Problem 102 - Algebra and Trigonometry Real Mathematics, Real People

Question

Determine whether the matrix below is in row-echelon form, reduced row-echelon form, or neither when it satisfies the given conditions. $$\left[\begin{array}{ll} 1 & b \\ c & 1 \end{array}\right]$$ (a) \(b=0, c=0\) (b) \(b \neq 0, c=0\) (c) \(b=0, c \neq 0\) (d) \(b \neq 0, c \neq 0\)

Step-by-Step Solution

Answer

(a) The matrix is in reduced row-echelon form. (b) The matrix is in reduced row-echelon form. (c) The matrix is neither in row-echelon form nor in reduced row-echelon form. (d) The matrix is neither in row-echelon form nor in reduced row-echelon form.

Step 1: Evaluate Case (a)

In this case, \(b=0\) and \(c=0\). Substituting this into the matrix yields: \[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \] The leading coefficient in each row is 1, and it is the only non-zero entry in its column. Thus, the matrix is in the reduced row-echelon form.

Step 2: Evaluate Case (b)

In this scenario, \(b\) is not equal to 0 and \(c = 0\). Substituting these values into the matrix gives: \[ \begin{array}{cc} 1 & b \\ 0 & 1 \end{array} \] Regardless of the value of \(b\), the first entry in the second row will always be zero, hence this matrix is in reduced row-echelon form.

Step 3: Evaluate Case (c)

Here, \(b = 0\) and \(c\) is not equal to 0. Inserting these values into the matrix results in: \[ \begin{array}{cc} 1 & 0 \\ c & 1 \end{array} \] Considering any value of \(c\), besides 0, it is seen that the first row has a nonzero entry that isn't below a leading 1, which contradicts the fact that for a matrix to be in row-echelon form or reduced row-echelon form, the leading coefficient in a non-zero row must always be to the right of the leading coefficient of the row above it. Hence, the matrix doesn't fulfil either of the form criteria.

Step 4: Evaluate Case (d)

In the final scenario, \(b\) and \(c\) are both non-zero. Substituting this into the matrix yields \[ \begin{array}{cc} 1 & b \\ c & 1 \end{array} \] For any non-zero value of \(c\), the first row has a non-zero element that is not below a leading 1, similiar to case (c). Hence, the matrix is neither of the two forms in this case.