Log In Start studying!

Select your language

Suggested languages for you:
Deutsch (DE)
Deutsch (UK)
Deutsch (US)

Americas

Europe

Answers without the blur. Sign up and see all textbooks for free! → Illustration

Q2.4-7Q

Expert-verified
Linear Algebra and its Applications
Found in: Page 93
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

Answers without the blur.

Just sign up for free and you're in.

Register for Free I’ll do it later
Illustration

Short Answer

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, find formulas for X, Y, and Z in terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint: Compute the product on the left, and set it equal to the right side.]

7. \[\left[ {\begin{array}{*{20}{c}}X&{\bf{0}}&{\bf{0}}\\Y&{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&Z\\{\bf{0}}&{\bf{0}}\\B&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\{\bf{0}}&I\end{array}} \right]\]

The formulas are \(X = {A^{ - 1}}\), \(Y = - B{A^{ - 1}}\), and \[Z = 0\].

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: 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:

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

Step 2: Obtain the product

Compute the product of the left part of the given equation by using the row-column rule, as shown below:

\[\begin{array}{c}\left[ {\begin{array}{*{20}{c}}X&0&0\\Y&0&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&Z\\0&0\\B&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{X\left( A \right) + 0\left( 0 \right) + 0\left( B \right)}&{X\left( Z \right) + 0\left( 0 \right) + 0\left( I \right)}\\{Y\left( A \right) + 0\left( 0 \right) + I\left( B \right)}&{Y\left( Z \right) + 0\left( 0 \right) + I\left( I \right)}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{XA + 0 + 0 \cdot B}&{XZ + 0 + 0 \cdot I}\\{YA + 0 + I \cdot B}&{YZ + 0 + {I^2}}\end{array}} \right]\end{array}\]

Use the matrix properties \(A \cdot I = A\), \(A \cdot 0 = 0\), and \({I^2} = I\).

\[\left[ {\begin{array}{*{20}{c}}X&0&0\\Y&0&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&Z\\0&0\\B&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{XA}&{XZ}\\{YA + B}&{YZ + I}\end{array}} \right]\]

Thus, \[\left[ {\begin{array}{*{20}{c}}X&0&0\\Y&0&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&Z\\0&0\\B&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{XA}&{XZ}\\{YA + B}&{YZ + I}\end{array}} \right]\].

Step 3: Equate both the sides

Equate both the matrices as shown below:

\[\left[ {\begin{array}{*{20}{c}}{XA}&{XZ}\\{YA + B}&{YZ + I}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&0\\0&I\end{array}} \right]\]

By comparing, the formulas become

\(\begin{array}{c}XA = I\\X\left( {{A^{ - 1}}A} \right) = {A^{ - 1}}\left( I \right)\\X \cdot I = {A^{ - 1}} \cdot I\\X = {A^{ - 1}}\end{array}\)

\[\begin{array}{c}XZ = 0\\{X^{ - 1}}\left( {XZ} \right) = {X^{ - 1}}\left( 0 \right)\\Z = 0\end{array}\]

and

\(\begin{array}{c}YA + B = 0\\YA = 0 - B\\YA = - B\end{array}\)

Solve further to get

\(\begin{array}{c}\left( {YA} \right){A^{ - 1}} = \left( { - B} \right){A^{ - 1}}\\Y \cdot \left( {A{A^{ - 1}}} \right) = - B{A^{ - 1}}\\Y \cdot I = - B{A^{ - 1}}\\Y = - B{A^{ - 1}}.\end{array}\)

Therefore, the formulas are \(X = {A^{ - 1}}\), \(Y = - B{A^{ - 1}}\), and \[Z = 0\].

Most popular questions for Math Textbooks

1. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{6}}\\{\bf{5}}&{\bf{4}}\end{aligned}} \right)\).

In Exercises 33 and 34, T is a linear transformation from \({\mathbb{R}^2}\) into \({\mathbb{R}^2}\). Show that T is invertible and find a formula for \({T^{ - 1}}\).

33. \(T\left( {{x_1},{x_2}} \right) = \left( { - 5{x_1} + 9{x_2},4{x_1} - 7{x_2}} \right)\)

Suppose the transfer function W(s) in Exercise 19 is invertible for some s. It can be showed that the inverse transfer function \(W{\left( s \right)^{ - {\bf{1}}}}\), which transforms outputs into inputs, is the Schur complement of \(A - BC - s{I_n}\) for the matrix below. Find the Sachur complement. See Exercise 15.

\(\left[ {\begin{array}{*{20}{c}}{A - BC - s{I_n}}&B\\{ - C}&{{I_m}}\end{array}} \right]\)

In Exercises 27 and 28, view vectors in \({\mathbb{R}^n}\) as \(n \times 1\) matrices. For \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^n}\), the matrix product \({{\mathop{\rm u}\nolimits} ^T}v\) is a \(1 \times 1\) matrix, called the scalar product, or inner product, of u and v. It is usually written as a single real number without brackets. The matrix product \({{\mathop{\rm uv}\nolimits} ^T}\) is an \(n \times n\) matrix, called the outer product of u and v. The products \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm uv}\nolimits} ^T}\) will appear later in the text.

27. Let \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\3\\{ - 4}\end{aligned}} \right)\) and \({\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}a\\b\\c\end{aligned}} \right)\). Compute \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \), \({{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \),\({{\mathop{\rm uv}\nolimits} ^T}\), and \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\).

How many rows does \(B\) have if \(BC\) is a \({\bf{3}} \times {\bf{4}}\) matrix?
Icon

Want to see more solutions like these?

Sign up for free to discover our expert answers
Get Started - It’s free

Recommended explanations on Math Textbooks

Decision Maths

Read explanation

Calculus

Read explanation

Probability and Statistics

Read explanation

Statistics

Read explanation

Mechanics Maths

Read explanation

Geometry

Read explanation

Pure Maths

Read explanation

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.