answering this question. To find the consistency conditions, we row-reduce the system

$$\left(\begin{array}{ccc}1&2&x_1\\ 3&-1&x_2\\ 1&1&x_3\end{array}\right)$$

to yield

$$\left(\begin{array}{ccc}1&2&x_1\\ 0&-7&x_2-3x_1\\ 0&0&x_3-\frac{4}{7} x_1 - \frac{1}{7} x_2\end{array}\right)$$

Remember, consistency conditions come from the row(s) of zeros in your linear system. The other rows simply tell you how to get the $\alpha$'s from the x's which is not the subject of the question. You may have more than one row of equations corresponding to more than one equation. Thus, the span on the right is a plane described by

$$x_3-\frac{4}{7}x_1 - \frac{1}{7} x_2 = 0.$$

Performing the same procedure on the right, we obtain

$$\left(\begin{array}{ccc}0&4&x_1\\ -7&-2&x_2\\ -1&2&x_3\end{array}\right).$$

After many happy row operations, we have

$$\left(\begin{array}{ccc}1&2&x_3\\ 0&12&x_2+7x_3\\ 0&0&x_1 + \frac{1}{4}x_2-\frac{7}{4}x_3\end{array}\right).$$

Thus,

$$x_1 + \frac{1}{4}x_2-\frac{7}{4}x_3 = 0.$$

If we multiply this equation by $-\frac{4}{7}$, we have the exact same equation as on the left, so we see that the two spans are the same.

3.

For what values of s does $A\mbox{\boldmath {$x$ }}=\mbox{\boldmath {$b$ }}$ have a solution for any $\mbox{\boldmath {$b$ }}$ when

$$A=\left(\begin{array}{ccc} 1&2&1\\ 1&0&1\\ 2&0&s\end{array}\right).$$

$A\mbox{\boldmath {$x$ }}=\mbox{\boldmath {$b$ }}$ has a solution for all $\mbox{\boldmath {$b$ }}$ if and only if $\mbox{\boldmath {$A$ }}$ is invertible. $\mbox{\boldmath {$A$ }}$ is invertible if and only if $\det \mbox{\boldmath {$A$ }} \neq 0$.

$$\begin{eqnarray*}\det \left(\begin{array}{ccc} 1&2&1\\ 1&0&1\\ 2&0&s\end{array}\... ...ft(\begin{array}{cc} 1&2\\ 1&0\end{array}\right) \\ & = & 4-2s. \end{eqnarray*}$$

Thus, there is a solution for any $\mbox{\boldmath {$b$ }}$ as long as $s \neq 2$. Note: Even when s=2, there may be (and are) solutions for certain $\mbox{\boldmath {$b$ }}$'s, but there are some $\mbox{\boldmath {$b$ }}$'s for which there is no solution.

4.

Let

$$T={\rm span}\left\{ \left(\begin{array}{c}1\\ 2\\ 3\end{array... ...,\; \left(\begin{array}{c}0\\ 2\\ 4\end{array}\right)\right\}.$$

Is the given set of vectors a basis for T and why? What is the dimension of T? If the set is not a basis, find a basis for T.

The quick way to get an answer to the first question is to take the determinant of the matrix whose columns are the vectors in the set above. The determinant is zero, so we would know that the vectors are not linearly independent and cannot form a basis.

Now that we know that it is not a basis, we need to figure out which vectors in the set are redundant. We need to know for values of $\alpha_1, \alpha_2, \alpha_3$ such that

$$\alpha_1 \left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)+ \... ...y}\right) = \left(\begin{array}{c}0\\ 0\\ 0\end{array}\right).$$

Again, we resort to Gaussian elimination on

$$\left(\begin{array}{ccc}1&-1&0\\ 2&0&2\\ 3&1&4\end{array}\right).$$

After many row operations, we obtain

$$\left(\begin{array}{ccc}1&0&1\\ 0&2&2\\ 0&0&0\end{array}\right).$$

Thus, we see that $\alpha_1 = -\alpha_3$ and $\alpha_2=-\alpha_3$. We have only one free variable; $\alpha_3$ is arbitrary. For instance, if $\alpha_3=1$,

$$\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)+ \left(\beg... ...ay}\right)= \left(\begin{array}{c}0\\ 2\\ 4\end{array}\right).$$

So, we can drop the last vector in the set to obtain a linearly independent spanning set.

$$T={\rm span}\left\{ \left(\begin{array}{c}1\\ 2\\ 3\end{array... ...\; \left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)\right\}.$$

Since there is only one free variable, one can only drop one vector, and we can see that $\dim T = 2$.

5.

(Hint: these questions can be answered without calculation by appealing to theorems from the class)

(a)

If A is a $7\times 5$ matrix, why can the rank of A not be 6?

The row rank is equivalent to the column rank, and there are only 5 columns, so the maximum rank is 5.

(b)

Is

$$\left\{\left(\begin{array}{c}1\\ 2\\ 7\end{array}\right),\; \... ...),\; \left(\begin{array}{c}3\\ 3\\ 3\end{array}\right)\right\}$$

a basis for ${\mathbb R}^3$ (why)?

No. ${\mathbb R}^3$ has three dimensions and so any spanning set should have three vectors in it.

(c)

Let

$$A=\left(\begin{array}{ccc} 1 & 2 & -1\\ 2 & 3 & 1\\ 1 & 4 &... ...c} -2&5&0&3\\ 0&10&2&9\\ -3&12&3&4\\ 1&4&1&5\end{array}\right)$$

Find B^TA^T?

We all know that B^T A^T =(AB)^T, so

$$B^TA^T = \left(\begin{array}{cccc} -2&0&-3&1\\ 5&10&12&4\\ 0&2&3&1\\ 3&9&4&5\end{array}\right)$$

6.

Expand the vector

$$\left(\begin{array}{c} 1\\ 1\\ 1\\ 1\end{array}\right)$$

in terms of the basis

$$\left\{ \left(\begin{array}{c}1\\ 0\\ 2\\ 0\end{array}\right... ...eft(\begin{array}{c}2\\ 0\\ -1\\ 1\end{array}\right) \right\}.$$

Essentially, you are trying to find $\alpha_1, \alpha_2, \alpha_3, \alpha_4$ such that

$$\left(\begin{array}{c} 1\\ 1\\ 1\\ 1\end{array}\right) = \al... ...alpha_4\left(\begin{array}{c}2\\ 0\\ -1\\ 1\end{array}\right).$$

Now, you could find all the $\alpha$'s by solving a linear system, but many inspected the basis and realized that it was orthogonal. This being the case, we do not have to solve a linear system because the basis vectors are all at right angles to one another, and so

$$\alpha_i = \frac{\mbox{\boldmath {$u$ }} \cdot \mbox{\boldmath {$e$ }}_i}{\vert \mbox{\boldmath {$e$ }}_i \vert^2}.$$

where the $\mbox{\boldmath {$e$ }}$'s are the basis vectors. If you compute the $\alpha$'s correctly, you will find that

$$\left(\begin{array}{c} 1\\ 1\\ 1\\ 1\end{array}\right) = \fr... ...c{1}{3}\left(\begin{array}{c}2\\ 0\\ -1\\ 1\end{array}\right).$$