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Chapter 4: Q28E (page 191)

Consider the following two systems of equations:

\(\begin{array}{c}5{x_1} + {x_2} - 3{x_3} = 0\\ - 9{x_1} + 2{x_2} + 5{x_3} = 1\\4{x_1} + {x_2} - 6{x_3} = 9\end{array}\) \(\begin{array}{c}5{x_1} + {x_2} - 3{x_3} = 0\\ - 9{x_1} + 2{x_2} + 5{x_3} = 5\\4{x_1} + {x_2} - 6{x_3} = 45\end{array}\)

It can be shown that the first system of a solution. Use this fact and the theory from this section to explain why the second system must also have a solution. (Make no row operations.)

Short Answer

Expert verified

The second system of equations must have a solution.

Step by step solution

01

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

\({\mathop{\rm Col}\nolimits} A = {\mathbb{R}^m}\)if and only if the equation \(Ax = {\mathop{\rm b}\nolimits} \) has a solution.

Consider \(A\) as the coefficient matrix of the system of equations. The constant vector \({\mathop{\rm b}\nolimits} = \left[ {\begin{array}{*{20}{c}}0\\1\\9\end{array}} \right]\) is in \({\mathop{\rm Col}\nolimits} A\) because the first system of the equation has a solution. It is closed under scalar multiplication because \({\mathop{\rm Col}\nolimits} A\) is a subspace of \({\mathbb{R}^3}\).

02

Explain that the second system must have a solution

Scalar multiplication of the constant vector is shown below:

\(\begin{array}{c}5{\mathop{\rm b}\nolimits} = 5\left[ {\begin{array}{*{20}{c}}0\\1\\9\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}0\\5\\{45}\end{array}} \right]\end{array}\)

Here, \(5{\mathop{\rm b}\nolimits} \) is also in \({\mathop{\rm Col}\nolimits} A\). Thus, the second system of equations must have a solution.

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Most popular questions from this chapter

A homogeneous system of twelve linear equations in eight unknowns has two fixed solutions that are not multiples of each other, and all other solutions are linear combinations of these two solutions. Can the set of all solutions be described with fewer than twelve homogeneous linear equations? If so, how many? Discuss.

In Exercise 5, find the coordinate vector \({\left( x \right)_{\rm B}}\) of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

5. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{5}}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{1}}\end{array}} \right)\)

Let H be an n-dimensional subspace of an n-dimensional vector space V. Explain why \(H = V\).

[M] Let \(A = \left[ {\begin{array}{*{20}{c}}7&{ - 9}&{ - 4}&5&3&{ - 3}&{ - 7}\\{ - 4}&6&7&{ - 2}&{ - 6}&{ - 5}&5\\5&{ - 7}&{ - 6}&5&{ - 6}&2&8\\{ - 3}&5&8&{ - 1}&{ - 7}&{ - 4}&8\\6&{ - 8}&{ - 5}&4&4&9&3\end{array}} \right]\).

  1. Construct matrices \(C\) and \(N\) whose columns are bases for \({\mathop{\rm Col}\nolimits} A\) and \({\mathop{\rm Nul}\nolimits} A\), respectively, and construct a matrix \(R\) whose rows form a basis for Row\(A\).
  2. Construct a matrix \(M\) whose columns form a basis for \({\mathop{\rm Nul}\nolimits} {A^T}\), form the matrices \(S = \left[ {\begin{array}{*{20}{c}}{{R^T}}&N\end{array}} \right]\) and \(T = \left[ {\begin{array}{*{20}{c}}C&M\end{array}} \right]\), and explain why \(S\) and \(T\) should be square. Verify that both \(S\) and \(T\) are invertible.

[M] Repeat Exercise 35 for a random integer-valued matrixwhose rank is at most 4. One way to makeis to create a random integ\(6 \times 7\)er-valued \(6 \times 4\) matrix \(J\) and a random integer-valued \(4 \times 7\) matrix \(K\), and set \(A = JK\). (See supplementary Exercise 12 at the end of the chapter; and see the study guide for the matrix-generating program.)
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