/*! 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 3 Given \\[ A=\left(\begin{arr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 3

Given \\[ A=\left(\begin{array}{lll} 1 & 2 & 1 \\ 0 & 4 & 3 \\ 1 & 2 & 2 \end{array}\right) \\] determine the (2,3) entry of \(A^{-1}\) by computing a quotient of two determinants.

Short Answer

Expert verified
The (2,3) entry of the inverse matrix \(A^{-1}\) is 0.

Step by step solution

01

Compute the determinant of the matrix \(A\)

We have the matrix \(A\): \[ A=\left(\begin{array}{lll} 1 & 2 & 1 \\ 0 & 4 & 3 \\ 1 & 2 & 2 \end{array}\right) \] Now, let's compute its determinant: \[ \text{det}(A) = (1)\left|\begin{array}{cc} 4 & 3 \\ 2 & 2 \end{array}\right| - (2)\left|\begin{array}{cc} 0 & 3 \\ 1 & 2 \end{array}\right| + (1)\left|\begin{array}{cc} 0 & 4 \\ 1 & 2 \end{array}\right| \] \[ \text{det}(A) = (1)(8 - 6) - (2)(6) + (1)(0 - 4) \] \[ \text{det}(A) = 2 - 12 - 4 = -14 \]
02

Replace the (2,3) entry by the corresponding cofactor

Now, let's replace the (2,3) entry \(a_{23}=3\) by its cofactor in the matrix \(A\). The cofactor is obtained by computing the determinant of the matrix formed by removing the row and the column containing the element we are replacing. In this case, we remove the second row and the third column: \[ C_{23} = \left|\begin{array}{cc} 1 & 2 \\ 1 & 2 \end{array}\right| \]
03

Compute the determinant of the new matrix

Now, compute the determinant of the matrix obtained above: \[ C_{23} = (1)(2) - (2)(1) = 0 \]
04

Compute the (2,3) entry of the inverse matrix \(A^{-1}\)

We have the formula for the (2,3) entry of the inverse matrix \(A^{-1}\): \((A^{-1})_{23} = \frac{C_{32}}{\text{det}(A)}\) Substitute the values of \(C_{32}\) and \(\text{det}(A)\) into the formula: \((A^{-1})_{23} = \frac{0}{-14} = 0\) So, the (2,3) entry of the inverse matrix \(A^{-1}\) is 0.

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

Use mathematical induction to prove that if \(A\) is an \((n+1) \times(n+1)\) matrix with two identical rows, then \(\operatorname{det}(A)=0\)

Let \(A\) be a nonsingular \(n \times n\) matrix with \(n>1\) Show that \\[ \operatorname{det}(\operatorname{adj} A)=(\operatorname{det}(A))^{n-1} \\]

Use determinants to determine whether the following \(2 \times 2\) matrices are nonsingular: (a) \(\left(\begin{array}{ll}3 & 5 \\ 2 & 4\end{array}\right)\) (b) \(\left(\begin{array}{ll}3 & 6 \\ 2 & 4\end{array}\right)\) (c) \(\left(\begin{array}{rr}3 & -6 \\ 2 & 4\end{array}\right)\)

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).

Evaluate the following determinants: (a) \(\left|\begin{array}{rr}3 & 5 \\ -2 & -3\end{array}\right|\) (b) \(\left|\begin{array}{rr}5 & -2 \\ -8 & 4\end{array}\right|\) (c) \(\left|\begin{array}{lll}3 & 1 & 2 \\ 2 & 4 & 5 \\ 2 & 4 & 5\end{array}\right|\) (d) \(\left|\begin{array}{rrr}4 & 3 & 0 \\ 3 & 1 & 2 \\ 5 & -1 & -4\end{array}\right|\) (e) \(\left|\begin{array}{rrr}1 & 3 & 2 \\ 4 & 1 & -2 \\ 2 & 1 & 3\end{array}\right|\) (f) \(\left|\begin{array}{rrr}2 & -1 & 2 \\ 1 & 3 & 2 \\ 5 & 1 & 6\end{array}\right|\) (g) \(\left|\begin{array}{llll}2 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 1 & 6 & 2 & 0 \\ 1 & 1 & -2 & 3\end{array}\right|\) (h) \(\left|\begin{array}{rrrr}2 & 1 & 2 & 1 \\ 3 & 0 & 1 & 1 \\ -1 & 2 & -2 & 1 \\ -3 & 2 & 3 & 1\end{array}\right|\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Geometry

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.