Let A denote an

$$m \times n$$

matrix. (a) Show that null A=null(UA) for every invertible

$$m \times n$$

matrix U. (b) Show that dim(null A)=dim(null(AV)) for every invertible

$$n \times n$$

matrix V. [Hint: If

$$\left\{ \mathbf { x } _ { 1 } , \mathbf { X } _ { 2 } , \dots , \mathbf { X } _ { k } \right\}$$

is a basis of null A, show that

$$\left\{ V ^ { - 1 } \mathbf { x } _ { 1 } , V ^ { - 1 } \mathbf { x } _ { 2 } , \dots , V ^ { - 1 } \mathbf { x } _ { k } \right\}$$

is a basis of null(AV).]

Find A + B.

$$A=\begin{bmatrix} 0 & 1&2\\ 1&-1& 0\\ 2&1& 1 \end{bmatrix}, \quad B=\begin{bmatrix} 0 & 1&-1\\ 2&1& 1\\ 0&1& 1 \end{bmatrix}$$

Find the lexicographic ordering of these n-tuples: a) (1, 1, 2), (1, 2, 1) b) (0, 1, 2, 3), (0, 1, 3, 2) c) (1, 0, 1, 0, 1), (0, 1, 1, 1, 0)

Show that every finite lattice has a least element and a greatest element.

Solve the boundary-value problem, if possible.

$$y’’+4y’+20y=0, y(0)=1, y(π)=2$$

Solve the boundary-value problem, if possible.

$$y’’+4y=0, y(0)=5, y(π/4)=3$$

Find the exact value of (a) $\sin ^ { - 1 } ( - 1 )\quad$ ( b ) $\cos ^ { - 1 } ( - 1 )$ (c) $tan ^ { - 1 } ( - 1 ) \quad$ (d) $\sec ^ { - 1 } ( 1 )$

How do you decide to shade above or below a boundary line? What does this shading represent?

Simplify. Give restrictions on the variables. $\frac{8}{5 m n}-\frac{2}{m}$

Determine the water of hydration for the following hydrates and write the chemical formula.

Ni(NO$_{3}$)$_{2}$ $\sdot$ XH$_{2}$O is found to contain $37.2$% water.

Find the rational number, expressed in lowest terms, represented by each of the following simple continued fractions. a) [2; 7], b) [1; 2, 3], c) [0; 5 , 6], d) [3; 7, 15, 1], e) [1; 1], f) [1; 1, 1], g) [1; 1, 1, 1], h) [1; 1, 1, 1, 1].

Using trapezoids instead of rectangles. That is, on the subinterval $(1 /(n+1), 1 / n)$, let $a_n$ denote the area of the trapezoid having heights $y=1 /(n+1)$ at $x=1 /(n+1)$ and $y=1 / n$ at $x=1 / n$.

For the divergent series $\sum_{k=1}^{\infty} \frac{1}{k}$, show that $\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=1$.

$$\frac { 3 } { 3 }$$

$$\frac { 1 } { 2 }$$

$$\frac { 1 } { 2 }$$

$$\frac { 1 } { 5 }$$

$$\frac { 1 } { 5 }$$

$$\frac { 1 } { 1 }$$

$$\frac { 1 } { 1 }$$

$$\frac { 1 } { 3 }$$

$$\frac { 1 } { 3 }$$

1