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Chapter 2: Q14SE (page 93)

Let \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}0\\0\\1\end{aligned}} \right)\) and \(x = \left( {\begin{aligned}{*{20}{c}}1\\5\\3\end{aligned}} \right)\). Determine \(P\) and \(Q\) as in Exercise 13, and compute \(Px\) and \(Qx\). The figure shows that \(Qx\) is the reflection of the x through \({x_1}{x_2}\)-plane.

Short Answer

Expert verified

\(P = \left( {\begin{aligned}{*{20}{c}}0&0&0\\0&0&0\\0&0&1\end{aligned}} \right),Q = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&{ - 1}\end{aligned}} \right)\), \(P{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}0\\0\\3\end{aligned}} \right)\), \(Q{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}1\\5\\{ - 3}\end{aligned}} \right)\).

Step by step solution

01

Determine \(P\) and \(Q\) as in Exercise 13

In Exercise 13, it is given that u is in \({\mathbb{R}^n}\) with \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm u}\nolimits} = 1\). Let \(P = {\mathop{\rm u}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\) (an outer product) and \(Q = I - 2P\).

Calculate \(P = {\mathop{\rm u}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\) as shown below.

\(\begin{aligned}{c}P = {\mathop{\rm u}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\\ = \left( {\begin{aligned}{*{20}{c}}0\\0\\1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}0&0&1\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}0&0&0\\0&0&0\\0&0&1\end{aligned}} \right)\end{aligned}\)

Calculate \(Q = I - 2P\) as shown below.

\(\begin{aligned}{c}Q = I - 2P\\ = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{aligned}} \right) - 2\left( {\begin{aligned}{*{20}{c}}0&0&0\\0&0&0\\0&0&1\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{aligned}} \right) - \left( {\begin{aligned}{*{20}{c}}0&0&0\\0&0&0\\0&0&2\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&{ - 1}\end{aligned}} \right)\end{aligned}\)

Thus, \(P = \left( {\begin{aligned}{*{20}{c}}0&0&0\\0&0&0\\0&0&1\end{aligned}} \right),Q = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&{ - 1}\end{aligned}} \right)\).

02

Determine \(Px\) and \(Qx\)

It is given that \(x = \left( {\begin{aligned}{*{20}{c}}1\\5\\3\end{aligned}} \right)\).

Calculate \(Px\) a shown below.

\(\begin{aligned}{c}P{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}0&0&0\\0&0&0\\0&0&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1\\5\\3\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{0 + 0 + 0}\\{0 + 0 + 0}\\{0 + 0 + 3}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}0\\0\\3\end{aligned}} \right)\end{aligned}\)

Calculate \(Qx\) as shown below.

\(\begin{aligned}{c}Q{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&{ - 1}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1\\5\\3\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{1 + 0 + 0}\\{0 + 5 + 0}\\{0 + 0 - 3}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1\\5\\{ - 3}\end{aligned}} \right)\end{aligned}\)

Thus, \(P{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}0\\0\\3\end{aligned}} \right)\),

\(Q{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}1\\5\\{ - 3}\end{aligned}} \right)\).

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Most popular questions from this chapter

If Ais an \(n \times n\) matrix and the equation \(A{\bf{x}} = {\bf{b}}\) has more than one solution for some b, then the transformation \({\bf{x}}| \to A{\bf{x}}\) is not one-to-one. What else can you say about this transformation? Justify your answer.

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.

Suppose A, B,and Care invertible \(n \times n\) matrices. Show that ABCis also invertible by producing a matrix Dsuch that \(\left( {ABC} \right)D = I\) and \(D\left( {ABC} \right) = I\).

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 a \(n \times n\) 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.

28. If u and v are in \({\mathbb{R}^n}\), how are \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \) related? How are \({{\mathop{\rm uv}\nolimits} ^T}\) and \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\) related?

Give a formula for \({\left( {ABx} \right)^T}\), where \({\bf{x}}\) is a vector and \(A\) and \(B\) are matrices of appropriate sizes.
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