Question

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\(A\).

Short answer

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

Step-by-step solution

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 \(b = \left[ {\begin{array}{*{20}{c}}1\\2\\3\end{array}} \right]\).

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.

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:

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

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.

Conclusion

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