Q15E Let \(A = \left( {\begin{array}{... [FREE SOLUTION]

91影视

Linear Algebra and its Applications

David C. Lay, Steven R. Lay and Judi J. McDonald

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 483 pages

ISBN: 978-03219822384

Chapter 1: Q15E (page 1)

Let $A = \left( {\begin{array}{{20}{c}}2&{ - 1}\\{ - 6}&3\end{array}} \right)$ and $b = \left( {\begin{array}{{20}{c}}{{b_1}}\\{{b_2}}\end{array}} \right)$. Show that the equation $Ax = {\mathop{\rm b}\nolimits} $ does not have a solution for all possible ${\mathop{\rm b}\nolimits} $, and describe the set of all b for which $Ax = {\mathop{\rm b}\nolimits} $ does have a solution.

Short Answer

Expert verified

The equation $Ax = b$ is inconsistent when $3{b_1} + {b_2}$ is non-zero. The set b for which the equation $Ax = b$ is consistent can be a line passing through the origin for all the set of points $\left( {{b_1},{b_2}} \right)$ satisfying${b_2} = - 3{b_1}$.

Step by step solution

Writing the matrix in the augmented form

Write the given matrix in the augmented form $\left( {\begin{array}{*{20}{c}}A&b\end{array}} \right)$.

$\left( {\begin{array}{*{20}{c}}2&{ - 1}&{{b_1}}\\{ - 6}&3&{{b_2}}\end{array}} \right)$

Step 2:Applying the row operation

Perform an elementary row operationto produce the first augmented matrix.

Write the sum of 3 times row one and row two in row two.

$\left( {\begin{array}{*{20}{c}}2&{ - 1}&{{b_1}}\\0&0&{{b_2} + 3{b_1}}\end{array}} \right)$

Converting the matrix into the equation form

To obtain the solution of the vector equation, convert the augmented matrix into vector equations.

Write the obtained matrix,$\left( {\begin{array}{*{20}{c}}2&{ - 1}&{{b_1}}\\0&0&{{b_2} + 3{b_1}}\end{array}} \right)$,in the equation notation.

$\begin{array}{c}2{x_1} - {x_2} = {b_1}\\0 = {b_2} + 3{b_1}\end{array}$

Showing the equation $Ax = b$ does not have a solution

The equation $Ax = b$ is inconsistent when $3{b_1} + {b_2}$ is non-zero.

Hence,the equation $Ax = b$ does not have a solution.

Determining whether the equation $Ax = b$ has a solution

The set ${\mathop{\rm b}\nolimits} $ for which the equation$Ax = b$ is consistent can be a line passing through the origin for all the set of points $\left( {{b_1},{b_2}} \right)$ that satisfy ${b_2} = - 3{b_1}$.

Thus, the equation $Ax = b$ is consistent for ${b_2} = - 3{b_1}$.