In Exercises 29 – 32, (a) does the equation \(A{\mathop{\rm x}\nolimits} = {\mathop{\rm b}\nolimits} \) have a nontrivial solution and (b) does the equation \(Ax = b\) have at least one solution for every possible \({\mathop{\rm b}\nolimits} \)?

30. \(A\) is a \(3 \times 3\) matrix with three pivot positions.

The variables corresponding to pivot columns in the matrix are called basic variables.The other variable is called a free variable.

\(3 \times 3\) matrix has two basic variables and one free variable.

(a)

The homogeneous equation \(Ax = 0\) will have anontrivial solutionif and only if the equation has at least one free variable.The system will have a nontrivial solution if a column in the coefficient matrix does not construct a pivot column.

\(A\) is a \(3 \times 3\) matrix with two pivot positions since the equation \(Ax = 0\) contains two basic variables and one free variable. Therefore, the equation \(Ax = 0\) has a nontrivial solution.

(b)

Theorem 4states that \(A\) is a \({\mathop{\rm m}\nolimits} \times n\) matrix. Hence, the following statement is equivalent. For a particular \(A\), either they are all true statements or false statements.

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 .
4. \(A\)has a pivot position in every row.

Since there are only two pivot positions in the matrix \(A\), so \(A\) cannot have a pivot position in every row. According to theorem 4, the equation \(Ax = b\) cannot have a solution for every value of \({\mathop{\rm b}\nolimits} \).

Thus, the equation \(Ax = b\) cannot have a solution for every value of \(b\).