Construct a \(3 \times 3\) matrix, not in echelon form, whose columns do not span \({\mathbb{R}^3}\). Show that the matrix you construct has the desired property.

Short Answer

Expert verified

The required matrix is \(A = \left[ {\begin{array}{*{20}{c}}1&0&1\\0&1&0\\1&1&1\end{array}} \right]\).

Step by step solution

01

Writing the conditions for the echelon form

The matrix is in echelon form if it satisfies the following conditions:

  • Non-zero rows should be positioned above zero rows.
  • Each row's leading entry should bein the column to the right of the row above its leading item.
  • In each column, all items below the leading entry must be zero.
02

Constructing a matrix in the non-echelon form

Assume that the matrix is \(A = \left[ {\begin{array}{*{20}{c}}1&0&1\\0&1&0\\1&1&1\end{array}} \right]\).

Here, the non-zero row is not positioned above zero rows.

03

Constructing an arbitrary vector in \({\mathbb{R}^3}\)

Consider \(b = \left[ {\begin{array}{*{20}{c}}0\\0\\1\end{array}} \right]\) is the required vector.

In \(\left( {A|b} \right)\) form, it can be written as shown below:

\(\left[ {\begin{array}{*{20}{c}}1&0&1&0\\0&1&0&0\\1&1&1&1\end{array}} \right]\)

04

Checking if the column matrix spans \({\mathbb{R}^3}\)

Apply the row operation in the matrix \(\left[ {\begin{array}{*{20}{c}}1&0&1&0\\0&1&0&0\\1&1&1&1\end{array}} \right]\).

Use rows one and two to eliminate \({x_1}\), \({x_2}\), and \({x_3}\) from the third equation.

\(\left[ {\begin{array}{*{20}{c}}1&0&1&0\\0&1&0&0\\0&0&0&1\end{array}} \right]\)

It shows that column \(b = \left[ {\begin{array}{*{20}{c}}0\\0\\1\end{array}} \right]\) of the assumed matrix does not span \({\mathbb{R}^3}\).

Therefore, the required matrix is \(A = \left[ {\begin{array}{*{20}{c}}1&0&1\\0&1&0\\1&1&1\end{array}} \right]\).

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Most popular questions from this chapter

In Exercise 19 and 20, choose \(h\) and \(k\) such that the system has

a. no solution

b. unique solution

c. many solutions.

Give separate answers for each part.

19. \(\begin{array}{l}{x_1} + h{x_2} = 2\\4{x_1} + 8{x_2} = k\end{array}\)

Let \({{\mathop{\rm a}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}1\\4\\{ - 2}\end{array}} \right],{{\mathop{\rm a}\nolimits} _2} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 3}\\7\end{array}} \right],\) and \({\rm{b = }}\left[ {\begin{array}{*{20}{c}}4\\1\\h\end{array}} \right]\). For what values(s) of \(h\) is \({\mathop{\rm b}\nolimits} \) in the plane spanned by \({{\mathop{\rm a}\nolimits} _1}\) and \({{\mathop{\rm a}\nolimits} _2}\)?

Find the elementary row operation that transforms the first matrix into the second, and then find the reverse row operation that transforms the second matrix into the first.

29. \(\left[ {\begin{array}{*{20}{c}}0&{ - 2}&5\\1&4&{ - 7}\\3&{ - 1}&6\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}1&4&{ - 7}\\0&{ - 2}&5\\3&{ - 1}&6\end{array}} \right]\)

Find the general solutions of the systems whose augmented matrices are given as

14. \(\left[ {\begin{array}{*{20}{c}}1&2&{ - 5}&{ - 6}&0&{ - 5}\\0&1&{ - 6}&{ - 3}&0&2\\0&0&0&0&1&0\\0&0&0&0&0&0\end{array}} \right]\).

Let \(A = \left[ {\begin{array}{*{20}{c}}1&0&{ - 4}\\0&3&{ - 2}\\{ - 2}&6&3\end{array}} \right]\) and \(b = \left[ {\begin{array}{*{20}{c}}4\\1\\{ - 4}\end{array}} \right]\). Denote the columns of \(A\) by \({{\mathop{\rm a}\nolimits} _1},{a_2},{a_3}\) and let \(W = {\mathop{\rm Span}\nolimits} \left\{ {{a_1},{a_2},{a_3}} \right\}\).

  1. Is \(b\) in \(\left\{ {{a_1},{a_2},{a_3}} \right\}\)? How many vectors are in \(\left\{ {{a_1},{a_2},{a_3}} \right\}\)?
  2. Is \(b\) in \(W\)? How many vectors are in W.
  3. Show that \({a_1}\) is in W.[Hint: Row operations are unnecessary.]
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