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Chapter 4: Q6E (page 191)

If a\({\bf{6}} \times {\bf{3}}\)matrix A has a rank 3, find dim Nul A, dim Row A, and rank\({A^T}\).

Short Answer

Expert verified

0, 3, and 3

Step by step solution

01

Find dim Nul A

Using the rank theorem,you get:

\(\begin{aligned} {\rm{rank}}\,A + \dim \,{\rm{Nul}}A &= n\\3 + \dim \;{\rm{Nul}}\,A &= 3\\\dim \;{\rm{Nul}}\,A &= 3 - 3\\ &= 0\end{aligned}\)

02

Find dim row A

The dim row A is equal to the rank of A i.e., 3.

03

Find the rank of \({A^T}\)

\(\dim \,\;{\rm{Row}}\;A = \dim \,{\rm{Col}}\,{A^T} = 3\)

The rank of \({A^T}\) is equal to dim Col \({A^T}\); so the rank of \({A^T}\) is 3.

Thus, dim Nul A =0, dim row A=3, and the rank of \({A^T}\)=3.

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Most popular questions from this chapter

If A is a \({\bf{6}} \times {\bf{4}}\) matrix, what is the smallest possible dimension of Null A?

Justify the following equalities:

a.\({\rm{dim Row }}A{\rm{ + dim Nul }}A = n{\rm{ }}\)

b.\({\rm{dim Col }}A{\rm{ + dim Nul }}{A^T} = m\)

(M) Show that \(\left\{ {t,sin\,t,cos\,{\bf{2}}t,sin\,t\,cos\,t} \right\}\) is a linearly independent set of functions defined on \(\mathbb{R}\). Start by assuming that

\({c_{\bf{1}}} \cdot t + {c_{\bf{2}}} \cdot sin\,t + {c_{\bf{3}}} \cdot cos\,{\bf{2}}t + {c_{\bf{4}}} \cdot sin\,t\,cos\,t = {\bf{0}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{5}} \right)\)

Equation (5) must hold for all real t, so choose several specific values of t (say, \(t = {\bf{0}},\,.{\bf{1}},\,.{\bf{2}}\)) until you get a system of enough equations to determine that the \({c_j}\) must be zero.

Suppose \({{\bf{p}}_{\bf{1}}}\), \({{\bf{p}}_{\bf{2}}}\), \({{\bf{p}}_{\bf{3}}}\), and \({{\bf{p}}_{\bf{4}}}\) are specific polynomials that span a two-dimensional subspace H of \({P_{\bf{5}}}\). Describe how one can find a basis for H by examining the four polynomials and making almost no computations.

Suppose the solutions of a homogeneous system of five linear equations in six unknowns are all multiples of one nonzero solution. Will the system necessarily have a solution for every possible choice of constants on the right sides of the equations? Explain.
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