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Expert-verified Found in: Page 132 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Express the line ${\mathbf{L}}$ in ${{\mathbf{ℝ}}}^{{\mathbf{3}}}$ spanned by the vector $\mathbf{\left[}\begin{array}{c}\mathbf{1}\\ \mathbf{1}\\ \mathbf{1}\end{array}\mathbf{\right]}$ as the image of a matrix ${\mathbit{A}}$and as the kernel of a matrix ${\mathbit{B}}$.

Matrix $A$

$\mathrm{A}=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$

Matrix $B$

$B=\left[\begin{array}{ccc}1& 1& -2\\ 1& -1& 0\end{array}\right]$

See the step by step solution

## Step 1: Spanned by vector

For a line to be an image of a matrix $A$ it has to be spanned by column vectors of $A$, and since it is spanned by a vector already we'll use it as a column vector

$A=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$

For a line to be a kernel of matrix $B$ the equality must be correct

$B=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=0$

## Step 2: Choose vector that gives 0 in dot product

Since we have no other conditions we can choose any vector that gives 0 in dot product with

$\stackrel{\to }{\mathrm{v}}=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$

We choose vector

$\stackrel{\to }{w}=\left[\begin{array}{c}1\\ 1\\ -2\end{array}\right]$

But we can construct another vector that gives 0 in dot product with $\stackrel{\to }{w}$ so our matrix is not complete, we know that kernel of must form a plane.

So we find another vector $\stackrel{\to }{k}$ that is non collinear with $\stackrel{\to }{w}$ and it's dot product with $\stackrel{\to }{v}$ is 0 let's say

$\stackrel{\to }{k}=\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]$

## Step 3: Construct matrix B

Now we have two non collinear vectors so they span a plane and now we construct matrix $B$

$B=\left[\begin{array}{ccc}1& 1& -2\\ 1& -1& 0\end{array}\right]$

## Step 4: The final answer

Matrix $A$

$A=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$

Matrix $B$

$B=\left[\begin{array}{ccc}1& 1& -2\\ 1& -1& 0\end{array}\right]$ ### Want to see more solutions like these? 