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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

Suppose the first two columns, \({{\bf{b}}_1}\) and \({{\bf{b}}_2}\), of Bare equal. What can you say about the columns of AB(if ABis defined)? Why?

Suppose \({A_{{\bf{11}}}}\) is an invertible matrix. Find matrices Xand Ysuch that the product below has the form indicated. Also,compute \({B_{{\bf{22}}}}\). [Hint:Compute the product on the left, and setit equal to the right side.]

\[\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\X&I&{\bf{0}}\\Y&{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{1}}1}}}&{{A_{{\bf{1}}2}}}\\{{A_{{\bf{2}}1}}}&{{A_{{\bf{2}}2}}}\\{{A_{{\bf{3}}1}}}&{{A_{{\bf{3}}2}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{B_{11}}}&{{B_{12}}}\\{\bf{0}}&{{B_{22}}}\\{\bf{0}}&{{B_{32}}}\end{array}} \right]\]

Let \(A = \left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\), and \(D = \left( {\begin{aligned}{*{20}{c}}2&0&0\\0&3&0\\0&0&5\end{aligned}} \right)\). Compute \(AD\) and \(DA\). Explain how the columns or rows of A change when A is multiplied by D on the right or on the left. Find a \(3 \times 3\) matrix B, not the identity matrix or the zero matrix, such that \(AB = BA\).

Suppose Tand Ssatisfy the invertibility equations (1) and (2), where T is a linear transformation. Show directly that Sis a linear transformation. [Hint: Given u, v in \({\mathbb{R}^n}\), let \[{\mathop{\rm x}\nolimits} = S\left( {\mathop{\rm u}\nolimits} \right),{\mathop{\rm y}\nolimits} = S\left( {\mathop{\rm v}\nolimits} \right)\]. Then \(T\left( {\mathop{\rm x}\nolimits} \right) = {\mathop{\rm u}\nolimits} \), \[T\left( {\mathop{\rm y}\nolimits} \right) = {\mathop{\rm v}\nolimits} \]. Why? Apply Sto both sides of the equation \(T\left( {\mathop{\rm x}\nolimits} \right) + T\left( {\mathop{\rm y}\nolimits} \right) = T\left( {{\mathop{\rm x}\nolimits} + y} \right)\). Also, consider \(T\left( {cx} \right) = cT\left( x \right)\).]

Suppose Tand U are linear transformations from \({\mathbb{R}^n}\) to \({\mathbb{R}^n}\) such that \(T\left( {U{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\) . Is it true that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\)? Why or why not?
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