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Construct a nonzero \(3 \times 3\) matrix A and a vector b such that b is not in Col A.

Short Answer

Expert verified

Write the echelon form of the \(3 \times 3\) matrix with only two pivots and consider b as any vector in \({\mathbb{R}^3}\) whose third entry is nonzero.

Step by step solution

01

Condition for Col A

Thecolumn spaceof matrix A is the set Col A of all linear combinationsof the columns of A.

02

Construct a nonzero \(3 \times 3\) matrix A and vector b

Construct the echelon form of the \(3 \times 3\) matrix with only two pivots and consider b as any vector in \({\mathbb{R}^3}\) whose third entry is nonzero, as shown below.

, \(b = \left[ {\begin{array}{*{20}{c}}0\\0\\1\end{array}} \right]\).

Thus, b is not in Col A.

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Most popular questions from this chapter

Exercises 15 and 16 concern arbitrary matrices A, B, and Cfor which the indicated sums and products are defined. Mark each statement True or False. Justify each answer.

15. a. If A and B are \({\bf{2}} \times {\bf{2}}\) with columns \({{\bf{a}}_1},{{\bf{a}}_2}\) and \({{\bf{b}}_1},{{\bf{b}}_2}\) respectively, then \(AB = \left( {\begin{aligned}{*{20}{c}}{{{\bf{a}}_1}{{\bf{b}}_1}}&{{{\bf{a}}_2}{{\bf{b}}_2}}\end{aligned}} \right)\).

b. Each column of ABis a linear combination of the columns of Busing weights from the corresponding column of A.

c. \(AB + AC = A\left( {B + C} \right)\)

d. \({A^T} + {B^T} = {\left( {A + B} \right)^T}\)

e. The transpose of a product of matrices equals the product of their transposes in the same order.

Let Abe an invertible \(n \times n\) matrix, and let B be an \(n \times p\) matrix. Show that the equation \(AX = B\) has a unique solution \({A^{ - 1}}B\).

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 the inverse found in Exercise 3 to solve the system

\(\begin{aligned}{l}{\bf{8}}{{\bf{x}}_{\bf{1}}} + {\bf{5}}{{\bf{x}}_{\bf{2}}} = - {\bf{9}}\\ - {\bf{7}}{{\bf{x}}_{\bf{1}}} - {\bf{5}}{{\bf{x}}_{\bf{2}}} = {\bf{11}}\end{aligned}\)

Give a formula for \({\left( {ABx} \right)^T}\), where \({\bf{x}}\) is a vector and \(A\) and \(B\) are matrices of appropriate sizes.

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