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Chapter 2: Q4Q (page 93)

In the rest of this exercise set and in those to follow, you should assume that each matrix expression is defined. That is, the sizes of the matrices (and vectors) involved match appropriately.

Compute \(A - {\bf{5}}{I_{\bf{3}}}\) and \(\left( {{\bf{5}}{I_{\bf{3}}}} \right)A\)

\(A = \left( {\begin{aligned}{*{20}{c}}{\bf{9}}&{ - {\bf{1}}}&{\bf{3}}\\{ - {\bf{8}}}&{\bf{7}}&{ - {\bf{6}}}\\{ - {\bf{4}}}&{\bf{1}}&{\bf{8}}\end{aligned}} \right)\)

Short Answer

Expert verified

\(\left( {\begin{aligned}{*{20}{c}}4&{ - 1}&3\\{ - 8}&2&{ - 6}\\{ - 4}&1&3\end{aligned}} \right)\), \(\left( {\begin{aligned}{*{20}{c}}{45}&{ - 5}&{15}\\{ - 40}&{35}&{ - 30}\\{ - 20}&5&{40}\end{aligned}} \right)\)

Step by step solution

01

Find the matrix \(A - 5{I_{\bf{3}}}\)

The value of \(A - 5{I_3}\) can be calculated as follows:

\(\begin{aligned}{c}A - 5{I_3} = \left( {\begin{aligned}{*{20}{c}}9&{ - 1}&3\\{ - 8}&7&{ - 6}\\{ - 4}&1&8\end{aligned}} \right) - 5\left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{9 - 5}&{ - 1 - 0}&{3 - 0}\\{ - 8 - 0}&{7 - 5}&{ - 6 - 0}\\{ - 4 - 0}&{1 - 0}&{8 - 5}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}4&{ - 1}&3\\{ - 8}&2&{ - 6}\\{ - 4}&1&3\end{aligned}} \right)\end{aligned}\)

02

Find the matrix \(\left( {5{I_3}} \right)A\)

The value of \(\left( {5{I_3}} \right)A\) can be calculated as follows:

\(\begin{aligned}{c}\left( {5{I_3}} \right)A = 5\left( {{I_3}A} \right)\\ = 5\left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}9&{ - 1}&3\\{ - 8}&7&{ - 6}\\{ - 4}&1&8\end{aligned}} \right)\\ = 5\left( {\begin{aligned}{*{20}{c}}9&{ - 1}&3\\{ - 8}&7&{ - 6}\\{ - 4}&1&8\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{45}&{ - 5}&{15}\\{ - 40}&{35}&{ - 30}\\{ - 20}&5&{40}\end{aligned}} \right)\end{aligned}\)

So, \(A - 5{I_3} = \left( {\begin{aligned}{*{20}{c}}4&{ - 1}&3\\{ - 8}&2&{ - 6}\\{ - 4}&1&3\end{aligned}} \right)\), and \(\left( {5{I_3}} \right)A = \left( {\begin{aligned}{*{20}{c}}{45}&{ - 5}&{15}\\{ - 40}&{35}&{ - 30}\\{ - 20}&5&{40}\end{aligned}} \right)\).

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Most popular questions from this chapter

If Ais an \(n \times n\) matrix and the equation \(A{\bf{x}} = {\bf{b}}\) has more than one solution for some b, then the transformation \({\bf{x}}| \to A{\bf{x}}\) is not one-to-one. What else can you say about this transformation? Justify your answer.

Suppose \({A_{{\bf{11}}}}\) is invertible. Find \(X\) and \(Y\) such that

\[\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{{A_{{\bf{12}}}}}\\{{A_{{\bf{21}}}}}&{{A_{{\bf{22}}}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\X&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{\bf{0}}\\{\bf{0}}&S\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&Y\\{\bf{0}}&I\end{array}} \right]\]

Where \(S = {A_{{\bf{22}}}} - {A_{21}}A_{{\bf{11}}}^{ - {\bf{1}}}{A_{{\bf{12}}}}\). The matrix \(S\) is called the Schur complement of \({A_{{\bf{11}}}}\). Likewise, if \({A_{{\bf{22}}}}\) is invertible, the matrix \({A_{{\bf{11}}}} - {A_{{\bf{12}}}}A_{{\bf{22}}}^{ - {\bf{1}}}{A_{{\bf{21}}}}\) is called the Schur complement of \({A_{{\bf{22}}}}\). Such expressions occur frequently in the theory of systems engineering, and elsewhere.

In the rest of this exercise set and in those to follow, you should assume that each matrix expression is defined. That is, the sizes of the matrices (and vectors) involved match appropriately.

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{1}}}\\{\bf{5}}&{ - {\bf{2}}}\end{aligned}} \right)\). Compute \({\bf{3}}{I_{\bf{2}}} - A\) and \(\left( {{\bf{3}}{I_{\bf{2}}}} \right)A\).

Use partitioned matrices to prove by induction that for \(n = 2,3,...\), the \(n \times n\) matrices \(A\) shown below is invertible and \(B\) is its inverse.

\[A = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\1&1&0&{}&0\\1&1&1&{}&0\\ \vdots &{}&{}& \ddots &{}\\1&1&1& \ldots &1\end{array}} \right]\]

\[B = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\{ - 1}&1&0&{}&0\\0&{ - 1}&1&{}&0\\ \vdots &{}& \ddots & \ddots &{}\\0&{}& \ldots &{ - 1}&1\end{array}} \right]\]

For the induction step, assume A and Bare \(\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrices, and partition Aand B in a form similar to that displayed in Exercises 23.

In Exercises 1鈥�9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1鈥�4.

4. \[\left[ {\begin{array}{*{20}{c}}I&0\\{ - X}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\]
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