Chapter 1: Q27E (page 1)

It can be shown that a solution of the system below is \({{\mathop{\rm x}\nolimits} _1} = 3\), \({{\mathop{\rm x}\nolimits} _2} = 2\), and \({{\mathop{\rm x}\nolimits} _3} = - 1\). Use this fact and the theory from this section to explain why another solution is \({{\mathop{\rm x}\nolimits} _1} = 30\), \({{\mathop{\rm x}\nolimits} _2} = 20\), and \({{\mathop{\rm x}\nolimits} _3} = - 10\). (Observe how the solutions are related, but make no other calculations.)
\(\begin{aligned}{c}{{\mathop{\rm x}\nolimits} _1} - 3{{\mathop{\rm x}\nolimits} _2} - 3{{\mathop{\rm x}\nolimits} _3} = 0\\ - 2{{\mathop{\rm x}\nolimits} _1} + 4{{\mathop{\rm x}\nolimits} _2} + 2{{\mathop{\rm x}\nolimits} _3} = 0\\ - {{\mathop{\rm x}\nolimits} _1} + 5{{\mathop{\rm x}\nolimits} _2} + 7{{\mathop{\rm x}\nolimits} _3} = 0\end{aligned}\)

Short Answer

Expert verified

\({{\mathop{\rm x}\nolimits} _1} = 30,{\rm{ }}{{\mathop{\rm x}\nolimits} _2} = 20,\) and \({{\mathop{\rm x}\nolimits} _3} = - 10\) is a solution to the system of equations.

Step by step solution

Nul A is a subspace of \({\mathbb{R}^3}\)

\({\mathop{\rm Nul}\nolimits} A = \left\{ 0 \right\}\)if and only if the equation \(Ax = 0\) has only a trivial solution. \({\mathop{\rm Nul}\nolimits} A\) is a subspaceof \({\mathbb{R}^n}\).

Consider \(A\) is the coefficient matrix of the given houmogeneos system of equations. \(x\) is in \({\mathop{\rm Nul}\nolimits} A\) since the equation \(Ax = 0\) for \(x = \left[ {\begin{array}{*{20}{c}}3\\2\\{ - 1}\end{array}} \right]\). It is closed under scalar multiplication because \({\mathop{\rm Nul}\nolimits} A\) is a subspace of \({\mathbb{R}^3}\).

Explain why another solution is \({{\mathop{\rm x}\nolimits} _1} = 30,{\rm{ }}{{\mathop{\rm x}\nolimits} _2} = 20,\) and \({{\mathop{\rm x}\nolimits} _3} = - 10\)

The scalar multiplication is shown below:

\(\begin{aligned}{c}10{\mathop{\rm x}\nolimits} = 10\left[ {\begin{aligned}{*{20}{c}}3\\2\\{ - 1}\end{aligned}} \right]\\ = \left[ {\begin{aligned}{*{20}{c}}{30}\\{20}\\{ - 10}\end{aligned}} \right]\end{aligned}\)

Therefore, \(\left[ {\begin{aligned}{*{20}{c}}{30}\\{20}\\{ - 10}\end{aligned}} \right]\) is also in \({\mathop{\rm Nul}\nolimits} A\) and \({{\mathop{\rm x}\nolimits} _1} = 30,{{\mathop{\rm x}\nolimits} _2} = 20,{{\mathop{\rm x}\nolimits} _3} = - 10\) is also the solution to the system of equations.Thus, \({{\mathop{\rm x}\nolimits} _1} = 30,{\rm{ }}{{\mathop{\rm x}\nolimits} _2} = 20,{\rm{ }}{{\mathop{\rm x}\nolimits} _3} = - 10\) is a solution to the system of equations