Question

the sum of the length, width, and height of a rectangular box is 16 cm; the width is twice the height; and twice the length exceeds the sum of the width and height by 5. find the length, width, and height of the box using matrices. a. length = 10 cm, width = 4 cm, height = 2 cm b. length = 7 cm, width = 6 cm, height = 3 cm c. length = 7 cm, width = 3 cm, height = 6 cm d. length = 10 cm, width = 2 cm, height = 4 cm

Answer 1

l + w + h = 16 . . . . . (1)
w = 2h
w - 2h = 0 . . . . . (2)
2l = w + h + 5
2l - w - h = 5 . . . . . (3)

$\left[\begin{array}{ccc}1&1&1\\0&1&-2\\2&-1&-1\end{array}\right] \left[\begin{array}{c}l&w&h\end{array}\right] = \left[\begin{array}{c}16&0&5\end{array}\right] \ \ -2R_1+R_3 \rightarrow R_3 \\ \left[\begin{array}{ccc}1&1&1\\0&1&-2\\0&-3&-3\end{array}\right] \left[\begin{array}{c}l&w&h\end{array}\right]=\left[\begin{array}{c}16&0&-27\end{array}\right] \ \ \frac{1}{3} R_3+R_1 \rightarrow R_1$
$\left[\begin{array}{ccc}1&0&0\\0&1&-2\\0&-3&-3\end{array}\right] \left[\begin{array}{c}l&w&h\end{array}\right]=\left[\begin{array}{c}7&0&-27\end{array}\right] \ \ 3 R_2+R_3 \rightarrow R_3 \\ \left[\begin{array}{ccc}1&0&0\\0&1&-2\\0&0&-9\end{array}\right] \left[\begin{array}{c}l&w&h\end{array}\right]=\left[\begin{array}{c}7&0&-27\end{array}\right] \ \ - \frac{1}{9} R_3 \rightarrow R_3$
$\left[\begin{array}{ccc}1&0&0\\0&1&-2\\0&0&1\end{array}\right] \left[\begin{array}{c}l&w&h\end{array}\right]=\left[\begin{array}{c}7&0&3\end{array}\right] \ \ 2R_3+R_2 \rightarrow R_2 \\ \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{c}l&w&h\end{array}\right]=\left[\begin{array}{c}7&6&3\end{array}\right]$

Therefore, length = 7 cm; width = 6 cm; height = 3 cm