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This Week

We asked the question “when is a matrix invertible” and we answered the question by saying that a matrix is invertible when its determinant is non-zero. We didn’t give a definition of  a determinant but did see how to calculate it. Then we introduced Cramer’s Rule as yet another way of solving linear systems. We have a handout with some exercises on determinants and the chain rule.

We also did quite a lot of work on Maple in terms of the basics.

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 so I’ve skipped some of them when they don’t compile straight away. For example, the answer below should have an extra line.

Exercise 4.41

Taking the Laplace expansion along the first column yields

$=1(-4-12)+6(14+3)=-16+102=86$.

Exercise 4.45

The rank is the number of non-zero rows when the matrix is in row echelon form. This matrix has rank 2. Thus it has determinant zero as it is not invertible.

Exercise 4.46

Determinant = 26.

Exercise 4.47

A $4\times 4$ matrix is calculated in a similar fashion to a $3\times 3$ but note that the sign matrix is $(-1)^{i+j}$ — where $i$ is the row and $j$ is the column. Thence, taking the Laplace expansion along the second column:

$-(1)\left|\begin{array}{ccc}3 & 0 & 2 \\ 0 & 2 & 1 \\ 5 & 0 & 7 \end{array}\right|+0(\dots)-(1)\left|\begin{array}{ccc} 0 & -1 & 0 \\ 3 & 0 & 2 \\ 5 & 0 & 7\end{array}\right|+0(\dots)$

$=(-1)\left(-0(\dots)+2\left|\begin{array}{cc} 3 & 2 \\ 5 & 7\end{array}\right|-0(\dots)\right)$

$+(-1)\left(-(-1)\left|\begin{array}{cc} 3 & 2 \\ 5 & 7\end{array}\right|+0(\dots)-0(\dots)\right)$

$=(-1)(2(21-10))-(1(21-10))=-22-11=-33$.

Exercise 4.48

We don’t really have the tools or need to look this.

Exercises 4.10

Question 1.

$0$$-1$$0$$-39$$2abc$$0$$-56$$abcd$.

Question 2

Doesn’t concern us.

“Calculus 3: Notes and Exercises on Cramer’s Rule” Exercises

1.

$x=\frac{Dx}{D}$$y=\frac{Dy}{D}$

where $D$ is the determinant of the coefficient matrix and

$Dx=\left|\begin{array}{cc}0 & -2\\ -1 & 1\end{array}\right|$ and $Dy=\left|\begin{array}{cc}1 & 0 \\ -1 & -1\end{array}\right|$.

Thus $x=2$ and $y=1$:

$(2)-2(1)=0$

$-(2)+(1)=-1$

$(x,y,z)=(1,1,1)$, no solutions, $(x,y)=(1+2t,t)$ as $t\in\mathbb{R}$.