MATH 251 Lecture 10

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How to calculate determinants

Basics

1. Go across the top row (left to right)
2. Cross off that item's row and column, then calculate the determinant of the resulting 2×2 Matrix

${\begin{aligned}{\begin{vmatrix}1&3&-1\\4&2&-2\\0&1&1\end{vmatrix}}&=1(2--2)-3(4-0)+-1(4-0)\\&=4-12-4\\&=-12\end{aligned}}$

Trick 1

You can go across any row OR column!

Be sure to follow the sign rules:

${\begin{pmatrix}+&-&+\\-&+&-\\+&-&+\end{pmatrix}}$

${\begin{aligned}{\begin{vmatrix}3&157&87\\1&15&73\\0&0&1\end{vmatrix}}&=0-0+1(45-157)\\&=-112\end{aligned}}$

Trick 2

Factor stuff from any row or column

${\begin{aligned}{\begin{vmatrix}x^{3}-3x^{2}&x^{3}&x^{2}+x^{4}\\x-3&1&0\\3-x&2&x^{2}+1\end{vmatrix}}&=x^{2}(x-3)(x^{2}+1){\begin{vmatrix}1&x&1\\1&1&0\\-1&2&1\end{vmatrix}}\\&=x^{2}(x-3)(x^{2}+1)\left[-1(x-2)+1(1--1)+0\right]\\&=x^{2}(x-3)(x^{2}+1)(4-x)\end{aligned}}$

Trick 3

Flipping any two rows or columns only flips the sign

... not very useful though

Trick 4

${\begin{aligned}{\begin{vmatrix}3y&2-xy&1+3y\\1&-2\end{vmatrix}}&={\begin{vmatrix}0&2&1\\1&-2&1\\2&1&5\end{vmatrix}}\\&=-2(5-2)+1(1--4)\\&=-6+5\\&=-1\end{aligned}}$

Notice in the first matrix that the $x$ and $y$ terms are just multiples of the second row added to the constants in the second matrix.

Looking at the geometry, doing this only shifts the vectors over, keeping the same base and height values.

Transformation Lattices

Drawing lattice lines w/r/t vector in "cartesian style"

Transform them into x-y lattice by holding one variable constant, then plotting the resulting shape. Hold the other variable constant, and plot those too.

Derivative

${\begin{cases}x=r\cos {\theta }\\y=r\sin {\theta }\end{cases}}$

${\begin{aligned}\mathrm {d} x&={\frac {\partial x}{\partial r}}\,\mathrm {d} r+{\frac {\partial x}{\partial \theta }}\,\mathrm {d} \theta =\cos {\theta }\mathrm {d} r+-r\sin {\theta }\mathrm {d} \theta \\\mathrm {d} y&={\frac {\partial y}{\partial r}}\,\mathrm {d} r+{\frac {\partial y}{\partial \theta }}\,\mathrm {d} \theta =\sin {\theta }\mathrm {d} r+r\cos {\theta }\mathrm {d} \theta \end{aligned}}$

So the derivative can be rewritten as ${\begin{pmatrix}\cos {\theta }&-r\sin {\theta }\\\sin {\theta }&r\cos {\theta }\end{pmatrix}}$

Jacobian

The determinant of the transformation matrix derivative (calculated above)

$r\cos ^{2}{\theta }+r\sin ^{2}{\theta }=r$

Multidimensional Limits

$f\left(x,y\right)={\frac {x^{2}+y^{2}}{{\sqrt {x^{2}+y^{2}+49}}-7}}$

Does $\lim _{(x,y)\to (0,0)}f(x,y)$ exist?

We approach (0,0) from all directions around it, and determine that they all converge to the same value.

All elementary functions are continuous at all points in their natural domain … In other words, if evaluating a limit at a point gives a real number, DO IT!!!

Trick: convert coordinates

let $x^{2}+y^{2}=r^{2}$ (distance to the origin), therefore $f(r)={\frac {r^{2}}{{\sqrt {r^{2}+49}}-7}}$ (L'Hospital's Rule)

Trick: approach along a plane

$\lim _{(x,y,z)\to (0,0,0)}{\frac {xy+7yz^{2}+8xz^{2}}{x^{2}+y^{2}+z^{4}}}$

Set $y=0$ and $z=0$ to approach along the $x$-axis.

$\lim _{x\to 0}{\frac {0}{x^{2}}}=0$

Try to approach it from a different direction: $y=x$, $z=0$

$\lim _{x\to 0}={\frac {x^{2}+0+0}{2x^{2}}}={\frac {1}{2}}$

Since $0\neq {\frac {1}{2}}$, the limit does not exist.