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Expert-verified Found in: Page 1 ### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384 # Use Theorem 7 in section 1.7 to explain why the columns of the matrix A are linearly independent.A = \left( {\begin{aligned}{*{20}{c}}1&0&0&0\\2&5&0&0\\3&6&8&0\\4&7&9&{10}\end{aligned}} \right)

The columns of matrix $$A$$ are linearly independent according to theorem 7.

See the step by step solution

## Step 1: Denote the columns of matrix A

The columns of matrix $$A$$ are denoted from right to left by vectors $${{\mathop{\rm v}\nolimits} _1},...,{{\mathop{\rm v}\nolimits} _4}$$.

## Step 2: Check if the columns of matrix A are linearly independent

Theorem 7 states that an indexed set $$S = \left\{ {{{\mathop{\rm v}\nolimits} _1},...,{v_p}} \right\}$$ of two or more vectors is linearly dependent if and only if at least one of the vectors in $$S$$ is a linear combination of the others. If $$S$$ is linearly dependent and $${{\mathop{\rm v}\nolimits} _1} \ne 0$$, then some $${{\mathop{\rm v}\nolimits} _j}$$ is a linear combination of the preceding vectors $${{\mathop{\rm v}\nolimits} _1},...,{v_{j - 1}}$$.

Vector $${{\mathop{\rm v}\nolimits} _1}$$ is non-zero; $${{\mathop{\rm v}\nolimits} _2}$$ is not a multiple of $${{\mathop{\rm v}\nolimits} _1}$$ (since the third entry of $${{\mathop{\rm v}\nolimits} _2}$$is non-zero), and $${{\mathop{\rm v}\nolimits} _3}$$ is not a linear combination of $${{\mathop{\rm v}\nolimits} _1}$$ and $${{\mathop{\rm v}\nolimits} _2}$$ (since the second entry of $${{\mathop{\rm v}\nolimits} _3}$$ is non-zero). Furthermore, $${{\mathop{\rm v}\nolimits} _4}$$ cannot be a linear combination of $${{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},$$ and $${{\mathop{\rm v}\nolimits} _3}$$, based on the first entry in the vector. Thus, the columns are linearly independent according to theorem 7. ### Want to see more solutions like these? 