Chapter 7.4, Problem 32E

To determine

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

Answer to Problem 32E

  $\left[\begin{array}{cccc}1& 2& 4& \frac{3}{2}\\ 0& -3& -7& \frac{1}{2}\\ 0& 2& -4& 6\end{array}\right]$ .

Explanation of Solution

Given information:

Given matrices are

  $\begin{array}{l}\left[\begin{array}{cccc}2& 4& 8& 3\\ 1& -1& -3& 2\\ 2& 6& 4& 9\end{array}\right]\\ \left[\begin{array}{cccc}1& -1& -3& 2\\ 2& & & \\ 2& 6& 4& 9\end{array}\right]\\ \left[\begin{array}{cccc}1& -1& -3& 2\\ 0& 6& & \\ 0& 8& & \end{array}\right]\end{array}$

Calculation:

Consider the following matrices,

  $\begin{array}{l}\left[\begin{array}{cccc}2& 4& 8& 3\\ 1& -1& -3& 2\\ 2& 6& 4& 9\end{array}\right]\mathrm{...................}(1)\\ \left[\begin{array}{cccc}1& -1& -3& 2\\ 2& & & \\ 2& 6& 4& 9\end{array}\right]\mathrm{....................}(2)\\ \left[\begin{array}{cccc}1& -1& -3& 2\\ 0& 6& & \\ 0& 8& & \end{array}\right]\mathrm{....................}(3)\end{array}$

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

Multiply first row of matrix (1) by $\frac{1}{2}$ that is $\frac{1}{2}{R}_1$

  $\left[\begin{array}{cccc}1& 2& 4& \frac{3}{2}\\ 1& -1& -3& 2\\ 2& 6& 4& 9\end{array}\right]$

Therefore second matrix after filling the missing elements will be,

  $\left[\begin{array}{cccc}1& 2& 4& \frac{3}{2}\\ 1& -1& -3& 2\\ 2& 6& 4& 9\end{array}\right]$ ……………..(4)

The missing terms of matrix (2) are shown in the box.

To converts matrix (1) to matrix (30 use following row operation

Multiply second row of matrix (4) by -1 that is $-R_1$

  $\left[\begin{array}{cccc}1& 2& 4& \frac{3}{2}\\ -1& 1& 3& -2\\ 2& 6& 4& 9\end{array}\right]$ …………..(5)

Add first row to second row for matrix (5) that is $R_1+R_2$

  $\left[\begin{array}{cccc}1& 2& 4& \frac{3}{2}\\ 0& 3& 7& -\frac{1}{2}\\ 2& 6& 4& 9\end{array}\right]$ ……………(6)

Multiply second row of matrix (6) by -1 that is $-R_1$

  $\left[\begin{array}{cccc}1& 2& 4& \frac{3}{2}\\ 0& -3& -7& \frac{1}{2}\\ 2& 6& 4& 9\end{array}\right]$ …………..(7)

Add -2 times of first row to third row for matrix (7) that is $-2R_1+R_3$

  $\left[\begin{array}{cccc}1& 2& 4& \frac{3}{2}\\ 0& -3& -7& \frac{1}{2}\\ 0& 2& -4& 6\end{array}\right]$ …………….(8)

The missing terms of matrix (3) are shown in the box.