Q62E Consider n脳n matrices聽A,B,C, a... [FREE SOLUTION]

91影视

Linear Algebra With Applications

Otto Bretscher

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 442 pages

ISBN: 9780321796974

Chapter 6: Q62E (page 293)

Consider $n 脳 n$ matrices $A, B, C$ , and $D$ such that $r a n k (A) = r a n k [\begin{matrix} A & B \\ C & D \end{matrix}] = n$ . Show that
a. $D = C A^{- 1} B$ , and
b. The $2 脳 2$ matrix $[\begin{matrix} d e t (A) & d e t (B) \\ d e t (C) & d e t (D) \end{matrix}]$ is noninvertible. Hint: Consider the product $[\begin{matrix} I_{n} & 0 \\ - C A^{- 1} & I_{n} \end{matrix}] [\begin{matrix} A & B \\ C & D \end{matrix}]$ .

Short Answer

Expert verified

Therefore,

a) It is proved that $D = C A^{- 1} B$ .

b) Yes, $[\begin{matrix} d e t A & d e t B \\ d e t C & d e t D \end{matrix}]$ it is non-invertible.

Step by step solution

Step by Step Solution: Step 1: Matrix Definition

Matrix is aset of numbers arranged in rowsand columns so as to form a rectangular array.

The numbers are called the elements, or entries, of the matrix.

If there are $m$ rows and $n$ columns, the matrix is said to be a 鈥� $m$ by $n$ 鈥� matrix, written 鈥� $m 脳 n$ .鈥�

Given

Given matrix,

$rank (A) = rank [\begin{matrix} A & B \\ C & D \end{matrix}] = n$

(a)Step 3: To show D=CA-1B

To show,

$[\begin{matrix} I_{n} & 0 \\ - C A^{- 1} & I_{n} \end{matrix}] [\begin{matrix} A & B \\ C & D \end{matrix}] = [\begin{matrix} A & B \\ 0 & - C A^{- 1} B + D \end{matrix}]$

Since $[\begin{matrix} I_{n} & 0 \\ - C A^{- 1} & I_{n} \end{matrix}]$

Invertible, then

$rank [\begin{matrix} A & B \\ 0 & - C A^{- 1} B + D \end{matrix}] = n$

This means that $C A^{- 1} B + D = 0$

So,

$D = C A^{- 1} B$

(b)Step 4: To show [det(A)det(B)det(C)det(D)] is non-invertible.

From $D = C A^{- 1} B$ ,

We have,

$\begin{array}{rcl} \det D & = & \det (C A^{- 1} B) \\ \det D & = & \det C \det A^{- 1} \det B \\ \det A \det D & = & \det B \det C \\ \det A \det D - \det B \det C & = & 0 . \end{array}$

So the matrix is $[\begin{matrix} d e t A & d e t B \\ d e t C & d e t D \end{matrix}]$ is non-invertible.

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Short Answer

Step by step solution

Step by Step Solution: Step 1: Matrix Definition

Given

(a)Step 3: To show D=CA-1B

(b)Step 4: To show [det(A)det(B)det(C)det(D)] is non-invertible.

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91影视

Consider n脳n matricesA,B,C, andD such that rank(A)=rank[ABCD]=n. Show thata. D=CA-1B, and b. The2脳2 matrix[det(A)det(B)det(C)det(D)] is noninvertible. Hint: Consider the product [In0-CA-1In][ABCD].

Short Answer

Step by step solution

Step by Step Solution: Step 1: Matrix Definition

Given

(a)Step 3: To show D=CA-1B

(b)Step 4: To show [det(A)det(B)det(C)det(D)] is non-invertible.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Geometry

Decision Maths

Statistics

Calculus

Pure Maths

Study anywhere. Anytime. Across all devices.