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1-6: Solve the equation Ax=b by using the LU factorization given for A. In Exercises 1 and 2, also solve \(Ax = b\) by ordinary row reduction.

2. \[A = \left[ {\begin{array}{*{20}{c}}4&3&{ - 5}\\{ - 4}&{ - 5}&7\\8&6&{ - 8}\end{array}} \right],{\mathop{\rm b}\nolimits} = \left[ {\begin{array}{*{20}{c}}2\\{ - 4}\\6\end{array}} \right]\]

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

Short Answer

Expert verified

\({\mathop{\rm x}\nolimits} = \left( {\frac{1}{4},2,1} \right)\) and \(y = \left[ {\begin{array}{*{20}{c}}2\\{ - 2}\\2\end{array}} \right]\)

Step by step solution

01

Solve the equation \(Ly = b\) for y

Acan be written in the form \(A = LU\), where Lis an \(m \times m\) lower triangular matrix with 1s on the diagonal, and U is an \(m \times n\) echelon form of A.

When \(A = LU\), the equation \(Ax = b\) can be written as \(L\left( {Ux} \right) = {\mathop{\rm b}\nolimits} \).

Write y for \(Ux\) and find x by solving the pair of equations

\(\begin{array}{l}Ly = b\\Ux = y.\end{array}\)

Here, \(L = \left[ {\begin{array}{*{20}{c}}1&0&0\\{ - 1}&1&0\\2&0&1\end{array}} \right]\,{\rm{, }}U = \left[ {\begin{array}{*{20}{c}}4&3&{ - 5}\\0&{ - 2}&2\\0&0&2\end{array}} \right]{\rm{, }}b = \left[ {\begin{array}{*{20}{c}}2\\{ - 4}\\6\end{array}} \right]\)

The matrix \(\left[ {\begin{array}{*{20}{c}}L&b\end{array}} \right]\) is shown below:

\(\left[ {\begin{array}{*{20}{c}}L&b\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0&0&2\\{ - 1}&1&0&{ - 4}\\2&0&1&6\end{array}} \right]\)

Perform an elementary row operation to produce the row-echelon form of the matrix.

At row two, multiply row one by 1 and add it to row two. At row three, multiply row one by 2 and subtract it from row three.

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

The arithmetic values take place only in the last column.

Thus, \(y = \left[ {\begin{array}{*{20}{c}}2\\{ - 2}\\2\end{array}} \right]\).

02

Use back-substitution to solve the equation \(Ux = y\) for x

The matrix \(\left[ {\begin{array}{*{20}{c}}U&y\end{array}} \right]\) is shown below:

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

Perform an elementary row operation to produce the row-echelon form of the matrix.

At row three, multiply row three by \(\frac{1}{2}\).

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

At row two, multiply row three by 1 and add it to row two. At row one, multiply row three by 5 and add it to row one.

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

At row two, multiply row two by \( - \frac{1}{2}\).

\[ \sim \left[ {\begin{array}{*{20}{c}}4&3&0&7\\0&1&0&2\\0&0&1&1\end{array}} \right]\]

At row one, multiply row two by \( - 3\) and add it to row one.

\[ \sim \left[ {\begin{array}{*{20}{c}}4&0&0&1\\0&1&0&2\\0&0&1&1\end{array}} \right]\]

At row one, multiply row one by \(\frac{1}{4}\).

\[ \sim \left[ {\begin{array}{*{20}{c}}1&0&0&{\frac{1}{4}}\\0&1&0&2\\0&0&1&1\end{array}} \right]\]

Thus, \({\mathop{\rm x}\nolimits} = \left( {\frac{1}{4},2,1} \right)\).

03

Solve the equation \(Ax = b\)

The matrix \(\left[ {\begin{array}{*{20}{c}}A&b\end{array}} \right]\) is shown below

\(\left[ {\begin{array}{*{20}{c}}A&b\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4&3&{ - 5}&2\\{ - 4}&{ - 5}&7&{ - 4}\\8&6&{ - 8}&6\end{array}} \right]\)

Perform an elementary row operation to produce the row-echelon form of the matrix.

At row two, multiply row one by 1 and add it to row two. At row three, multiply row one by 2 and subtract it from row three.

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

Thus, the row reduction is the same as \(\left[ {\begin{array}{*{20}{c}}U&y\end{array}} \right]\), yielding the same result.

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Most popular questions from this chapter

Let \({{\bf{r}}_1} \ldots ,{{\bf{r}}_p}\) be vectors in \({\mathbb{R}^{\bf{n}}}\), and let Qbe an\(m \times n\)matrix. Write the matrix\(\left( {\begin{aligned}{*{20}{c}}{Q{{\bf{r}}_1}}& \cdots &{Q{{\bf{r}}_p}}\end{aligned}} \right)\)as a productof two matrices (neither of which is an identity matrix).

Let \(X\) be \(m \times n\) data matrix such that \({X^T}X\) is invertible., and let \(M = {I_m} - X{\left( {{X^T}X} \right)^{ - {\bf{1}}}}{X^T}\). Add a column \({x_{\bf{0}}}\) to the data and form

\(W = \left[ {\begin{array}{*{20}{c}}X&{{x_{\bf{0}}}}\end{array}} \right]\)

Compute \({W^T}W\). The \(\left( {{\bf{1}},{\bf{1}}} \right)\) entry is \({X^T}X\). Show that the Schur complement (Exercise 15) of \({X^T}X\) can be written in the form \({\bf{x}}_{\bf{0}}^TM{{\bf{x}}_{\bf{0}}}\). It can be shown that the quantity \({\left( {{\bf{x}}_{\bf{0}}^TM{{\bf{x}}_{\bf{0}}}} \right)^{ - {\bf{1}}}}\) is the \(\left( {{\bf{2}},{\bf{2}}} \right)\)-entry in \({\left( {{W^T}W} \right)^{ - {\bf{1}}}}\). This entry has a useful statistical interpretation, under appropriate hypotheses.

In the study of engineering control of physical systems, a standard set of differential equations is transformed by Laplace transforms into the following system of linear equations:

\(\left[ {\begin{array}{*{20}{c}}{A - s{I_n}}&B\\C&{{I_m}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\bf{x}}\\{\bf{u}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{y}}\end{array}} \right]\)

Where \(A\) is \(n \times n\), \(B\) is \(n \times m\), \(C\) is \(m \times n\), and \(s\) is a variable. The vector \({\bf{u}}\) in \({\mathbb{R}^m}\) is the โ€œinputโ€ to the system, \({\bf{y}}\) in \({\mathbb{R}^m}\) is the โ€œoutputโ€ and \({\bf{x}}\) in \({\mathbb{R}^n}\) is the โ€œstateโ€ vector. (Actually, the vectors \({\bf{x}}\), \({\bf{u}}\) and \({\bf{v}}\) are functions of \(s\), but we suppress this fact because it does not affect the algebraic calculations in Exercises 19 and 20.)

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 \({I_n}\).

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be an invertible linear transformation. Explain why T is both one-to-one and onto \({\mathbb{R}^n}\). Use equations (1) and (2). Then give a second explanation using one or more theorems.

Use matrix algebra to show that if A is invertible and D satisfies \(AD = I\) then \(D = {A^{ - {\bf{1}}}}\).

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