LAC Exercises (with solutions)

Section 1.4 Exercises (with solutions)

Subsection 1.4.1 Linear equations — Mechanics

Exercises Exercises

1.

Let

\begin{equation*} M=\ba{rrrrr} 0 \amp 0 \amp 0 \amp 1 \amp 2 \\ 1 \amp -5 \amp -1 \amp 0 \amp -3 \\ -2 \amp 10 \amp 3 \amp 5 \amp 17 \\ -2 \amp 10 \amp 3 \amp 2 \amp 11 \ea \end{equation*}

be the augmented matrix of a linear system given by the matrix equation \(Ax=b,\) and let

\begin{equation*} R=\ba{rrrrr} 1 \amp -5 \amp 0 \amp 0 \amp -2 \\ 0 \amp 0 \amp 1 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \amp 1 \amp 2 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \ea \end{equation*}

be its reduced row-echelon form.

(a)

How many equations and how many variables does the linear system have?

Solution.

The matrix \(M = [A|b]\) is \(4\times5\) which means there are four equations and four unknowns.

(b)

Is the linear system consistent>

Solution.

Yes, the last column is not a pivot column, see Insight 1.2.1.

(c)

What are the pivot positions?

Solution.

They are boxed:

\begin{equation*} R=\ba{rrrrr} \fbox{1} \amp -5 \amp 0 \amp 0 \amp -2 \\ 0 \amp 0 \amp \fbox{1} \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \amp \fbox{1} \amp 2 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \ea \end{equation*}

(d)

What are the free and constrained variables (if any)?

Solution.

\(x_1, x_3\) and \(x_4\) are constrained (corresponding to pivots), while \(x_2\) is unconstrained or free.

(e)

Describe the solutions.

Solution.

\begin{equation*} \begin{cases}x_1 = -2+5x_2\\x_2 \mbox{ is free}\\x_3 = 1\\x_4=2 \end{cases} \end{equation*}

is one way to describe the solutions. Perhaps a better way is in vector form:

\begin{equation*} \ba{r}x_1\\x_2\\x_3\\x_4\ea = \ba{r}-2\\0\\1\\2\ea + x_2\ba{r}5\\1\\0\\0\ea. \end{equation*}

(f)

Describe the solutions to the homogeneous system \(Ax=0.\)

Solution.

We just throw away the last column of \(M\) and of its reduced row-echelon form \(R.\) Then we read them from the answer above:

\begin{equation*} \begin{cases}x_1 = 5x_2\\x_2 \mbox{ is free}\\x_3 = 0\\x_4=0 \end{cases} \end{equation*}

is one way to describe the solutions. Perhaps a better way is in vector form:

\begin{equation*} \ba{r}x_1\\x_2\\x_3\\x_4\ea = x_2\ba{r}5\\1\\0\\0\ea. \end{equation*}

2.

Let

\begin{equation*} M=\ba{rrrrrr} 1 \amp 4 \amp 3 \amp -5 \amp -11 \amp 10 \\ 0 \amp 0 \amp 1 \amp -1 \amp -3 \amp 3 \\ -1 \amp -4 \amp -5 \amp 7 \amp 17 \amp -16 \\ 1 \amp 4 \amp 2 \amp -4 \amp -8 \amp 7 \ea \end{equation*}

be the augmented matrix of a linear system given by the matrix equation \(Ax=b,\) and let

\begin{equation*} R=\ba{rrrrrr} 1 \amp 4 \amp 0 \amp -2 \amp -2 \amp 1 \\ 0 \amp 0 \amp 1 \amp -1 \amp -3 \amp 3 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \ea \end{equation*}

be its reduced row-echelon form.

(a)

How many equations and how many variables does the linear system have?

Solution.

The augmented matrix \(M = [A|b]\) is \(4\times6\) which means there are four equations and five unknowns.

(b)

Is the linear system consistent?

Solution.

Yes, the last column is not a pivot column, see Insight 1.2.1.

(c)

What are the pivot positions?

Solution.

They are boxed:

\begin{equation*} R=\ba{rrrrrr} \fbox{1} \amp 4 \amp 0 \amp -2 \amp -2 \amp 1 \\ 0 \amp 0 \amp \fbox{1} \amp -1 \amp -3 \amp 3 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \ea \end{equation*}

(d)

What are the free and constrained variables (if any)?

Solution.

\(x_1\) and \(x_3\) are constrained (corresponding to pivots), so \(x_2,x_4\) and \(x_5\) are unconstrained or free.

(e)

Describe the solutions.

