Chapter 7.4, Problem 28E

To determine

To fill: the blank using elementary row operations to form a row equivalent matrix.

Answer to Problem 28E

Hence missing element is 2.

Explanation of Solution

Given information:

Given matrix

  $\left[\begin{array}{ccc}3& 6& 8\\ 4& -3& 6\end{array}\right]$

And

  $\left[\begin{array}{ccc}1& & \frac{8}{3}\\ 4& -3& 6\end{array}\right]$

Calculation:

Let us consider the following row equivalent matrices

Given matrix

  $\left[\begin{array}{ccc}3& 6& 8\\ 4& -3& 6\end{array}\right]$

And

  $\left[\begin{array}{ccc}1& & \frac{8}{3}\\ 4& -3& 6\end{array}\right]$

Observe both the matrices. Here second row of these matrices are same and given elements of first row in the matrix (2) are one third of that of matrix (1).

Therefore to convert matrix (1) into matrix (2) use following row operation,

Multiply first row of matrix (1) by $\frac{1}{3}$ ,

That is,

  $\left[\begin{array}{ccc}1& 2& \frac{8}{3}\\ 4& -3& 6\end{array}\right]$

Hence missing element is 2.