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Let \(A = \left( {\begin{aligned}{{}{}}{\bf{3}}&{\bf{4}}\\{ - {\bf{2}}}&{\bf{1}}\\{\bf{3}}&{\bf{4}}\end{aligned}} \right)\), \({\bf{b}} = \left( {\begin{aligned}{{}{}}{{\bf{11}}}\\{ - {\bf{9}}}\\{\bf{5}}\end{aligned}} \right)\), \({\bf{u}} = \left( {\begin{aligned}{{}{}}{\bf{5}}\\{ - {\bf{1}}}\end{aligned}} \right)\), and \({\bf{v}} = \left( {\begin{aligned}{{}{}}{\bf{5}}\\{ - {\bf{2}}}\end{aligned}} \right)\). Compute \(A{\bf{u}}\) and \(A{\bf{v}}\), and compare them with b. Could u possibly be a least-squares solution of \(A{\bf{x}} = {\bf{b}}\)?

Short Answer

Expert verified

The vector u cannot be a least-squares solution for the equation \(A{\bf{x}} = {\bf{b}}\).

Step by step solution

01

Find the value of \(\left\| {{\bf{b}} - A{\bf{u}}} \right\|\)

The value of \({\bf{b}} - A{\bf{u}}\) can be calculated as follows:

\(\begin{aligned}{}{\bf{b}} - A{\bf{u}} &= \left[ {\begin{aligned}{{}{}}{11}\\{ - 9}\\5\end{aligned}} \right] - \left[ {\begin{aligned}{{}{}}3&4\\{ - 2}&1\\3&4\end{aligned}} \right]\left[ {\begin{aligned}{{}{}}5\\{ - 1}\end{aligned}} \right]\\ & = \left[ {\begin{aligned}{{}{}}{11}\\{ - 9}\\5\end{aligned}} \right] - \left[ {\begin{aligned}{{}{}}{11}\\{ - 11}\\{11}\end{aligned}} \right]\\ & = \left[ {\begin{aligned}{{}{}}0\\2\\{ - 6}\end{aligned}} \right]\end{aligned}\)

Find the value of \(\left\| {{\bf{b}} - A{\bf{u}}} \right\|\).

\(\begin{aligned}{}\left\| {{\bf{b}} - A{\bf{u}}} \right\| & = \sqrt {{{\left( 0 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( { - 6} \right)}^2}} \\ & = \sqrt {0 + 4 + 36} \\ & = \sqrt {40} \\ & = 2\sqrt {10} \end{aligned}\)

02

Find the value of \(\left\| {{\bf{b}} - A{\bf{v}}} \right\|\)

The value of \({\bf{b}} - A{\bf{v}}\) can be calculated as follows:

\(\begin{aligned}{}{\bf{b}} - A{\bf{v}} & = \left( {\begin{aligned}{{}{}}{11}\\{ - 9}\\5\end{aligned}} \right) - \left( {\begin{aligned}{{}{}}3&4\\{ - 2}&1\\3&4\end{aligned}} \right)\left( {\begin{aligned}{{}{}}5\\{ - 2}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{11}\\{ - 9}\\5\end{aligned}} \right) - \left( {\begin{aligned}{{}{}}7\\{ - 12}\\7\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}4\\3\\{ - 2}\end{aligned}} \right)\end{aligned}\)

Find the value of \(\left\| {{\bf{b}} - A{\bf{u}}} \right\|\).

\(\begin{aligned}{}\left\| {{\bf{b}} - A{\bf{v}}} \right\| & = \sqrt {{{\left( 4 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( { - 2} \right)}^2}} \\ & = \sqrt {16 + 9 + 4} \\ & = \sqrt {29} \end{aligned}\)

It can be observed that \(A{\bf{v}}\) is closer to bas compared to \(A{\bf{u}}\).

Thus, u cannot be a least-squares solution for the equation \(A{\bf{x}} = {\bf{b}}\).

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Most popular questions from this chapter

Let \({\mathbb{R}^{\bf{2}}}\) have the inner product of Example 1. Show that the Cauchy-Schwarz inequality holds for \({\bf{x}} = \left( {{\bf{3}}, - {\bf{2}}} \right)\) and \({\bf{y}} = \left( { - {\bf{2}},{\bf{1}}} \right)\). (Suggestion: Study \({\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right|^{\bf{2}}}\).)

Suppose radioactive substance A and B have decay constants of \(.02\) and \(.07\), respectively. If a mixture of these two substances at a time \(t = 0\) contains \({M_A}\) grams of \(A\) and \({M_B}\) grams of \(B\), then a model for the total amount of mixture present at time \(t\) is

\(y = {M_A}{e^{ - .02t}} + {M_B}{e^{ - .07t}}\) (6)

Suppose the initial amounts \({M_A}\) and are unknown, but a scientist is able to measure the total amounts present at several times and records the following points \(\left( {{t_i},{y_i}} \right):\left( {10,21.34} \right),\left( {11,20.68} \right),\left( {12,20.05} \right),\left( {14,18.87} \right)\) and \(\left( {15,18.30} \right)\).

a.Describe a linear model that can be used to estimate \({M_A}\) and \({M_B}\).

b. Find the least-squares curved based on (6).

In Exercises 9-12, find a unit vector in the direction of the given vector.

11. \(\left( {\begin{aligned}{*{20}{c}}{\frac{7}{4}}\\{\frac{1}{2}}\\1\end{aligned}} \right)\)

Find an orthogonal basis for the column space of each matrix in Exercises 9-12.

11. \(\left( {\begin{aligned}{{}{}}1&2&5\\{ - 1}&1&{ - 4}\\{ - 1}&4&{ - 3}\\1&{ - 4}&7\\1&2&1\end{aligned}} \right)\)

In Exercises 17 and 18, all vectors and subspaces are in \({\mathbb{R}^n}\). Mark each statement True or False. Justify each answer.

a. If \(W = {\rm{span}}\left\{ {{x_1},{x_2},{x_3}} \right\}\) with \({x_1},{x_2},{x_3}\) linearly independent,

and if \(\left\{ {{v_1},{v_2},{v_3}} \right\}\) is an orthogonal set in \(W\) , then \(\left\{ {{v_1},{v_2},{v_3}} \right\}\) is a basis for \(W\) .

b. If \(x\) is not in a subspace \(W\) , then \(x - {\rm{pro}}{{\rm{j}}_W}x\) is not zero.

c. In a \(QR\) factorization, say \(A = QR\) (when \(A\) has linearly

independent columns), the columns of \(Q\) form an

orthonormal basis for the column space of \(A\).

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