1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# 2x − 3z + w = 1 x − y + 2w = 1 − 3y + z + w = 1 x + y + z = 1

Open in App
Solution

## $D=\left|\begin{array}{cccc}2& 0& -3& 1\\ 1& -1& 0& 2\\ 0& -3& 1& 1\\ 1& 1& 1& 0\end{array}\right|\phantom{\rule{0ex}{0ex}}2\left|\begin{array}{ccc}-1& 0& 2\\ -3& 1& 1\\ 1& 1& 0\end{array}\right|-0-3\left|\begin{array}{ccc}1& -1& 2\\ 0& -3& 1\\ 1& 1& 0\end{array}\right|-1\left|\begin{array}{ccc}1& -1& 0\\ 0& -3& 1\\ 1& 1& 1\end{array}\right|\phantom{\rule{0ex}{0ex}}=2\left[-1\left(0-1\right)-0\left(0-1\right)+2\left(-3-1\right)\right]-3\left[1\left(0-1\right)+1\left(0-1\right)+2\left(0+3\right)\right]-1\left[1\left(-3-1\right)+1\left(0-1\right)+0\left(0+3\right)\right]\phantom{\rule{0ex}{0ex}}=-21\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{D}_{1}=\left|\begin{array}{cccc}1& 0& -3& 1\\ 1& -1& 0& 2\\ 1& -3& 1& 1\\ 1& 1& 1& 0\end{array}\right|\phantom{\rule{0ex}{0ex}}1\left|\begin{array}{ccc}-1& 0& 2\\ -3& 1& 1\\ 1& 1& 0\end{array}\right|-0-3\left|\begin{array}{ccc}1& -1& 2\\ 1& -3& 1\\ 1& 1& 0\end{array}\right|-1\left|\begin{array}{ccc}1& -1& 0\\ 1& -3& 1\\ 1& 1& 1\end{array}\right|\phantom{\rule{0ex}{0ex}}=1\left[-1\left(0-1\right)-0\left(0-1\right)+2\left(-3-1\right)\right]-3\left[1\left(0-1\right)+1\left(0-1\right)+2\left(1+3\right)\right]-1\left[1\left(-3-1\right)+1\left(0-1\right)+2\left(1+3\right)\right]\phantom{\rule{0ex}{0ex}}=-21\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{D}_{2}=\left|\begin{array}{cccc}2& 1& -3& 1\\ 1& 1& 0& 2\\ 0& 1& 1& 1\\ 1& 1& 1& 0\end{array}\right|\phantom{\rule{0ex}{0ex}}=2\left|\begin{array}{ccc}1& 0& 2\\ 1& 1& 1\\ 1& 1& 0\end{array}\right|-1\left|\begin{array}{ccc}1& 0& 2\\ 0& 1& 1\\ 1& 1& 0\end{array}\right|+\left(-3\right)\left|\begin{array}{ccc}1& 1& 2\\ 0& 1& 1\\ 1& 1& 0\end{array}\right|-1\left|\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 1& 1& 1\end{array}\right|\phantom{\rule{0ex}{0ex}}2\left[1\left(0-1\right)+2\left(1-1\right)\right]-1\left[1\left(0-1\right)+2\left(0-1\right)\right]-3\left[1\left(0-1\right)-1\left(0-1\right)+2\left(0-1\right)\right]-1\left[1\left(1-1\right)-1\left(0-1\right)\right]\phantom{\rule{0ex}{0ex}}=6\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{D}_{3}=\left|\begin{array}{cccc}2& 0& 1& 1\\ 1& -1& 1& 2\\ 0& -3& 1& 1\\ 1& 1& 1& 0\end{array}\right|\phantom{\rule{0ex}{0ex}}=2\left|\begin{array}{ccc}-1& 1& 2\\ -3& 1& 1\\ 1& 1& 0\end{array}\right|-0+1\left|\begin{array}{ccc}1& -1& 2\\ 0& -3& 1\\ 1& 1& 0\end{array}\right|-1\left|\begin{array}{ccc}1& -1& 1\\ 0& -3& 1\\ 1& 1& 1\end{array}\right|\phantom{\rule{0ex}{0ex}}=2\left[-1\left(0-1\right)-1\left(0-1\right)+2\left(-3-1\right)\right]+1\left[1\left(0-1\right)+1\left(0-1\right)+2\left(0+3\right)\right]-1\left[1\left(-3-1\right)+1\left(0-1\right)+1\left(0+3\right)\right]\phantom{\rule{0ex}{0ex}}=-6\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{D}_{4}=\left|\begin{array}{cccc}2& 0& -3& 1\\ 1& -1& 0& 1\\ 0& -3& 1& 1\\ 1& 1& 1& 1\end{array}\right|\phantom{\rule{0ex}{0ex}}=2\left|\begin{array}{ccc}-1& 0& 1\\ -3& 1& 1\\ 1& 1& 1\end{array}\right|-0-3\left|\begin{array}{ccc}1& -1& 1\\ 0& -3& 1\\ 1& 1& 1\end{array}\right|-1\left|\begin{array}{ccc}1& -1& 0\\ 0& -3& 1\\ 1& 1& 1\end{array}\right|\phantom{\rule{0ex}{0ex}}=2\left[-1\left(1-1\right)+1\left(-3-1\right)\right]-3\left[1\left(-3-1\right)+1\left(0-1\right)+1\left(0+3\right)\right]-1\left[1\left(-3-1\right)+1\left(0-1\right)\right]\phantom{\rule{0ex}{0ex}}=3\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{by}\mathrm{Cramer}\text{'}\mathrm{s}\mathrm{rule},\mathrm{we}\mathrm{obtain}\phantom{\rule{0ex}{0ex}}x=\frac{{D}_{1}}{D}=\frac{21}{21}=1\phantom{\rule{0ex}{0ex}}y=\frac{{D}_{2}}{D}=\frac{6}{-21}=-\frac{2}{7}\phantom{\rule{0ex}{0ex}}z=\frac{{D}_{3}}{D}=\frac{-6}{-21}=\frac{2}{7}\phantom{\rule{0ex}{0ex}}w=\frac{{D}_{4}}{D}=\frac{3}{-21}=-\frac{1}{7}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},x=1,y=-\frac{2}{7},z=\frac{2}{7},w=-\frac{1}{7}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Integration by Partial Fractions
MATHEMATICS
Watch in App
Join BYJU'S Learning Program