Exercise $5\mathrm{.}1$

A$.$A$,$ how$-$

ever, which gives an entirely different answer, can only be performed if

what condition is true?

Exercise $5\mathrm{.}2$

In the second week of the Justice League fundraiser discussed in

the lesson, contributions were as follows:

For the space station: from each hero $$500,$ each sidekick $$300,$ and

each villain $$200$ : a total of $$7100$

For the Fortress of Solitude: from each hero $$1000,$ each sidekick

$$100,$ each villain $$300$ : a total of $$8300$

For the Batcave: from each hero $$200,$ each sidekick $$200,$ each

villain $$0($really$,$ nobody likes Batman$)$: a total of $$3400$

Prove that the number of heroes, sidekicks, and villains is the same as

found in the lesson $(6$ heroes, $11$ sidekicks, $4$ villains$).$

Exercise $5\mathrm{.}3$

Create a $33$ matrix A$,$ with any numerical content. Use

MATLAB to determine its inverse, then manually perform the matrix

multiplication $A^{**}{A}^{-1}$ to prove that the answer is $1.$

Exercise $5\mathrm{.}4$

In Germany, annual professional fees are set by law for engineers,

doctors, and lawyers. In Frankfurt, the total fees charged in a single

year by $23$ doctors, $18$ lawyers, and $12$ engineers were equivalent to

$$76,500,000$ US$.$ In that same year in Berlin, $25$ doctors, $20$ lawyers, and

$16$ engineers charged $$85,500,000.$ In Munich, $18$ doctors, $9$ lawyers, and

$20$ engineers charged $$55,000,000.$

In MATLAB, use the formula $B=A^{-1**}C$ to determine the annual fee

charged by each doctor, each lawyer, and each engineer. Tip: Use $76\mathrm{.}5$

rather than $76,500,000($etc$.)$ in constructing that matrix.

Exercise $5\mathrm{.}5$

Using the matrices A and B from Exercise $5\mathrm{.}4,$ in a single

MATLAB command, calculate a new $33$ matrix $D$ that contains the

specific amounts charged by each profession in each city. For example,

$(1,1)$ will be the total fees charged by doctors in Frankfurt, $(2,2)$ the

total fees charged by doctors in Berlin, and so on$.$