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
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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In Exercise 1, compute \(u + v\) and \(u - 2v\).
Let \(A\) be a \(3 \times 3\) matrix with the property that the linear transformation \({\bf{x}} \mapsto A{\bf{x}}\) maps \({\mathbb{R}^3}\) into \({\mathbb{R}^3}\). Explain why transformation must be one-to-one.
Explain why a set \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\) in \({\mathbb{R}^5}\) must be linearly independent when \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\) is linearly independent and \({{\mathop{\rm v}\nolimits} _4}\) is not in Span \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\).
In a grid of wires, the temperature at exterior mesh points is maintained at constant values, as shown in the accompanying figure. When the grid is in thermal equilibrium, the temperature Tat each interior mesh point is the average of the temperatures at the four adjacent points. For example,
Find the temperatures andwhen the grid is in thermal equilibrium.
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