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Chapter 2: Q3Q (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.

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 \(\left( {{\bf{3}}{I_{\bf{2}}}} \right)A\).

Short Answer

Expert verified

\(\left( {\begin{aligned}{*{20}{c}}{ - 1}&1\\{ - 5}&5\end{aligned}} \right)\), \(\left( {\begin{aligned}{*{20}{c}}{12}&{ - 3}\\{15}&{ - 6}\end{aligned}} \right)\)

Step by step solution

01

Find the matrix \(3{I_{\bf{2}}} - A\)

The value of \(3{I_2} - A\) can be calculated as follows:

\(\begin{aligned}{c}3{I_2} - A = 3\left( {\begin{aligned}{*{20}{c}}1&0\\0&1\end{aligned}} \right) - \left( {\begin{aligned}{*{20}{c}}4&{ - 1}\\5&{ - 2}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}3&0\\0&3\end{aligned}} \right) - \left( {\begin{aligned}{*{20}{c}}4&{ - 1}\\5&{ - 2}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 1}&1\\{ - 5}&5\end{aligned}} \right)\end{aligned}\)

02

Find the matrix \(\left( {{\bf{3}}{I_{\bf{2}}}} \right)A\)

The value of \(\left( {3{I_2}} \right)A\) can be calculated as follows:

\(\begin{aligned}{c}\left( {3{I_2}} \right)A = 3\left( {{I_2}A} \right)\\ = 3\left( {\begin{aligned}{*{20}{c}}1&0\\0&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}4&{ - 1}\\5&{ - 2}\end{aligned}} \right)\\ = 3\left( {\begin{aligned}{*{20}{c}}{1 \times 4 + 0}&{1 \times \left( { - 1} \right) + 0}\\{0 + 1 \times 5}&{0 \times \left( { - 1} \right) + 1 \times \left( { - 2} \right)}\end{aligned}} \right)\\ = 3\left( {\begin{aligned}{*{20}{c}}4&{ - 1}\\5&{ - 2}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{12}&{ - 3}\\{15}&{ - 6}\end{aligned}} \right)\end{aligned}\)

So, \(3{I_2} - A = \left( {\begin{aligned}{*{20}{c}}{ - 1}&1\\{ - 5}&5\end{aligned}} \right)\), and \(\left( {3{I_2}} \right)A = \left( {\begin{aligned}{*{20}{c}}{12}&{ - 3}\\{15}&{ - 6}\end{aligned}} \right)\).

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Most popular questions from this chapter

Suppose A, B, and Care \(n \times n\) matrices with A, X, and \(A - AX\) invertible, and suppose

\({\left( {A - AX} \right)^{ - 1}} = {X^{ - 1}}B\) …(3)

  1. Explain why B is invertible.
  2. Solve (3) for X. If you need to invert a matrix, explain why that matrix is invertible.

Suppose \(\left( {B - C} \right)D = 0\), where Band Care \(m \times n\) matrices and \(D\) is invertible. Show that B = C.

[M] Suppose memory or size restrictions prevent your matrix program from working with matrices having more than 32 rows and 32 columns, and suppose some project involves \(50 \times 50\) matrices A and B. Describe the commands or operations of your program that accomplish the following tasks.

a. Compute \(A + B\)

b. Compute \(AB\)

c. Solve \(Ax = b\) for some vector b in \({\mathbb{R}^{50}}\), assuming that \(A\) can be partitioned into a \(2 \times 2\) block matrix \(\left[ {{A_{ij}}} \right]\), with \({A_{11}}\) an invertible \(20 \times 20\) matrix, \({A_{22}}\) an invertible \(30 \times 30\) matrix, and \({A_{12}}\) a zero matrix. [Hint: Describe appropriate smaller systems to solve, without using any matrix inverse.]

(M) Read the documentation for your matrix program, and write the commands that will produce the following matrices (without keying in each entry of the matrix).

  1. A \({\bf{5}} \times {\bf{6}}\) matrix of zeros
  2. A \({\bf{3}} \times {\bf{5}}\) matrix of ones
  3. The \({\bf{6}} \times {\bf{6}}\) identity matrix
  4. A \({\bf{5}} \times {\bf{5}}\) diagonal matrix, with diagonal entries 3, 5, 7, 2, 4

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}\\{\bf{5}}&{{\bf{12}}}\end{aligned}} \right),{b_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}\\{\bf{3}}\end{aligned}} \right),{b_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{ - {\bf{5}}}\end{aligned}} \right),{b_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{6}}\end{aligned}} \right),\) and \({b_{\bf{4}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{5}}\end{aligned}} \right)\).

  1. Find \({A^{ - {\bf{1}}}}\), and use it to solve the four equations \(Ax = {b_{\bf{1}}},\)\(Ax = {b_2},\)\(Ax = {b_{\bf{3}}},\)\(Ax = {b_{\bf{4}}}\)\(\)
  2. The four equations in part (a) can be solved by the same set of row operations, since the coefficient matrix is the same in each case. Solve the four equations in part (a) by row reducing the augmented matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{b_{\bf{1}}}}&{{b_{\bf{2}}}}&{{b_{\bf{3}}}}&{{b_{\bf{4}}}}\end{aligned}} \right)\).
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