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Chapter 2: Q2.5-12Q (page 93)

Find an LU factorization of the matrices in Exercise 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.

\[\left[ {\begin{array}{*{20}{c}}{\bf{2}}&{ - {\bf{4}}}&{\bf{2}}\\{\bf{1}}&{\bf{5}}&{ - {\bf{4}}}\\{ - {\bf{6}}}&{ - {\bf{2}}}&{\bf{4}}\end{array}} \right]\]

Short Answer

Expert verified

\(\left[ {\begin{array}{*{20}{c}}1&0&0\\{\frac{1}{2}}&1&0\\{ - 3}&{ - 2}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2&{ - 4}&2\\0&7&{ - 5}\\0&0&0\end{array}} \right]\)

Step by step solution

01

Apply the row operation in the given matrix

Let \(A = \left[ {\begin{array}{*{20}{c}}2&{ - 4}&2\\1&5&{ - 4}\\{ - 6}&{ - 2}&4\end{array}} \right]\).

Apply row operation to reduce the matrix into an upper triangular matrix.

At row 3, multiply row 1 by 3 and add it to row 3, i.e., \({R_3} \to {R_3} + 3{R_1}\).

\(\left[ {\begin{array}{*{20}{c}}2&{ - 4}&2\\1&5&{ - 4}\\0&{ - 14}&{10}\end{array}} \right]\)

At row 2, divide row 1 by 2 and subtract it from row 2, i.e., \({R_2} \to {R_2} - \frac{1}{2}{R_1}\).

\(\left[ {\begin{array}{*{20}{c}}2&{ - 4}&2\\0&7&{ - 5}\\0&{ - 14}&{10}\end{array}} \right]\)

02

Apply row operation in the given matrix

At row 3, multiply row 2 by 2 and add it to row 3, i.e., \({R_3} \to {R_3} + 2{R_2}\).

\(\left[ {\begin{array}{*{20}{c}}2&{ - 4}&2\\0&7&{ - 5}\\0&0&0\end{array}} \right]\)

03

Calculate matrix L using the pivoted column of U

Using matrix \(A\), matrix Lcan be written as

\(\left[ {\begin{array}{*{20}{c}}2&0&0\\1&7&0\\{ - 6}&{ - 14}&1\end{array}} \right]\).

Divide column 1 by 2 and column 2 by 7.

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

04

Write the product \(LU\)

The product of lower and upper triangular matricescan be written as

\(LU = \left[ {\begin{array}{*{20}{c}}1&0&0\\{\frac{1}{2}}&1&0\\{ - 3}&{ - 2}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2&{ - 4}&2\\0&7&{ - 5}\\0&0&0\end{array}} \right]\).

So, the product \(LU\) is \(\left[ {\begin{array}{*{20}{c}}1&0&0\\{\frac{1}{2}}&1&0\\{ - 3}&{ - 2}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2&{ - 4}&2\\0&7&{ - 5}\\0&0&0\end{array}} \right]\).

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Most popular questions from this chapter

Explain why the columns of an \(n \times n\) matrix Aare linearly independent when Ais invertible.

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be an invertible linear transformation, and let Sand U be functions from \({\mathbb{R}^n}\) into \({\mathbb{R}^n}\) such that \(S\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) and \(\)\(U\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\). Show that \(U\left( v \right) = S\left( v \right)\) for all v in \({\mathbb{R}^n}\). This will show that Thas a unique inverse, as asserted in theorem 9. [Hint: Given any v in \({\mathbb{R}^n}\), we can write \({\mathop{\rm v}\nolimits} = T\left( {\mathop{\rm x}\nolimits} \right)\) for some x. Why? Compute \(S\left( {\mathop{\rm v}\nolimits} \right)\) and \(U\left( {\mathop{\rm v}\nolimits} \right)\)].

Suppose Ais an \(m \times n\) matrix and there exist \(n \times m\) matrices C and D such that \(CA = {I_n}\) and \(AD = {I_m}\). Prove that \(m = n\) and \(C = D\). (Hint: Think about the product CAD.)

In Exercise 9 mark each statement True or False. Justify each answer.

9. a. In order for a matrix B to be the inverse of A, both equations \(AB = I\) and \(BA = I\) must be true.

b. If A and B are \(n \times n\) and invertible, then \({A^{ - {\bf{1}}}}{B^{ - {\bf{1}}}}\) is the inverse of \(AB\).

c. If \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\) and \(ab - cd \ne {\bf{0}}\), then A is invertible.

d. If A is an invertible \(n \times n\) matrix, then the equation \(Ax = b\) is consistent for each b in \({\mathbb{R}^{\bf{n}}}\).

e. Each elementary matrix is invertible.

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