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## This Week

We said that a homogeneous linear system was one where all the constants are zero. We then proved the following:

### Proposition

Let $A\mathbf{x}=\mathbf{0}$ is a homogenous linear system of $n$ equations in $n$ unknowns where $A$ is the coefficient matrix and the variables are $x_1,\,x_2,\,\dots,\,x_n$ ( i.e. $\mathbf{x}=(x_1\,\,x_2\,\,\cdots\,\,x_n)^T$. Then the following holds:

1. If $\det A=0$ then the system has (an infinite number of) non-trivial solutions.
2. If $\det A\neq =0$ then the system has (a unique solution given by) the trivial solution ($x_1=x_2=\cdots=x_n=0$).

Once this was established we looked at all the linear/matrix algebra  from Summer 2009Autumn 2009Summer 2011 and Autumn 2011. Although we can have problems via little mistakes we should be able to do all of these questions.

## Next Week

We will continue our work with Maple next week after we begin statistics.

## Test

The test will take place [Week 7] Wednesday 14 March and will be on linear algebra. Expect a sample test by next week. I said that I will put in a bonus question that will be on the theory to save those who make small calculation errors but do know their stuff.

## Exercise Solutions

I will give the solution to the first part of a question and then the answers to remaining parts. Some of code to write the matrices is very buggy and I’m afraid I’m not going to do up the whole Gauss and Gauss-Jordan work.

### Autumn 2009 Q. 1(ii)

Apply Gaussian elimination to the augmented matrix to get:

$\left[\begin{array}{ccc|c}1&-2&4&4\\ 0&1&5&0\\ 0&0&1&4\end{array}\right]$.

Now go further to do the full Gauss-Jordan algorithm:

$\left[\begin{array}{ccc|c}1&0&0&-52\\ 0&1&0&20\\ 0&0&1&4\end{array}\right]$.

Hence the solutions are $x=-52,\,y=-20$ and $z=4$.

### Autumn 2009 Q. 2(i)

Starting with

$\left[\begin{array}{ccc|ccc}1&0&1&1&0&0\\ 1&-1&4&0&1&0\\ -3&-1&2&0&0&1\end{array}\right]$,

apply the Gauss-Jordan algorithm until we have

$\left[\begin{array}{ccc|ccc}1&0&0&-1&1/2&-1/2\\ 0&1&0&7&-5/2&3/2\\ 0&0&1&2&-1/2&1/2\end{array}\right]$,

The answer is thus given by

$A^{-1}=\left(\begin{array}{ccc}-1&1/2&-1/2\\7&-5/2&3/2\\2&-1/2&1/2\end{array}\right)$.

### Autumn 2009 Q. 2(ii)

We calculate the determinant of the coefficient matrix and then use the above proposition. Expanding along the first column using $\sum (\text{sign})(\text{entry})(\text{block})$:

$\det A=(+)(3)\left|\begin{array}{cc}1&-3\\ -5&-1\end{array}\right|+(-1)(2)\left|\begin{array}{cc}-4&-2\\ -5&-1\end{array}\right|+(+)(2)\left|\begin{array}{cc}-4&-2\\ 1 & -3\end{array}\right|$

$=3(-1-15)-2(4-10)+2(12+2)=-48+12+28\neq 0$.

Hence the only solution is the trivial solution.

### Autumn 2009 Q. 2(iii)

First off there are solutions because there is no row of the form $[0\,\,0\,\,\cdots\,\,0|k]$ with $k\neq 0$. Now we calculate the number of parameters:

# parameters = # variables  –  # non-zero rows in row echelon form

Here the number of variables is 4, the number of non-zero rows is 3 so there is one parameter. Usually we like to set the ‘lowest’ variable equal to the parameter (e.g. $x_4=t$), but this is untenable in this example as the third row actually reads

$0x_1+0x_2+0x_3+x_4=3\Rightarrow x_4=3$,

so it can’t be the parameter. Let $x_3=t$ be the parameter (it can take on any value $t\in\mathbb{R}$) and thus $x_3$ is the free variable. Now we can read off:

$x_2=9+3x_3=9+2t$, and

$x_1=-4-8x_3=-4-8t$.

So the solution set, $S$, is given by: $S=\{(-4-8t,9+2t,t,3):t\in\mathbb{R}\}$.

### Summer 2009 Q. 1(iii)

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

### Summer 2009 Q. 2(i)

Answer is $(x,y,z)=(13,4,-1)$.

### Summer 2009 Q. 2(ii)

$\det A=-29$ so only the trivial solution.

### Summer 2009 Q. 2(iii)

The free variable is $x_3=t$ and the solution set is $S=\{(-2-6t,5+8t,t,2)\}$.

In notes.

### Summer 2011 Question 1(c)

The answer is $\displaystyle \left(\begin{array}{ccc}9&16&-40\\ -3&-5&13\\ -1&-2&5\end{array}\right)$.

In notes.

### Summer 2011 Question 3(b)

$\det A=49$ so only the trivial solution.

### Autumn 2011 Question 1(a)

The answer is $\displaystyle\left(\begin{array}{ccc}1&1&1\\ 3&5&4\\ 3&6&5\end{array}\right)$.

### Autumn 2011 Question 2(a)

The answer to the linear system is $(x,y,z)=(4,3,-2)$.

Cramer’s Rule says that, in this case, $\displaystyle y=\frac{D_y}{D}$ where $D$ is the determinant of the coefficient matrix (this is equal to -1). $D_y$ is the determinant got by replacing the ‘$y$-column’ of  the coefficient matrix with the column of constants:

$D_y=\left|\begin{array}{ccc}-1&-3&1\\ -2&2&1\\ 2&1&-1\end{array}\right|=-3$ and

$\displaystyle \frac{D_y}{D}=\frac{-3}{-1}=3$ as required (to match up with the above).