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Chapter 2: Q22Q (page 93)

Show that if the columns of Bare linearly dependent, then so are the columns of AB.

Short Answer

Expert verified

The columns of ABare linearly dependent.

Step by step solution

01

The columns of B are linearly dependent

When the columns of Bare linearly dependent, there exists a nonzero vector \(x\) such that \(Bx = 0\).

02

Show the columns of AB are linearly dependent

Therefore,

\(\begin{aligned}{l}A\left( {Bx} \right) = A \times 0\\\left( {AB} \right)x = 0\,\left( {By\,Associative\,law} \right)\end{aligned}\)

The columns of ABmust be linearly dependent since \(x\) is nonzero.

Thus, the columns of ABare linearly dependent.

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Most popular questions from this chapter

Let \(A = \left( {\begin{aligned}{*{20}{c}}3&{ - 6}\\{ - 1}&2\end{aligned}} \right)\). Construct a \({\bf{2}} \times {\bf{2}}\) matrix Bsuch that ABis the zero matrix. Use two different nonzero columns for B.

Use partitioned matrices to prove by induction that the product of two lower triangular matrices is also lower triangular. [Hint: \(A\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrix \({A_1}\) can be written in the form below, where \[a\] is a scalar, v is in \({\mathbb{R}^k}\), and Ais a \(k \times k\) lower triangular matrix. See the study guide for help with induction.]

\({A_1} = \left[ {\begin{array}{*{20}{c}}a&{{0^T}}\\0&A\end{array}} \right]\).

Prove the Theorem 3(d) i.e., \({\left( {AB} \right)^T} = {B^T}{A^T}\).

(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

Solve the equation \(AB = BC\) for A, assuming that A, B, and C are square and Bis invertible.
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