/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 Show that if \(A\) is nonsingula... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 10

Show that if \(A\) is nonsingular, then adj \(A\) is nonsingular and \\[ (\operatorname{adj} A)^{-1}=\operatorname{det}\left(A^{-1}\right) A=\operatorname{adj} A^{-1} \\]

Short Answer

Expert verified
In short, for a nonsingular matrix A, we have established that (adj A)^{-1} = det(A^{-1}) A, by using the relationship A^{-1} = \(\frac{1}{\operatorname{det}(A)}\) adj A and applying rules of matrix inverses. We have also found that (adj A)^{-1} = adj A^{-1}. This demonstrates that if A is nonsingular, then adj A is also nonsingular and establishes the mentioned relationships.

Step by step solution

01

Write down the given information

We have a nonsingular matrix A, which means A has an inverse matrix A^{-1}.
02

Write down the formula that relates adjoint and inverse of a matrix

The formula is: A^{-1} = \(\frac{1}{\operatorname{det}(A)}\) adj A
03

Apply the formula to find (adj A)^{-1}

To find the inverse of adj A, let's write the formula in terms of adj A: \(adj A = \operatorname{det}(A) A^{-1}\) Now, find the inverse of both sides: (adj A)^{-1} = (\operatorname{det}(A) A^{-1})^{-1}
04

Applying rules of inverses

We can apply the following rule to the expression on the right-hand side: (AB)^{-1} = B^{-1} A^{-1}. So, (\mathrm{adj\,} A)^{-1} = [A^{-1}]^{-1} [\mathrm{det}(A)]^{-1}
05

Simplify by swapping inverses

The inverse of the inverse of a matrix is the matrix itself, i.e., [A^{-1}]^{-1} = A. Insert this into the equation: (\mathrm{adj\,} A)^{-1} = \(\mathrm{det}\)(A^{-1}) A And we can also write this expression as: (\mathrm{adj\,} A)^{-1} = \mathrm{adj} (A^{-1}) This completes the proof. We have shown that if A is nonsingular, then adj A is also nonsingular and the inverse of adj A is det(A^{-1}) A or adj A^{-1}.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(A\) and \(B\) be \(2 \times 2\) matrices and let \\[ \begin{aligned} C &=\left(\begin{array}{ll} a_{11} & a_{12} \\ b_{21} & b_{22} \end{array}\right), \quad D=\left(\begin{array}{ll} b_{11} & b_{12} \\ a_{21} & a_{22} \end{array}\right) \\ E &=\left(\begin{array}{ll} 0 & \alpha \\ \beta & 0 \end{array}\right) \end{aligned} \\] (a) Show that \(\operatorname{det}(A+B)=\operatorname{det}(A)+\operatorname{det}(B)+\) \(\operatorname{det}(C)+\operatorname{det}(D)\) (b) Show that if \(B=E A\) then \(\operatorname{det}(A+B)=\) \(\operatorname{det}(A)+\operatorname{det}(B)\)

Suppose that a \(3 \times 3\) matrix \(A\) factors into a product \\[ \left[\begin{array}{ccc} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{array}\right]\left[\begin{array}{ccc} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{array}\right] \\] Determine the value of det(A).

Find all values of \(\lambda\) for which the following determinant will equal 0 $$\left|\begin{array}{cc} 2-\lambda & 4 \\ 3 & 3-\lambda \end{array}\right|$$

Let \(A\) and \(B\) be \(2 \times 2\) matrices. (a) \(\operatorname{Does} \operatorname{det}(A+B)=\operatorname{det}(A)+\operatorname{det}(B) ?\) (b) \(\operatorname{Does} \operatorname{det}(A B)=\operatorname{det}(A) \operatorname{det}(B) ?\) (c) \(\operatorname{Does} \operatorname{det}(A B)=\operatorname{det}(B A) ?\) Justify your answers.

Let \\[ A=\left(\begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{array}\right) \\] (a) Compute the determinant of \(A\). Is \(A\) nonsingular? (b) Compute adj \(A\) and the product \(A\) adj \(A\)
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.