13. For A as in exercise 11 i.e., \[A = \left[ {\begin{array}{*{20}{c}}{\bf{3}}&{\bf{2}}&{\bf{1}}&{ - {\bf{5}}}\\{ - {\bf{9}}}&{ - {\bf{4}}}&{\bf{1}}&{\bf{7}}\\{\bf{9}}&{\bf{2}}&{ - {\bf{5}}}&{\bf{1}}\end{array}} \right]\], find a nonzero vector in Nul A and a nonzero vector in Col A.

Short Answer

Expert verified

One of the nonzero vectors in Nul A is \[\left( {1,2, - 1,6} \right)\], and one of the nonzero vectors in Col A is \[\left( {3, - 9,9} \right)\].

Step by step solution

01

Use row reduction

First, solve the equation \[Ax = 0\].

Its augmented matrix is:

\[\left[ {\begin{array}{*{20}{c}}A&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}3&2&1&{ - 5}&0\\{ - 9}&{ - 4}&1&7&0\\9&2&{ - 5}&1&0\end{array}} \right]\]

At row 2, multiply row 1 by 3 and add it to row 2, i.e., , and at row 3, multiply row 1 by 3 and subtract it from row 3, i.e., \[{R_3} \to {R_3} - 3{R_1}\]. Then,

\[ \sim \left[ {\begin{array}{*{20}{c}}3&2&1&{ - 5}&0\\0&2&4&{ - 8}&0\\0&{ - 4}&{ - 8}&{16}&0\end{array}} \right]\]

At row 3, multiply row 2 by 2 and add it to row 3, i.e., \[{R_3} \to {R_3} + 2{R_2}\].

\[ \sim \left[ {\begin{array}{*{20}{c}}3&2&1&{ - 5}&0\\0&2&4&{ - 8}&0\\0&0&0&0&0\end{array}} \right]\]

At row 1, subtract row 2 from row 1, i.e., \[{R_1} \to {R_1} - {R_2}\].

\[ \sim \left[ {\begin{array}{*{20}{c}}3&0&{ - 3}&3&0\\0&2&4&{ - 8}&0\\0&0&0&0&0\end{array}} \right]\]

At row 1, divide row 1 by 3, and at row 2, divide row 2 by 2.

\[\left[ {\begin{array}{*{20}{c}}A&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&{ - 1}&1&0\\0&1&2&{ - 4}&0\\0&0&0&0&0\end{array}} \right]\]

This implies that

\[\begin{array}{c}{x_1} - {x_3} + {x_4} = 0\\{x_2} + 2{x_3} - 4{x_4} = 0\\0 = 0\end{array}\]

Thus, the system is consistent.

02

Find a nonzero vector in Nul A

The general solution is \[{x_1} = {x_3} - {x_4}\], and with \[{x_3},\] and are free. So, the nonzero vector in Nul A is given by \[{x_3},\] and .

Choose and then , and \[{x_2} = 6\] to obtain a nonzero vector \[\left( {1,2, - 1,6} \right)\] in Nul A.

03

Find a nonzero vector in Col A

By definition of column space, Col A is the span of \[\left\{ {\left( {3, - 9,9} \right),\left( {2, - 4,2} \right),\left( {1,1, - 5} \right),\left( { - 5,7,1} \right)} \right\}\].

So, every column of A belongs to Col A. Choose any nonzero column of A. So, is a nonzero vector in Col A.

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Most popular questions from this chapter

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be an invertible linear transformation, and let Sand U be functions from \({\mathbb{R}^n}\) into \({\mathbb{R}^n}\) such that \(S\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) and \(\)\(U\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\). Show that \(U\left( v \right) = S\left( v \right)\) for all v in \({\mathbb{R}^n}\). This will show that Thas a unique inverse, as asserted in theorem 9. [Hint: Given any v in \({\mathbb{R}^n}\), we can write \({\mathop{\rm v}\nolimits} = T\left( {\mathop{\rm x}\nolimits} \right)\) for some x. Why? Compute \(S\left( {\mathop{\rm v}\nolimits} \right)\) and \(U\left( {\mathop{\rm v}\nolimits} \right)\)].

Show that block upper triangular matrix \(A\) in Example 5is invertible if and only if both \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\) are invertible. [Hint: If \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\) are invertible, the formula for \({A^{ - {\bf{1}}}}\) given in Example 5 actually works as the inverse of \(A\).] This fact about \(A\) is an important part of several computer algorithims that estimates eigenvalues of matrices. Eigenvalues are discussed in chapter 5.

In exercise 5 and 6, compute the product \(AB\) in two ways: (a) by the definition, where \(A{b_{\bf{1}}}\) and \(A{b_{\bf{2}}}\) are computed separately, and (b) by the row-column rule for computing \(AB\).

\(A = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{2}}}\\{ - {\bf{3}}}&{\bf{0}}\\{\bf{3}}&{\bf{5}}\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{3}}\\{\bf{2}}&{ - {\bf{1}}}\end{aligned}} \right)\)

Suppose the third column of Bis the sum of the first two columns. What can you say about the third column of AB? Why?

Let Ube the \({\bf{3}} \times {\bf{2}}\) cost matrix described in Example 6 of Section 1.8. The first column of Ulists the costs per dollar of output for manufacturing product B, and the second column lists the costs per dollar of output for product C. (The costs are categorized as materials, labor, and overhead.) Let \({q_1}\) be a vector in \({\mathbb{R}^{\bf{2}}}\) that lists the output (measured in dollars) of products B and C manufactured during the first quarter of the year, and let \({q_{\bf{2}}}\), \({q_{\bf{3}}}\) and \({q_{\bf{4}}}\) be the analogous vectors that list the amounts of products B and C manufactured in the second, third, and fourth quarters, respectively. Give an economic description of the data in the matrix UQ, where \(Q = \left( {\begin{aligned}{*{20}{c}}{{{\bf{q}}_1}}&{{{\bf{q}}_2}}&{{{\bf{q}}_3}}&{{{\bf{q}}_4}}\end{aligned}} \right)\).

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