Question

Iry to hhel ieal 4 Suppose that the 3 x 2 matrix A has rank 2 and we want to solve Ax b. a) (10 pts) If there exists a solution x ()l show that 0 0 b) (5 pts) Is the 3 x 3 augmented matrix (Alb) invertible? Why or why not? c) (10 pts) Suppose that you found the solution below 2 (A | b) 30 0 Can you compute the solution to Ax = b? If yes do it. (5 ntslFor the directed graph below give the incidence matrix A. Iry to hhel ieal 4 Suppose that the 3 x 2 matrix A has rank 2 and we want to solve Ax b. a) (10 pts) If there exists a solution x ()l show that 0 0 b) (5 pts) Is the 3 x 3 augmented matrix (Alb) invertible? Why or why not? c) (10 pts) Suppose that you found the solution below 2 (A | b) 30 0 Can you compute the solution to Ax = b? If yes do it. (5 ntslFor the directed graph below give the incidence matrix A.

Answer 1

4 a) Let us write

b2 2 122 3/

Since

2

is a solution to $Ax=b$, we get

T1 a21 a22 b3 431 a32 0 a31 a32 ba 0 0 0

b) The augmented matrix
B= (A | b)
is $3\times 3$. This would be invertible if and only if the system
Bu=0
has trivial (all zero) solution. But we have seen in part a) that a solution to
Bu=0
is

121

which is not trivial. Hence, is not invertible.

B= (A | b)

c) We have

0 0 2a11 3a12 5bi 0 0 5b1 2a11 + 3aj2 2a31 3a32 5bz 11 12 d21 22 d31 132 11 12 21 223- 02 d31 a32

So, we have computed solution to $Ax=b$; this is

$\begin{align*}\begin{pmatrix}{-\frac25}\\ {-\frac35}\end{pmatrix}\end{align*}$