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

1. \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\E&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\)

Short Answer

Expert verified

The product is \(\left[ {\begin{array}{*{20}{c}}A&B\\{EA + C}&{EB + D}\end{array}} \right]\).

Step by step solution

01

State the row-column rule

If the sum of the products of matching entries from row\(i\)of matrix A and column\(j\)of matrix B equals the item in row\(i\)and column\(j\)of AB, then it can be said that product AB is defined.

The product is shown below:

\({\left( {AB} \right)_{ij}} = {a_{i1}}{b_{1j}} + {a_{i2}}{b_{2j}} + ... + {a_{in}}{b_{nj}}\)

02

Obtain the product

Compute the product by using the row-column rule, as shown below:

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}I&0\\E&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{I\left( A \right) + 0\left( C \right)}&{I\left( B \right) + 0\left( D \right)}\\{E\left( A \right) + I\left( C \right)}&{E\left( B \right) + I\left( D \right)}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{IA}&{IB}\\{EA + IC}&{EB + ID}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}A&B\\{EA + C}&{EB + D}\end{array}} \right]\end{array}\)

Thus, \(\left[ {\begin{array}{*{20}{c}}I&0\\E&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}A&B\\{EA + C}&{EB + D}\end{array}} \right]\).

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Most popular questions from this chapter

Suppose Aand Bare \(n \times n\), Bis invertible, and ABis invertible. Show that Ais invertible. (Hint: Let C=AB, and solve this equation for A.)

Suppose the third column of Bis the sum of the first two columns. What can you say about the third column of AB? Why?

Show that the transformation in Exercise 7 is equivalent to a rotation about the origin followed by a translation by p. Find p.

In exercise 11 and 12, the matrices are all \(n \times n\). Each part of the exercise is an implication of the form 鈥淚f 鈥渟tatement 1鈥 then 鈥渟tatement 2鈥.鈥滿ark the implication as True if the truth of 鈥渟tatement 2鈥漚lways follows whenever 鈥渟tatement 1鈥 happens to be true. An implication is False if there is an instance in which 鈥渟tatement 2鈥 is false but 鈥渟tatement 1鈥 is true. Justify each answer.

a. If the equation \[A{\bf{x}} = {\bf{0}}\] has only the trivial solution, then \(A\) is row equivalent to the \(n \times n\) identity matrix.

b. If the columns of \(A\) span \({\mathbb{R}^n}\), then the columns are linearly independent.

c. If \(A\) is an \(n \times n\) matrix, then the equation \(A{\bf{x}} = {\bf{b}}\) has at least one solution for each \({\bf{b}}\) in \({\mathbb{R}^n}\).

d. If the equation \[A{\bf{x}} = {\bf{0}}\] has a non trivial solution, then \[A\] has fewer than \(n\) pivot positions.

e. If \({A^T}\) is not invertible, then \(A\) is not invertible.

Let T be a linear transformation that maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\). Is \({T^{ - 1}}\) also one-to-one?

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