Question: Exercises 17-20 refer to the matrices \(A\) and \(B\) below. Make appropriate calculations that justify your answers and mention an appropriate theorem.

\(A = \left( {\begin{array}{*{20}{c}}1&3&0&3\\{ - 1}&{ - 1}&{ - 1}&1\\0&{ - 4}&2&{ - 8}\\2&0&3&{ - 1}\end{array}} \right)\)\(B = \left( {\begin{array}{*{20}{c}}1&3&{ - 2}&2\\0&1&1&{ - 5}\\1&2&{ - 3}&7\\{ - 2}&{ - 8}&2&{ - 1}\end{array}} \right)\)

20. Can every vector in \({\mathbb{R}^4}\) be written as a linear combination of the columns of the matrix \(B\) above? Do the columns of \(B\) span \({\mathbb{R}^3}\)?

Perform an elementary row operation to produce the first augmented matrix.

Write the sum of \( - 1\) time row one and row three in row three and the sum of \(2\)times row one and row four in row four.

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

Perform an elementary row operation to produce the second augmented matrix.

Write the sum of \(1\) time row two and row three in row three and the sum of \(2\)times row two and row four in row four.

\(\left( {\begin{array}{*{20}{c}}1&3&{ - 2}&2\\0&1&1&{ - 5}\\0&0&0&0\\0&0&0&{ - 7}\end{array}} \right)\)

Perform an elementary row operationto produce the third augmented matrix.

Interchange rows three and four.

\(\left( {\begin{array}{*{20}{c}}1&3&{ - 2}&2\\0&1&1&{ - 5}\\0&0&0&{ - 7}\\0&0&0&0\end{array}} \right)\)

For a particular value of matrix \(A\), all the statementsare either true or false.

1. For each \({\mathop{\rm b}\nolimits} \) in \({\mathbb{R}^m}\), the equation \(Ax = b\) has a solution.
2. Each \({\mathop{\rm b}\nolimits} \) in \({\mathbb{R}^m}\) is a linear combination of the columns of A.
3. The columns of \(A\) span \(\;{\mathbb{R}^m}\).
4. \(A\) has a pivot position in every row.

\(\left( {\begin{array}{*{20}{c}}1&3&{ - 2}&2\\0&1&1&{ - 5}\\0&0&0&{ - 7}\\0&0&0&0\end{array}} \right)\)

Since not every row of \(B\) contains a pivot position, it means statement (d) is false. As a result, all four statements in theorem 4 are false. Consequently, not all vectors in \({\mathbb{R}^4}\) can be represented as a linear combination of the columns of \(B\). The columns of \(B\) cannot span \({\mathbb{R}^3}\) since each column is in \({\mathbb{R}^4}\) and not in \({\mathbb{R}^3}\) .