Question

In (a) and (b), suppose the vectors are linearly independent. What can you say about the numbers \(a,....,f\) ? Justify your answers. (Hint: Use a theorem for (b).)

1. \(\left( {\begin{aligned}{*{20}{c}}a\\0\\0\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}b\\c\\d\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}d\\e\\f\end{aligned}} \right)\)
2. \(\left( {\begin{aligned}{*{20}{c}}a\\1\\0\\0\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}b\\c\\1\\0\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}d\\e\\f\\1\end{aligned}} \right)\)

Short answer

1. The pivot position is exactly where \(a,c\) and \(f\) are located because \(A\) is a \(3 \times 3\) matrix. When the three vectors are linearly independent, \(a,c,\) and \(f\) are all non-zero.
2. Any values can be assigned to the numbers \(a,...,f\).

Step-by-step solution

Explain the numbers \(a,...,f\) if the vectors are linearly independent

(a)

The columns of matrix \(A\) arelinearly independentif and only if the equation \(Ax = 0\) has a trivial solution.

When the three vectors are linearly independent, \(a,c,\) and \(f\) must be non-zero. Consider matrix \(A\) whose columns are three linearly independent vectors.

Matrix \(A\) must contain three pivot columns. It means there can be no free variables in the system of equations because the equation \(Ax = 0\) has only a trivial solution. The pivot position is exactly where \(a,c\) and \(f\) are located because \(A\) is a \(3 \times 3\) matrix.

Explain the numbers \(a,...,f\) if the vectors are linearly independent

(b)

Theorem 7 states that an indexed set \(S = \left\{ {{{\mathop{\rm v}\nolimits} _1},...,{v_p}} \right\}\) of two or more vectors islinearly dependentif and only if at least one of the vectors in \(S\) is a linear combination of the others.

Any values can be assigned to the numbers \(a,...,f\). The columns are denoted by \({{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}\), and \({{\mathop{\rm v}\nolimits} _3}\). Note that \({{\mathop{\rm v}\nolimits} _1}\) is not a zero vector. Furthermore, \({{\mathop{\rm v}\nolimits} _2}\) is not a multiple of \({{\mathop{\rm v}\nolimits} _1}\) since the third entry in \({{\mathop{\rm v}\nolimits} _2}\) is non-zero. Also, \({{\mathop{\rm v}\nolimits} _3}\) is not a linear combination of \({{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _2}\)because there is a non-zero value in the fourth entry of \({{\mathop{\rm v}\nolimits} _3}\). Therefore, \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\) are linearly independent according to theorem 7.