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Exercises 25 and 26 prove Theorem 4 for $A=\left[\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right]$ Show that if $a d-b c=0,$ then the equation $A \mathbf{x}=0$ has more than one solution. Why does this imply that $A$ is not invertible? $[\text { Hint: First, consider } a=b=0 . \text { Then, if } a \text { and }$ $b$ are not both zero, consider the vector $\mathbf{x}=\left[\begin{array}{r}{-b} \\ {a}\end{array}\right] . ]$

$\left[ \begin{array} { l l l } { a } & { b } & { 0 } \\ { 0 } & { 0 } & { 0 } \end{array} \right] \not ^ { } \mathcal { H } I _ { n }$

Algebra

Chapter 2

Matrix Algebra

Section 2

The Inverse of a Matrix

Introduction to Matrices

Missouri State University

Oregon State University

University of Michigan - Ann Arbor

Lectures

01:32

In mathematics, the absolu…

01:11

03:10

Exercises 25 and 26 prove …

02:48

Show that if $A B$ is inve…

01:26

In Exercises 31–36, mentio…

01:43

Use determinants to prove …

01:37

00:50

Suppose that $A, B,$ and $…

02:03

Let $a, b,$ and $c$ be rea…

01:44

Consider the 2 by 2 square…

03:04

Use part (c) of the Invert…

01:52

Suppose that $B$ is any ma…

the mad tricks, eh? Is given by a B C. The we know the formula for a embers is one over a d minus. BC and then we moved, applied by the metric something from a by swapping elements on the mend. I argo so d A and negating the elements on the secondary diagonal so minus B and minus e. So we want to verify that this formula does indeed give the inverse matrix away, assuming that a D minus BC is non zero so that the ratio is, ah, make Maxence and so to verify that we just want to show that a embers time's A should give the identity matrix as a result. So we just compute this product and verify that we indeed get the identity matrix. So let's explicit right? A inverse, eh? So this is one over determinant away. So I d minus BC. Then the money's being on a C. Hey, that multiplies a B c d. Now we compute this product girl by colon. So one over a demon's BC and then the product in blue. It's going to be D a minors B C baby minus maybe minus C A plus a seat and minus see me plus 80 now teases one over a D minus BC and in blue, we have a D minus. BC top left Stop, right, We just have zero bottom left. We also have just zero and bottom, right? We have a demon is BC. And since a D minus BC is assumed to be known zero. We can divide by the expression. And so the final result is just one zero zero one, which is the identity matrix, as we wanted to see.

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In mathematics, the absolute value or modulus |x| of a real number x is its …

Exercises 25 and 26 prove Theorem 4 for $A=\left[\begin{array}{ll}{a} & …

Show that if $A B$ is invertible, so is $A$ . You cannot use Theorem 6$(\tex…

In Exercises 31–36, mention an appropriate theorem in your explanation.S…

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In a certain region, about 6% of a city’s population moves to the surroundin…

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In Exercises $25-28$ , determine if the specified linear transformation is (…

02:16

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03:18

[M] Read the documentation for your matrix program, and write the commands t…

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Let $T : \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ be the transformation th…

05:06

Unless otherwise specified, assume that all matrices in these exercises are …

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A useful way to test new ideas in matrix algebra, or to make conjectures, is…

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Use the vectors $\mathbf{u}=\left(u_{1}, \ldots, u_{n}\right), \mathbf{v}=\l…

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