Question

b. - 2 -1 1 and b Let A = Show that the equation Ax =b does not have a solution for all possible b, and -3 0 3 4-2 2 b3 describe the set of all b for which Ax b does have a solution How can it be shown that the equation Ax = b does not have a solution for all possible b? Choose the correct answer below. O A. Find a vector b for which the solution to Ax = b is the zero vector. O B. Row reduce the matrix A to demonstrate that A does not have a pivot position in every row. C. Row reduce the augmented matrix has a pivot position in every row. to demonstrate that A b A b O D. Row reduce the matrix A to demonstrate that A has a pivot position in every row. O E. Find a vector x for which Ax b is the zero vector Describe the set of all b for which Ax b does have a solution. 0 = (Type an expression using b,, b2, and bå as the variables and 1 as the coefficient of ba.)

Answer 1

to prove that the system does not have a solution for all possible b, row reduce the matrix A to demonstrate that A does not have a pivot position in every row

option B is correct

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augmented matrix is

$\begin{pmatrix}1&-2&-1&b_1\\ -3&3&0&b_2\\ 4&-2&2&b_3\end{pmatrix}$

$R_1\:\leftrightarrow \:R_3$

$=\begin{pmatrix}4&-2&2&b_3\\ -3&3&0&b_2\\ 1&-2&-1&b_1\end{pmatrix}$

$R_2\:\leftarrow \:R_2+\frac{3}{4}\cdot \:R_1$

$=\begin{pmatrix}4&-2&2&b_3\\ 0&\frac{3}{2}&\frac{3}{2}&\frac{4b_2+3b_3}{4}\\ 1&-2&-1&b_1\end{pmatrix}$

$R_3\:\leftarrow \:R_3-\frac{1}{4}\cdot \:R_1$

$=\begin{pmatrix}4&-2&2&b_3\\ 0&\frac{3}{2}&\frac{3}{2}&\frac{4b_2+3b_3}{4}\\ 0&-\frac{3}{2}&-\frac{3}{2}&\frac{4b_1-b_3}{4}\end{pmatrix}$

$R_3\:\leftarrow \:R_3+1\cdot \:R_2$

$=\begin{pmatrix}4&-2&2&b_3\\ 0&\frac{3}{2}&\frac{3}{2}&\frac{4b_2+3b_3}{4}\\ 0&0&0&\frac{b_3+2b_1+2b_2}{2}\end{pmatrix}$

$R_3 \leftarrow R_3\cdot 2$

$=\begin{pmatrix}4&-2&2&b_3\\ 0&\frac{3}{2}&\frac{3}{2}&\frac{4b_2+3b_3}{4}\\ 0&0&0&b_3+2b_1+2b_2\end{pmatrix}$

reduced matrix A is

$\begin{pmatrix}{\color{Red} 4}&-2&2\\ 0&{\color{Red} \frac{3}{2}}&\frac{3}{2}\\ 0&0&0\end{pmatrix}$

as we can see there is no pivot entry at the third row

so the system does not have a solution for all possible b

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reduced augmented matrix is

$=\begin{pmatrix}4&-2&2&b_3\\ 0&\frac{3}{2}&\frac{3}{2}&\frac{4b_2+3b_3}{4}\\ 0&0&0&{\color{Red} b_3+2b_1+2b_2}\end{pmatrix}$

when $b_3+2b_1+2b_2=0$ then system has a solutions

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so consition is

${\color{Red} 0=2b_1+2b_2+b_3}$

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