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Suppose that all the entries in $A$ are integers and det $A=1 .$ Explain why all the entries in $A^{-1}$ are integers.$$(-2,0),(0,3),(1,3),(-1,0)$$

3

Algebra

Chapter 3

Determinants

Section 3

Cramer’s Rule, Volume, and Linear Transformations

Introduction to Matrices

Campbell University

Oregon State University

McMaster University

Idaho State University

Lectures

01:32

In mathematics, the absolu…

01:11

02:08

Suppose that all the entri…

02:29

02:21

01:37

01:45

CHALLENGE If $a_{1}$ and $…

00:27

Explain why the whole numb…

02:17

Without performing the act…

16:29

Describe the elements of t…

01:57

If $A=\left[\begin{array}{…

01:49

Show that if $a | b$ and $…

were given the Vergis ease of a parallelogram, and we were asked to find the area of this parallelogram. The Vergis is our negative. 20 03 13 and negative 10 So, first of all, we noticed that none of the vergis ease air the origin, so I wanted to shift parallelogram to the origin. We can do this by subtracting. One of the vergis is from each of the other. Vergis is So for example, we could do this by subtracting the Vertex negative 10 So we end up with a parallelogram with new vergis is negative two minus negative. One is negative. One. It's going to be negative. 10 and then 03 minus negative. 10 says 13 and 13 minus negative one. Well, that's me too. Three. And of course, the origin. 00 So visualizing the new parallelogram, we see that we have a point. Vertex it negative 10 as well as the vertex at 13 and at to three. So parallelogram as the same area as the original and moreover, is determined by the columns of the Matrix A in this as the column vectors which are the Vergis, is negative 10 and 23 So it follows that the area is equal to the absolute value of the determinant of this matrix, which is the absolute value of negative three, which is simply three.

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