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Chapter 2: Q2.5-8Q (page 93)

Find a LU factorization of the matrices in Exercises 7-16 (with L unit lower triangular). Note that MATLAB will usually produce a permuted LU factorization because it uses partial pivoting for numerical accuracy.

8. \(\left[ {\begin{array}{*{20}{c}}6&9\\4&5\end{array}} \right]\)

Short Answer

Expert verified

The LU factorization of the matrices is \(L = \left[ {\begin{array}{*{20}{c}}1&0\\{\frac{2}{3}}&1\end{array}} \right]\).

Step by step solution

01

Apply the row operation to determine matrix U

Place the first pivot column of \(\left[ {\begin{array}{*{20}{c}}6&9\\4&5\end{array}} \right]\) in the first column of L after dividing the column by top pivot entry 6. Then at row two, multiply row one by \(\frac{2}{3}\) and subtract it from row two to produce matrix U.

\[A = \left[ {\begin{array}{*{20}{c}}6&9\\4&5\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}6&9\\0&{ - 1}\end{array}} \right] = U\]

02

Determine matrix L

The first pivot column of L is the first column of Adivided by the top pivot entry.

Divide the first column of \(\left[ {\begin{array}{*{20}{c}}6&9\\4&5\end{array}} \right]\)by pivot entry 6.

Also, divide row two of matrix U by \( - 1\), as shown below.

\[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{\frac{6}{6}}\\{\frac{4}{6}}\end{array}} \right]\,\left[ {\frac{{ - 1}}{{ - 1}}} \right]\\\,\,\, \downarrow \,\,\,\,\, \downarrow \\\left[ {\begin{array}{*{20}{c}}1&{}\\{\frac{2}{3}}&1\end{array}} \right]\end{array}\]

Place the result in L.

\(L = \left[ {\begin{array}{*{20}{c}}1&0\\{\frac{2}{3}}&1\end{array}} \right]\)

Thus, the LU factorization of the matrices is \(L = \left[ {\begin{array}{*{20}{c}}1&0\\{\frac{2}{3}}&1\end{array}} \right]\).

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Most popular questions from this chapter

Describe in words what happens when you compute \({A^{\bf{5}}}\), \({A^{{\bf{10}}}}\), \({A^{{\bf{20}}}}\), and \({A^{{\bf{30}}}}\) for \(A = \left( {\begin{aligned}{*{20}{c}}{1/6}&{1/2}&{1/3}\\{1/2}&{1/4}&{1/4}\\{1/3}&{1/4}&{5/12}\end{aligned}} \right)\).

Generalize the idea of Exercise 21(a) [not 21(b)] by constructing a \(5 \times 5\) matrix \(M = \left[ {\begin{array}{*{20}{c}}A&0\\C&D\end{array}} \right]\) such that \({M^2} = I\). Make C a nonzero \(2 \times 3\) matrix. Show that your construction works.

Let \(A = \left[ {\begin{array}{*{20}{c}}B&{\bf{0}}\\{\bf{0}}&C\end{array}} \right]\), where \(B\) and \(C\) are square. Show that \(A\)is invertible if an only if both \(B\) and \(C\) are invertible.

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 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?
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