Question

12 3-5 2 U 0 0 0 0 3 (2) A matri A is no1 0 (Thi is not the matris A) (2) A matrix A iownuivalent to This is nohe matrix A! 11 pts] Give the rank and nullity of Λ. rank(A)--null(.)-- 4 pts Does Ar have a solution for every rigt-haud-side ector BYes or No Justify your aswer 2 pts Give a gemetric description for the set all veetrswih the property that A has a solution 4 ptsl Suppose A ive the general solution to A- 1:3 19

Answer 1

a) Since is row equivalent to the $4\times 5$ matrix

$B:=\begin{pmatrix}1&2&3&-1&5\\ 0&1&2&0&4\\ 0&0&0&0&3\\ 0&0&0&0&0\end{pmatrix}$

they have the same null-space, and hence, the same nullity. By rank-nullity theorem, we have

$5=\mbox{rank}(A)+\mbox{nullity}(A)=\mbox{rank}(A)+\mbox{nullity}(B)$

which implies

$\mbox{rank}(A)=\mbox{rank}(B),~~~~~~\mbox{nullity}(A)=5-\mbox{rank}(B)$

Now, the matrix has pivot rows/columns; first, second, and third row (respectively, first, second, and fifth column); therefore, $\mbox{rank}(A)=\mbox{rank}(B)=3$ , which implies $\mbox{nullity}(A)=5-\mbox{rank}(B)=5-3=2$ . Hence,

$\mbox{rank}(A)=3,~~~~~~\mbox{nullity}(A)=2$

b) Since is row equivalent to the $4\times 5$ matrix

$B:=\begin{pmatrix}1&2&3&-1&5\\ 0&1&2&0&4\\ 0&0&0&0&3\\ 0&0&0&0&0\end{pmatrix}$

there is a product of elementary (hence invertible) matrices, , such that $A=RB$ . Thus, $Ax=b$ has a solution if and only if $RBx=b~\Rightarrow~ Bx=R^{-1}b$ has a solution. But the fourth row of is all-zero; therefore, $R^{-1}b$ need to have its fourth entry zero, in order to have a solution to $Ax=b$ .

Hence, the system $Ax=b$ does not have a solution for every vector .

c) If $Ax=b$ has no solution, then the solution set is just $\emptyset$ , the empty set.

Suppose that $Ax=b$ has a solution $x_0$ ; then, for any other solution , we have

$A(x-x_0)=Ax-Ax_0=b-b=0$

Thus, for every solution of $Ax=b$ , we get a solution of $Az=0$ , and vice-versa. Hence, the solution set of $Ax=b$ is given by

$x_0+\{z:Az=0\}$

Now, is geometrically the intersection of the hyperplanes whose normal vectors are the rows of the matrix . Therefore, $x_0+\{z:Az=0\}$ is a translation of the intersection of all the hyperplanes that have the rows of as their normal vectors.

d) As we have seen above, the solution set to $Ax=b$ is given by

$\begin{pmatrix}1\\2\\3\\2\\1\end{pmatrix}+\begin{Bmatrix}z:=\begin{pmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{pmatrix}\mid Az=0\end{Bmatrix}$

As noted in part a), we have

$\begin{align*} Az=0&~\mbox{if and only if}~Bz=0\\ &~\mbox{if and only if}~\begin{pmatrix}1&2&3&-1&5\\0&1&2&0&4\\0&0&0&0&3\\0&0&0&0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{pmatrix} =\begin{pmatrix}0\\0\\0\\0\end{pmatrix}\\ &~\mbox{if and only if}~\begin{pmatrix}x_1+2x_2+3x_3-x_4+5x_5\\x_2+2x_3+4x_5\\3x_5\\0\end{pmatrix} =\begin{pmatrix}0\\0\\0\\0\end{pmatrix}\end{align*}$

From the last equality, we get

$\begin{align*}x_5&=0\\ x_2&=-2x_3\\ x_1&=x_3+x_4\end{align*}$

Therefore,

$\begin{align*} Az=0&~\mbox{if and only if}~z=\begin{pmatrix}x_3+x_4\\-2x_3\\x_3\\x_4\\0\end{pmatrix}=x_3\begin{pmatrix}1\\-2\\1\\0\\0\end{pmatrix}+x_4\begin{pmatrix}1\\0\\0\\1\\0\end{pmatrix}\end{align*}$

which shows that the solutions to $Ax=b$ are

${\bf x}= \begin{pmatrix}1\\2\\3\\2\\1\end{pmatrix}+x_3\begin{pmatrix}1\\-2\\1\\0\\0\end{pmatrix}+x_4\begin{pmatrix}1\\0\\0\\1\\0\end{pmatrix}$