Chapter 2: Q19Q (page 93)
If the columns of a \({\bf{7}} \times {\bf{7}}\) matrix \(D\) are linearly independent, what can you say about solutions of \(D{\bf{x}} = {\bf{b}}\)? Why?
Short Answer
The given equation has a unique solution.
Chapter 2: Q19Q (page 93)
If the columns of a \({\bf{7}} \times {\bf{7}}\) matrix \(D\) are linearly independent, what can you say about solutions of \(D{\bf{x}} = {\bf{b}}\)? Why?
The given equation has a unique solution.
All the tools & learning materials you need for study success - in one app.
Get started for freeSuppose P is invertible and \(A = PB{P^{ - 1}}\). Solve for Bin terms of A.
In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1–4.
2. \[\left[ {\begin{array}{*{20}{c}}E&{\bf{0}}\\{\bf{0}}&F\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\]
Prove Theorem 2(d). (Hint: The \(\left( {i,j} \right)\)- entry in \(\left( {rA} \right)B\) is \(\left( {r{a_{i1}}} \right){b_{1j}} + ... + \left( {r{a_{in}}} \right){b_{nj}}\).)
In Exercises 27 and 28, view vectors in \({\mathbb{R}^n}\) as \(n \times 1\) matrices. For \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^n}\), the matrix product \({{\mathop{\rm u}\nolimits} ^T}v\) is a \(1 \times 1\) matrix, called the scalar product, or inner product, of u and v. It is usually written as a single real number without brackets. The matrix product \({{\mathop{\rm uv}\nolimits} ^T}\) is an \(n \times n\) matrix, called the outer product of u and v. The products \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm uv}\nolimits} ^T}\) will appear later in the text.
27. Let \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\3\\{ - 4}\end{aligned}} \right)\) and \({\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}a\\b\\c\end{aligned}} \right)\). Compute \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \), \({{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \),\({{\mathop{\rm uv}\nolimits} ^T}\), and \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\).
Let \(A = \left( {\begin{aligned}{*{20}{c}}3&{ - 6}\\{ - 1}&2\end{aligned}} \right)\). Construct a \({\bf{2}} \times {\bf{2}}\) matrix Bsuch that ABis the zero matrix. Use two different nonzero columns for B.
What do you think about this solution?
We value your feedback to improve our textbook solutions.