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Use the concept of volume to explain why the determinant of a $3 \times 3$ matrix $A$ is zero if and only if $A$ is not invertible. Do not appeal to Theorem 4 in Section $3.2 .[\text { Hint: Think about }$the columns of $A . ]$

$\operatorname{det} A=0$

Algebra

Chapter 3

Determinants

Section 3

Cramer’s Rule, Volume, and Linear Transformations

Introduction to Matrices

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Lectures

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In mathematics, the absolu…

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Prove that a 3 by 3 determ…

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were asked to use the concept of volume to explain why the determinant of the three by three matrix A is zero if and only if the Matrix A is not in vertebral. We were asked to avoid appealing to fear um four in section 3.2. So we know by the invariable matrix the're, um that a three by three matrix A is not in vertebral. Yes, and only it It's columns are linearly dependent. And this happens if and only if one of the columns is a linear combination of the others. In other words, if one of the vectors is in the plane spanned by the other two vectors now this is equivalent to the condition. The parallel pipe head, determined by the three vectors, has zero volume or, in other words, that the determinant of the matrix a representing this parallel pipe head is

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