Solution.

\begin{equation*} \begin{cases} x_1\amp= 1 -4x_2 +2x_4+2x_5\\ x_2\amp\mbox{is free}\\ x_3\amp= 3 + x_4 +3x_5\\ x_4\amp\mbox{is free}\\ x_5\amp\mbox{is free} \end{cases} \end{equation*}

is one way to describe the solutions. Perhaps a better way is in vector form:

\begin{equation*} \ba{r}x_1\\x_2\\x_3\\x_4\\x_5\ea = \ba{r}1\\0\\3\\0\\0\ea + x_2\ba{r}-4\\0\\0\\0\\0\ea+ x_4\ba{r}2\\0\\1\\0\\0\ea + x_5\ba{r}2\\0\\3\\0\\0\ea. \end{equation*}

(f)

Describe the solutions to the homogeneous system \(Ax=0.\)

Solution.

We just throw away the last column of \(M\) and of its reduced row-echelon form \(R.\) Then we read them from the answer above:

\begin{equation*} \ba{r}x_1\\x_2\\x_3\\x_4\\x_5\ea = x_2\ba{r}-4\\0\\0\\0\\0\ea+ x_4\ba{r}2\\0\\1\\0\\0\ea + x_5\ba{r}2\\0\\3\\0\\0\ea. \end{equation*}

Subsection 1.4.2 Linear equations — Theory

Exercises Exercises

1.

Let

\begin{equation*} A=\ba{rrr}1\amp 2\amp 3\\3\amp 6\amp 9\\5\amp 10\amp 15\ea \text{ and } b=\ba{r}1\\2\\3\ea. \end{equation*}

Is the system \(Ax=b\) solvable?

Hint.

Can you guess the answer first and a reason why?

Solution.

No it is not solvable. We know by Corollary 1.3.3 that \(Ax=b\) is solvable iff \(b\) is in the column space of \(A\text{.}\) By inspection, we see that columns 2 and 3 of the matrix \(A\) are multiples of the first column, which says the column space of \(A\) is spanned by its first column. So the only way for the system to be solvable is if

\begin{equation*} b = \ba{r}1\\2\\3\ea = \lambda \ba{r}1\\3\\5\ea. \end{equation*}

Comparing first coordinates says that \(\lambda=1,\) but comparing second coordinates requires that \(\lambda = 2/3,\) so there can be no solution.

Of course if the observation above escaped us, we could row reduced the augmented matrix associated to the system yielding

\begin{equation*} [A|b] = \ba{rrrr}1\amp 2\amp 3\amp3\\3\amp 6\amp 9\amp2\\5\amp 10\amp 15\amp3\ea\mapsto R= \ba{rrrr}1\amp2\amp3\amp0\\0\amp0\amp0\amp1\\0\amp0\amp0\amp0\ea. \end{equation*}

We observe the last column is a pivot column (Insight 1.2.1) so the system is not solvable.

2.

Let

\begin{equation*} A=\ba{rrrrrr} 1 \amp 4 \amp 3 \amp -5 \amp -11 \amp 10 \\ 0 \amp 0 \amp 1 \amp -1 \amp -3 \amp 3 \\ -1 \amp -4 \amp -5 \amp 7 \amp 17 \amp -16 \\ 1 \amp 4 \amp 2 \amp -4 \amp -8 \amp 7 \ea \end{equation*}

and let

be its reduced row-echelon form.

(a)

What are the pivot columns of \(A\text{?}\)

Solution.

From the RREF, we see the pivot columns are the first and third.

(b)

The column space is guaranteed to be spanned by which two columns of \(A\text{?}\)

Solution.

The first and third columns of \(A\) span its column space since they are the pivot columns, but so do the first and fourth, fifth, or sixth. Why?

(c)

Since we know the column space is spanned by the columns from the previous part, all the other columns of \(A\) are linear combinations of those two. Is it possible to write down the particular combinations from the information we have so far?

Solution.

By Theorem 1.3.1, we know that the solutions to \(Ax=0\) and \(Rx=0\) are exactly the same, but that means that if \(A\) has columns \(a_i\text{,}\) and \(R\) has columns \(r_i\text{,}\) then we have equal linear combination of columns:

\begin{equation*} Ax=0=x_1a_1 + x_2 a_2 + \cdots + x_6a_6 = x_1r_1+x_2r_2+\cdots + x_6r_6=Rx. \end{equation*}

In looking at

it is easy to see that (for example) the fifth column of \(R\text{,}\) is equal to

\begin{equation*} r_5 = -2r_1-3r_3, \text{ giving } -2r_1 -3r_3 -r_5 = 0. \end{equation*}

But that means that

\begin{equation*} -2a_1-3a_3-a_5=0 \text{ giving } a_5 = -2a_1 -3a_3. \end{equation*}

So we can indeed read of how to write each column of \(A\) as a linear combination of the pivot columns.

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