solve the problem

Question:

$x+y+z+w=2$

$x-2 y+2 z+2 w=-6$

$2 x+y-2 z+2 w=-5$

$3 x-y+3 z-3 w=-3$

Solution:

$D=\left|\begin{array}{cccc}1 & 1 & 1 & 1 \\ 1 & -2 & 2 & 2 \\ 2 & 1 & -2 & 2 \\ 3 & -1 & 3 & -3\end{array}\right|$

$1\left|\begin{array}{ccc}-2 & 2 & 2 \\ 1 & -2 & 2 \\ -1 & 3 & -3\end{array}\right|-1\left|\begin{array}{ccc}1 & 2 & 2 \\ 2 & -2 & 2 \\ 3 & 3 & -3\end{array}\right|+1\left|\begin{array}{ccc}1 & -2 & 2 \\ 2 & 1 & 2 \\ 3 & -1 & -3\end{array}\right|-1\left|\begin{array}{ccc}1 & -2 & 2 \\ 2 & 1 & -2 \\ 3 & -1 & 3\end{array}\right|$

$=1[-2(6-6)-2(-3+2)+2(3-2)]-1[1(6-6)-2(-6-6)+2(6+6)]+1[1(-3+2)+2(-6-6)+2(-2-3)]-$

$1[1(3-2)+2(6+6)+2(-2-3)]$

$=4-48-35-15$

$=-94$

$D_{1}=\left|\begin{array}{cccc}2 & 1 & 1 & 1 \\ -6 & -2 & 2 & 2 \\ -5 & 1 & -2 & 2 \\ -3 & -1 & 3 & -3\end{array}\right|$

$2\left|\begin{array}{ccc}-2 & 2 & 2 \\ 1 & -2 & 2 \\ -1 & 3 & -3\end{array}\right|-1\left|\begin{array}{ccc}-6 & 2 & 2 \\ -5 & -2 & 2 \\ -3 & 3 & -3\end{array}\right|+1\left|\begin{array}{ccc}-6 & -2 & 2 \\ -5 & 1 & 2 \\ -3 & -1 & -3\end{array}\right|-1\left|\begin{array}{ccc}-6 & -2 & 2 \\ -5 & 1 & -2 \\ -3 & -1 & 3\end{array}\right|$

$=2[-2(6-6)-2(-3+2)+2(3-2)]-1[-6(6-6)-2(15+6)+2(-15-6)]+1[-6(-3+2)+2(15+6)+2(5+3)]$

$-1[-6(3-2)+2(-15-6)+2(5+3)]$

$=188$

$D_{2}=\left|\begin{array}{cccc}1 & 2 & 1 & 1 \\ 1 & -6 & 2 & 2 \\ 2 & -5 & -2 & 2 \\ 3 & -3 & 3 & -3\end{array}\right|$

$1\left|\begin{array}{ccc}-6 & 2 & 2 \\ -5 & -2 & 2 \\ -3 & 3 & -3\end{array}\right|-2\left|\begin{array}{ccc}1 & 2 & 2 \\ 2 & -2 & 2 \\ 3 & 3 & -3\end{array}\right|+1\left|\begin{array}{ccc}1 & -6 & 2 \\ 2 & -5 & 2 \\ 3 & -3 & -3\end{array}\right|-1\left|\begin{array}{ccc}1 & -6 & 2 \\ 2 & -5 & -2 \\ 3 & -3 & 3\end{array}\right|$

$1[-6(6-6)-2(15+6)+2(-15-6)]-2[1(6-6)-2(-6-6)+2(6+6)]+1[1(15+6)+6(-6-6)+2(-6+15)]$

$-1[1(-15-6)-6(6+6)+2(-6+15)]$

$=1$

$D_{3}=\left|\begin{array}{cccc}1 & 1 & 2 & 1 \\ 1 & -2 & -6 & 2 \\ 2 & 1 & -5 & 2 \\ 3 & -1 & -3 & -3\end{array}\right|$

$1\left|\begin{array}{ccc}-2 & -6 & 2 \\ 1 & -5 & 2 \\ -1 & -3 & -3\end{array}\right|-1\left|\begin{array}{ccc}1 & -6 & 2 \\ 2 & -5 & 2 \\ 3 & -3 & -3\end{array}\right|+2\left|\begin{array}{ccc}1 & -2 & 2 \\ 2 & 1 & 2 \\ 3 & -1 & -3\end{array}\right|-1\left|\begin{array}{ccc}1 & -2 & -6 \\ 2 & 1 & -5 \\ 3 & -1 & -3\end{array}\right|$

$=1[-2(15+6)+6(-3+2)+2(-3-5)]-1[1(15+6)+6(-6-6)+2(-6+15)]$

$+2[1(-3+2)+2(-6-6)+2(-2-3)]-1[1(-3-5)+2(-6+15)-6(-2-3)]$

$=-141$

$D_{4}=\left|\begin{array}{cccc}1 & 1 & 1 & 2 \\ 1 & -2 & 2 & -6 \\ 2 & 1 & -2 & -5 \\ 3 & -1 & 3 & -3\end{array}\right|$

$1\left|\begin{array}{ccc}-2 & 2 & -6 \\ 1 & -2 & -5 \\ -1 & 3 & -3\end{array}\right|-1\left|\begin{array}{ccc}1 & 2 & -6 \\ 2 & -2 & -5 \\ 3 & 3 & -3\end{array}\right|+1\left|\begin{array}{ccc}1 & -2 & -6 \\ 2 & 1 & -5 \\ 3 & -1 & -3\end{array}\right|-2\left|\begin{array}{ccc}1 & -2 & 2 \\ 2 & 1 & -2 \\ 3 & -1 & -3\end{array}\right|$

$1[-2(6+15)-2(-3-5)-6(3-2)]-1[1(6+15)-2(-6+15)-6(6+6)]+1[1(-3-5)+2(-6+15)-6(-2-3)]$

$-2[1(-3-2)+2(-6+6)+2(-2-3)]$

$=47$

Thus,

$x=\frac{D_{1}}{D}=\frac{188}{-94}=-2$

$y=\frac{D_{2}}{D}=\frac{-282}{-94}=3$

$z=\frac{D_{3}}{D}=\frac{-141}{-94}=1.5$

$w=\frac{D_{4}}{D}=\frac{47}{-94}=-0.5$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now