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Q22E

Expert-verifiedFound in: Page 1

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

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

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.

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:

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

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.

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

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