Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 300

If $$ \mathrm{A}=\left|\begin{array}{ccc} 1 & -1 & 1 \\ 0 & 2 & -3 \\ 2 & 1 & 0 \end{array}\right| $$ and \(\mathrm{B}=(\operatorname{adj} \mathrm{A}), \mathrm{C}=5 \mathrm{~A}\) then \([(\mid\) adj \(\mathrm{B} \mid) /|\mathrm{C}|]=\) (a) 5 (b) 1 (c) 3 \((1 / 5)\)

Short Answer

Expert verified
The short answer is: \((c) 3(\frac{1}{5})\).

Step by step solution

01

Find the adjoint of matrix A

To find the adjoint of matrix A, we need to find the matrix of cofactors of A and then transpose that matrix. The matrix A is given as: \(A = \begin{pmatrix} 1 & -1 & 1 \\ 0 & 2 & -3 \\ 2 & 1 & 0 \end{pmatrix} \) First, we find the matrix of cofactors of A: \(Cof(A) = \begin{pmatrix} (2)(0)-(-3)(1) & -[(0)(0)-(-3)(2)] & (0)(2)-(2)(-3) \\ -(1)(1)-(-1)(2) & (1)(0)-(-1)(2) & (-1)(2)-(1)(1) \\ -(1)(-3)-(-1)(2) & (1)(1)-(0)(-1) & (1)(2)-(0)(1) \end{pmatrix} = \begin{pmatrix} 2 & 6 & 6 \\ 1 & 2 & -1 \\ 5 & 1 & 2 \end{pmatrix} \) Now, we transpose the matrix of cofactors to obtain the adjoint of A: \(adj(A) = B = \begin{pmatrix} 2 & 1 & 5 \\ 6 & 2 & 1 \\ 6 & -1 & 2 \end{pmatrix} \)
02

Find the adjoint of matrix B

We will now follow the same process to find the adjoint of matrix B: \(B = \begin{pmatrix} 2 & 1 & 5 \\ 6 & 2 & 1 \\ 6 & -1 & 2 \end{pmatrix} \) First, we find the matrix of cofactors of B: \(Cof(B) = \begin{pmatrix} (2)(2)-(-1)(1) & -[(2)(1)-(6)(-1)] & (1)(-1)-(6)(2) \\ -[(1)(1)-(-1)(6)] & (1)(2)-(5)(6) & (-1)(5)-(-1)(6) \\ -[(1)(1)-(2)(5)] & (2)(5)-(1)(6) & (1)(6)-(2)(2) \end{pmatrix} = \begin{pmatrix} 4 & -8 & -13 \\ 5 & -28 & 1 \\ -9 & 4 & 4 \end{pmatrix} \) Now, we transpose the matrix of cofactors to obtain the adjoint of B: \(adj(B) = \begin{pmatrix} 4 & 5 & -9 \\ -8 & -28 & 4 \\ -13 & 1 & 4 \end{pmatrix} \)
03

Find the determinant of adjoint of matrix B

Now, we will find the determinant of the adjoint of matrix B: \(|adj(B)| = \begin{vmatrix} 4 & 5 & -9 \\ -8 & -28 & 4 \\ -13 & 1 & 4 \end{vmatrix} = 4((-28)(4)-(4)(1))+5((-8)(4)+(9)(1))+(-9)(-8-(-13)(1)) = 4(-112+4)+5(-32+9)-9(-8+13) = 4(-108)+5(-23)-9(5)= (-432)-(-115)-45 = -432+115-45= -362 \)
04

Calculate matrix C

Matrix C is given as 5 times matrix A \(C = 5A = 5\begin{pmatrix} 1 & -1 & 1 \\ 0 & 2 & -3 \\ 2 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 5 & -5 & 5 \\ 0 & 10 & -15 \\ 10 & 5 & 0 \end{pmatrix} \)
05

Find the determinant of matrix C

Now, we will find the determinant of matrix C: \(|C| = \begin{vmatrix} 5 & -5 & 5 \\ 0 & 10 & -15 \\ 10 & 5 & 0 \end{vmatrix} = 5(10)(0)-(5)(-15)-5(0)(-15) = (-75)=-75 \)
06

Calculate the final result

Finally, we need to divide the determinant of adjoint of matrix B by the determinant of matrix C \([|\operatorname{adj}(B)|/|C|] = \frac{-362}{-75} = \frac{362}{75} = \frac{362}{5} * \frac{1}{15} = 72.4 * \frac{1}{15} \) Therefore, the correct answer is (c) 3(1/5).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

If $$ \mathrm{A}=\left|\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right| $$ then \(\mathrm{A}^{2013}=\) (a) \(2^{2012} \mathrm{~A}\) (b) \(2^{1006} \mathrm{~A}\) (c) \(-2^{2013} \mathrm{~A}\) (d) I

If \(\mathrm{A}=\left|\begin{array}{ll}\alpha & 0 \\ 1 & 1\end{array}\right|\) and \(\mathrm{B}=\left|\begin{array}{ll}1 & 0 \\ 5 & 1\end{array}\right|\) then \(\mathrm{A}^{2}=\mathrm{B}\) for (a) \(\alpha=4\) (b) \(\alpha=1\) (c) \(\alpha=-1\) (d) no \(\alpha\)

If the system of equations \(x+k y+3 z=0,3 x+k y-2 z=0\) \(2 \mathrm{x}+3 \mathrm{y}-4 \mathrm{z}=0\) has non trivial solution, then \(\left(\mathrm{xy} / \mathrm{z}^{2}\right)=\ldots\) (a) \((5 / 6)\) (b) \(-(5 / 6)\) (c) \((6 / 5)\) (d) \(-(6 / 5)\)

\(\left|\begin{array}{ccc}(b+c)^{2} & a^{2} & a^{2} \\ b^{2} & (c+a)^{2} & b^{2} \\ c^{2} & c^{2} & (a+b)^{2}\end{array}\right|=k(a b c)(a+b+c)^{3}\), then \(k=\ldots\) (a) (b) - 1 (c) \(-2\) (d) 2

If $$ A=\left|\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \alpha & -\sin \alpha \\ 0 & \sin \alpha & \cos \alpha \end{array}\right| $$ then \(|(\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj} \mathrm{A}))))|=\) (a) 1 (b) 2 (c) \(2^{4}\) (d) \(2^{12}\)
See all solutions

Recommended explanations on English Textbooks

Argumentative Essay

Read Explanation

Language Acquisition

Read Explanation

Rhetorical Analysis Essay

Read Explanation

Single Paragraph Essay

Read Explanation

Syntax

Read Explanation

5 Paragraph Essay

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.

Sign-up for free