Chapter 1: Q28E (page 1)
If \({\mathop{\rm b}\nolimits} \ne 0\), can the solution set of \(Ax = b\) be a plane through the origin? Explain.
Short Answer
When \(b \ne 0\), the solution set of \(Ax = b\) cannot form a plane through the origin
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Chapter 1: Q28E (page 1)
If \({\mathop{\rm b}\nolimits} \ne 0\), can the solution set of \(Ax = b\) be a plane through the origin? Explain.
When \(b \ne 0\), the solution set of \(Ax = b\) cannot form a plane through the origin
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Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be an invertible linear transformation. Explain why T is both one-to-one and onto \({\mathbb{R}^n}\). Use equations (1) and (2). Then give a second explanation using one or more theorems.
Let \(A = \left[ {\begin{array}{*{20}{c}}1&0&{ - 4}\\0&3&{ - 2}\\{ - 2}&6&3\end{array}} \right]\) and \(b = \left[ {\begin{array}{*{20}{c}}4\\1\\{ - 4}\end{array}} \right]\). Denote the columns of \(A\) by \({{\mathop{\rm a}\nolimits} _1},{a_2},{a_3}\) and let \(W = {\mathop{\rm Span}\nolimits} \left\{ {{a_1},{a_2},{a_3}} \right\}\).
Suppose an experiment leads to the following system of equations:
\(\begin{aligned}{c}{\bf{4}}.{\bf{5}}{x_{\bf{1}}} + {\bf{3}}.{\bf{1}}{x_{\bf{2}}} = {\bf{19}}.{\bf{249}}\\1.6{x_{\bf{1}}} + 1.1{x_{\bf{2}}} = 6.843\end{aligned}\) (3)
\(\begin{aligned}{c}{\bf{4}}.{\bf{5}}{x_{\bf{1}}} + {\bf{3}}.{\bf{1}}{x_{\bf{2}}} = {\bf{19}}.{\bf{25}}\\1.6{x_{\bf{1}}} + 1.1{x_{\bf{2}}} = 6.8{\bf{4}}\end{aligned}\) (4)
In Exercises 10, write a vector equation that is equivalent tothe given system of equations.
10. \(4{x_1} + {x_2} + 3{x_3} = 9\)
\(\begin{array}{c}{x_1} - 7{x_2} - 2{x_3} = 2\\8{x_1} + 6{x_2} - 5{x_3} = 15\end{array}\)
Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^m}\) be a linear transformation, and suppose \(T\left( u \right) = {\mathop{\rm v}\nolimits} \). Show that \(T\left( { - u} \right) = - {\mathop{\rm v}\nolimits} \).
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