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.

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

Short Answer

Expert verified

The LU factorization of the matrices is \(L = \left[ {\begin{array}{*{20}{c}}1&0\\{ - \frac{3}{2}}&1\end{array}} \right]\)and \(U = \left[ {\begin{array}{*{20}{c}}2&5\\0&{\frac{7}{2}}\end{array}} \right]\).

Step by step solution

01

Apply row operation to determine matrix U

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

Place the first pivot column of \(\left[ {\begin{array}{*{20}{c}}2&5\\{ - 3}&{ - 4}\end{array}} \right]\) in the first column of L after dividing the column by top pivot entry 2.

At row two, multiply row one by \(\frac{3}{2}\) and add it to row two to produce matrix U.

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

02

Determine matrix L

Divide the first column of \(\left[ {\begin{array}{*{20}{c}}2&5\\{ - 3}&{ - 4}\end{array}} \right]\)by pivot entry 2.

And divide row two of U by \(\frac{7}{2}\), as shown below:

\(\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{\frac{2}{2}}\\{ - \frac{3}{2}}\end{array}} \right]\,\left[ {\frac{{\frac{7}{2}}}{{\frac{7}{2}}}} \right]\\\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\, \downarrow \\\left[ {\begin{array}{*{20}{c}}1&{}\\{\frac{{ - 3}}{2}}&1\end{array}} \right]\end{array}\)

Place the result in matrix L.

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

Thus, the LU factorization of the matrices is \(L = \left[ {\begin{array}{*{20}{c}}1&0\\{ - \frac{3}{2}}&1\end{array}} \right]\)and \(U = \left[ {\begin{array}{*{20}{c}}2&5\\0&{\frac{7}{2}}\end{array}} \right]\).

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Most popular questions from this chapter

Let Ube the \({\bf{3}} \times {\bf{2}}\) cost matrix described in Example 6 of Section 1.8. The first column of Ulists the costs per dollar of output for manufacturing product B, and the second column lists the costs per dollar of output for product C. (The costs are categorized as materials, labor, and overhead.) Let \({q_1}\) be a vector in \({\mathbb{R}^{\bf{2}}}\) that lists the output (measured in dollars) of products B and C manufactured during the first quarter of the year, and let \({q_{\bf{2}}}\), \({q_{\bf{3}}}\) and \({q_{\bf{4}}}\) be the analogous vectors that list the amounts of products B and C manufactured in the second, third, and fourth quarters, respectively. Give an economic description of the data in the matrix UQ, where \(Q = \left( {\begin{aligned}{*{20}{c}}{{{\bf{q}}_1}}&{{{\bf{q}}_2}}&{{{\bf{q}}_3}}&{{{\bf{q}}_4}}\end{aligned}} \right)\).

In Exercise 10 mark each statement True or False. Justify each answer.

10. a. A product of invertible \(n \times n\) matrices is invertible, and the inverse of the product of their inverses in the same order.

b. If A is invertible, then the inverse of \({A^{ - {\bf{1}}}}\) is A itself.

c. If \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\) and \(ad = bc\), then A is not invertible.

d. If A can be row reduced to the identity matrix, then A must be invertible.

e. If A is invertible, then elementary row operations that reduce A to the identity \({I_n}\) also reduce \({A^{ - {\bf{1}}}}\) to \({I_n}\).

In Exercises 33 and 34, Tis a linear transformation from \({\mathbb{R}^2}\) into \({\mathbb{R}^2}\). Show that T is invertible and find a formula for \({T^{ - 1}}\).

33. \(T\left( {{x_1},{x_2}} \right) = \left( { - 5{x_1} + 9{x_2},4{x_1} - 7{x_2}} \right)\)

Suppose the last column of ABis entirely zero but Bitself has no column of zeros. What can you sayaboutthe columns of A?

Prove Theorem 2(b) and 2(c). Use the row-column rule. The \(\left( {i,j} \right)\)- entry in \(A\left( {B + C} \right)\) can be written as \({a_{i1}}\left( {{b_{1j}} + {c_{1j}}} \right) + ... + {a_{in}}\left( {{b_{nj}} + {c_{nj}}} \right)\) or \(\sum\limits_{k = 1}^n {{a_{ik}}\left( {{b_{kj}} + {c_{kj}}} \right)} \).

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