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### 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

# Construct a $$3 \times 3$$ matrix$$A$$, with nonzero entries, and a vector $$b$$ in $${\mathbb{R}^3}$$ such that $$b$$ is not in the set spanned by the columns of$$A$$.

$$A = \left[ {\begin{array}{*{20}{c}}1&1&1\\{ - 1}&1&1\\{ - 1}&{ - 1}&{ - 1}\end{array}} \right]$$ and $$b = \left[ {\begin{array}{*{20}{c}}1\\2\\3\end{array}} \right]$$

See the step by step solution

## Step 1: Idea to construct the matrices

The idea is to consider a matrix $$A$$ with nonzero entries such that the system is inconsistent.

Take $$A = \left[ {\begin{array}{*{20}{c}}1&1&1\\{ - 1}&1&1\\{ - 1}&{ - 1}&{ - 1}\end{array}} \right]$$ and $$b = \left[ {\begin{array}{*{20}{c}}1\\2\\3\end{array}} \right]$$.

## Step 2: Determine the objective to be shown

To prove $$b$$ is not in the set spanned by the columns of$$A$$, show $$x\left[ {\begin{array}{*{20}{c}}1\\{ - 1}\\{ - 1}\end{array}} \right] + y\left[ {\begin{array}{*{20}{c}}1\\1\\{ - 1}\end{array}} \right] + z\left[ {\begin{array}{*{20}{c}}1\\1\\{ - 1}\end{array}} \right] = b$$ has no solution.

## Step 3: Convert the linear combination to the system

This linear combination can be written as:

\begin{aligned}{c}\left[ {\begin{array}{*{20}{c}}x\\{ - x}\\{ - x}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}y\\y\\{ - y}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}z\\z\\{ - z}\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}1\\2\\3\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{x + y + z}\\{ - x + y + z}\\{ - x - y - z}\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}1\\2\\3\end{array}} \right]\end{aligned}

This implies that the system is:

$$\begin{array}{c}x + y + z = 1\\ - x + y + z = 2\\ - x - y - z = 3\end{array}$$

## Step 4: Determine the inconsistency of the system

The row echelon form of the augmented matrix of this system is given as follows:

Apply row operations $${R_2} \to {R_2} + {R_1}$$ and $${R_3} \to {R_3} + {R_1}$$.

$$\left[ {\begin{array}{*{20}{c}}1&1&1&1\\{ - 1}&1&1&2\\{ - 1}&{ - 1}&{ - 1}&3\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&1&1&1\\0&2&2&3\\0&0&0&4\end{array}} \right]\,$$

The equivalent system of equations using the echelon form is:

\begin{aligned}{c}x + y + z &= 1\\2y + 2z &= 3\\0 &= 4\end{aligned}

The last equation is not possible.

Therefore, the system is inconsistent.

## Step 5: Conclusion

Hence, the system has no solution. This ensures that $$b$$ is not in the set spanned by the columns of$$A$$.