Question

7. Let A be a 4 x 3 matrix, and let b and y be two arbitrary vectors in R. We are told that the system Ax- b has a unique solution. What can you say about the number of solutions of the system Ax - y? Explain your answer. 8. Let u. v, w, b be arbitrary vectors in R". Suppose that b = x1u+xy+23w for some scalars i, r23. Show that Span u, v, w, b Span u, v, w Hint: Take an arbitrary vector y-clu+cv+w+cb and show that y for some scalars di,d2,ds dud2v +djw

Answer 1

(7). Note that the rank of the $4\times 3$ matrix $A$ is $\le 3$. Now, the existence of the unique solution of $Ax=b$ implies that $\mbox{rank }A=3.$

Now, if $[A:y]$ is the augmented matrix of the equation $Ax=y$, then

$\mbox{rank }[A:y]\neq \mbox{rank }A=3$

implies there is no solution of the equation $Ax=y.$

And, if $\mbox{rank }[A:y]=\mbox{rank }A=3$ , then the equation $Ax=y$ also has unique solution.

(8). Since $\{u,v,w\}\subseteq \{u,v,w,b\},$ it follows that

$\mbox{span }\{u,v,w\}\subseteq\mbox{span } \{u,v,w,b\}...........(1)$.

Now for the reverse, let $y\in\mbox{span } \{u,v,w,b\}.$ Then, there exists $c_1,c_2,c_3,c_4\in\mathbb{R}$ such that

$y=c_1u+c_2v+c_3w+c_4b............(2)$

Since $b=x_1u+x_2v+x_3w$ , it follows from (2) that

$y=c_1u+c_2v+c_3w+c_4(x_1u+x_2v+x_3w)\\=(c_1+c_4x_1)u+(c_2+c_4x_2)v+(c_3+c_4x_3)w\\ =d_1u+d_2v+d_3w$

and hence $y\in\mbox{span } \{u,v,w\}$ and

$\mbox{span } \{u,v,w,b\}\subseteq\mbox{span }\{u,v,w\}................(3)$

Thus, it follows from (1) and (3) that $\mbox{span }\{u,v,w\}=\mbox{span } \{u,v,w,b\}.$