Question

In Exercises 23-26, describe the possible echelon forms of the matrix. Use the notation of Example 1 in Section 1.2

26. \(A\) is a \(4 \times 3\) matrix, \(A = \left[ {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}\end{array}} \right]\) such that \(\left\{ {{{\mathop{\rm a}\nolimits} _1},{a_2}} \right\}\) is linearly independent and \({a_3}\) is not in Span \(\left\{ {{a_1},{a_2}} \right\}\).

Short answer

The echelon form of the \(4 \times 3\) matrix is .

Step-by-step solution

Recall the notation of example 1 used for matrices in the echelon form

In example 1, the following matrices are in the echelon form. The leading entries may have any non-zero value, and the starred entries \(\left( * \right)\) may have any value (including zero).

Use the above notation to determine the echelon forms of the matrix

It is given that \(A = \left[ {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}\end{array}} \right]\) is a \(4 \times 3\) matrix, such that \(\left\{ {{a_1},{a_2}} \right\}\) is linearly independent and \({a_3}\) is not in the span \(\left\{ {{a_1},{a_2}} \right\}\).

Use the leading entries and starred entries to construct the echelon form of the \(4 \times 3\)matrix.

Determine whether the column of the matrix is linearly independent

Theorem 7 states that an indexed set \(S = \left\{ {{v_1},...,{v_p}} \right\}\) of two or more vectors is linearly dependent if and only if at least one of the vectors in \(S\) is a linear combination of the others. If \(S\) is linearly dependent and \[{{\mathop{\rm v}\nolimits} _1} \ne 0\], some \({v_j}\) is a linear combination of the preceding vectors \[{{\mathop{\rm v}\nolimits} _1},...,{v_{j - 1}}\].

According to theorem 7, the column of matrix \(A\) is linearly independent since the first column is not zero; the second column is not a multiple of the first column, and the third column is not a linear combination of the first two columns.

Thus, the echelon form of the \(4 \times 3\) matrix is .