91Ó°ÊÓ

Find the rank of the matrix

$\left[\begin{array}{ccc}a& b& c\\ 0& d& e\\ 0& 0& f\end{array}\right]$,

where a, d and f are nonzero, and b,c and e are arbitrary numbers.

Question:A lower triangular 3x3 matrix has rank 3 if (and only if) the product of its diagonal entries is nonzero.

Kyle is getting some flowers for Olivia, his Valentine. Being of a precise analytical mind, he plans to spend exactly \(24 on a bunch of exactly two dozen flowers. At the flower market they have lilies (\)3 each), roses (\(2 each), and daisies (\)0.50 each). Kyle knows that Olivia loves lilies; what is he to do?

Consider the function $\mathbf{T}\left(A\right)\left(\stackrel{â‡\P Ä}{X}\right)\mathbf{=}{\stackrel{\mathbf{â‡\P Ä}}{\mathbf{x}}}^{\mathbf{T}}\mathbf{A}\stackrel{\mathbf{â‡\P Ä}}{\mathbf{x}}\mathbf{}\mathbf{from}\mathbf{}{\mathbf{R}}^{\mathbf{nxn}}\mathbf{}\mathbf{to}\mathbf{}{\mathbf{Q}}_{\mathbf{n}}$. Show that T is a linear transformation. Find the image, kernel, rank, and nullity of T.

Determine whether the statements that follow are true or false, and justify your answer.

47.If $\mathit{a}\mathit{d}\mathbf{-}\mathit{b}\mathit{c}\mathbf{≠}\mathbf{0}$, then the matrix$\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ must have rank 2.

Consider a linear transformation T from ${\mathit{R}}^{\mathbf{2}}\mathbf{}\mathbf{to}\mathbf{}{\mathit{R}}^{\mathbf{2}}$. We are told that the matrix T with respect to the basis $\left[\begin{array}{c}0\\ 1\end{array}\right]\mathbf{,}\left[\begin{array}{c}1\\ 0\end{array}\right]\mathbf{}\mathbf{is}\mathbf{}\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$

Find the standard matrix of T in terms of a, b, c and d.

Consider the equations

$\left|\begin{array}{c}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{3}\mathbf{z}\mathbf{=}\mathbf{4}\\ \mathbf{x}\mathbf{+}\mathbf{ky}\mathbf{+}\mathbf{4}\mathbf{z}\mathbf{=}\mathbf{6}\\ \mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{(}\mathbf{k}\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{z}\mathbf{=}\mathbf{6}\end{array}\right|$

where$\mathit{k}$is an arbitrary constant.

a. For which values of the constant does this system have a unique solution?

b. When is there no solution?

c. When are there infinitely many solutions?

Question:A linear system of the form$\mathit{A}\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{=}\mathbf{0}$ is called homogeneous. Justify the following facts:

a.All homogeneous systems are consistent.

b.A homogeneous system with fewer equations than unknowns has infinitely many solutions.

c.If$\stackrel{\mathbf{→}}{{\mathbf{x}}_{\mathbf{1}}}\mathbf{}\mathbf{and}\mathbf{}\stackrel{\mathbf{→}}{{\mathbf{x}}_{\mathbf{2}}}$ are solutions of the homogeneous system$\mathit{A}\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{=}\mathbf{0}$, then$\stackrel{\mathbf{→}}{{\mathbf{x}}_{\mathbf{1}}}\mathbf{+}\stackrel{\mathbf{→}}{{\mathbf{x}}_{\mathbf{2}}}$ is a solution as well.

d.If$\stackrel{\mathbf{→}}{\mathbf{x}}$ is a solution of the homogeneous system$\mathit{A}\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{=}\mathbf{0}$ andkis an arbitrary constant, then$k\stackrel{→}{x}$ is a solution as well.

Consider the equations

$\mathbf{\left|}\begin{array}{c}\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{k}\mathbf{z}\mathbf{=}\mathbf{0}\\ \mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{6}\mathbf{z}\mathbf{=}\mathbf{2}\\ \mathbf{k}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{z}\mathbf{=}\mathbf{1}\end{array}\mathbf{\right|}$

where$\mathit{k}$ is an arbitrary constant.

a. For which values of the constant $\mathit{k}$does this system have a unique solution?

b. When is there no solution?

c. When are there infinitely many solutions?

Consider a solution$\stackrel{\mathbf{→}}{{\mathbf{x}}_{\mathbf{1}}}$of the linear system$\mathit{A}\stackrel{\mathbf{→}}{\mathbf{x}}=\stackrel{\mathbf{→}}{b}$. Justify the facts stated in parts (a) and (b):

a. If${\stackrel{\mathbf{→}}{\mathbf{x}}}_{\mathbf{h}}$is a solution of the system$\mathit{A}\stackrel{\mathbf{→}}{\mathbf{x}}=\stackrel{\mathbf{→}}{0}$, then$\stackrel{\mathbf{→}}{{\mathbf{x}}_{\mathbf{1}}}\mathbf{+}\stackrel{\mathbf{→}}{{\mathbf{x}}_{\mathbf{h}}}$ is a solution of the system$\mathbf{A}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{b}}$.

b. If$\stackrel{\mathbf{→}}{{\mathbf{x}}_{\mathbf{2}}}$is another solution of the system$\mathit{A}\stackrel{\mathbf{→}}{\mathbf{x}}=\stackrel{\mathbf{→}}{b}$, then$\stackrel{\mathbf{→}}{{\mathbf{x}}_{\mathbf{1}}}\mathbf{+}\stackrel{\mathbf{→}}{{\mathbf{x}}_{\mathbf{h}}}$is a solution of the system $\mathbf{A}\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{+}\stackrel{\mathbf{→}}{0}$.

c. Now suppose A is a$\mathbf{2}\mathbf{×}\mathbf{2}$matrix. A solution vector$\stackrel{\mathbf{→}}{{\mathbf{x}}_{\mathbf{1}}}$of the system$\mathbf{A}\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{+}\stackrel{\mathbf{→}}{b}$is shown in the accompanying figure. We are told that the solutions of the system$\mathbf{A}\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{0}}$form the line shown in the sketch. Draw the line consisting of all solutions of the system$\mathbf{A}\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{=}\stackrel{\mathbf{→}}{b}$.

If you are puzzled by the generality of this problem, think about an example first:

$\mathbf{A}\mathbf{=}\left(\begin{array}{l}1â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots 2\\ 3â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots 6\end{array}\right)\mathbf{,}\stackrel{\mathbf{→}}{\mathbf{b}}\mathbf{=}\left[\begin{array}{l}3\\ 9\end{array}\right]\mathbf{and}\stackrel{\mathbf{→}}{{\mathbf{x}}_{\mathbf{1}}}\mathbf{=}\left[\begin{array}{l}1\\ 1\end{array}\right]$