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.
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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.
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Give a formula for \({\left( {ABx} \right)^T}\), where \({\bf{x}}\) is a vector and \(A\) and \(B\) are matrices of appropriate sizes.
In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1–4.
1. \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\E&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\)
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 \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\).
Suppose Aand Bare \(n \times n\), Bis invertible, and ABis invertible. Show that Ais invertible. (Hint: Let C=AB, and solve this equation for A.)
Suppose \(\left( {B - C} \right)D = 0\), where Band Care \(m \times n\) matrices and \(D\) is invertible. Show that B = C.
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