written 4.6 years ago by
teamques10
★ 64k
|
modified 23 months ago
by
pedsangini276
• 4.7k
|
|
written 4.6 years ago by
teamques10
★ 64k
|
Solution:
$D = \quad \begin{vmatrix} 1 & 1 & 1\\ 1 & -1 & 1\\ 1 & 1 & -1 \end{vmatrix}$
=1(1-1)-1(-1-1)+1(1+1)= 4
$D_{V1} = \quad \begin{vmatrix} 9 & 1 & 1\\ 3 & -1 & 1\\ 1 & 1 & -1 \end{vmatrix}$
= 9(1-1)-1(-3-1)+1(3+1) = 8
$V_{1} = \frac{D_{V1}}{D} = \frac{8}{4} = 2$
$D_{V2} = \quad \begin{vmatrix} 1 & 9 & 1\\ 1 & 3 & 1\\ 1 & 1 & -1 \end{vmatrix}$
= 1(-3-1)-9(-1-1)+1(1-3) = 12
$V_{2} = \frac{D_{V2}}{D} = \frac{12}{4} = 3$
$D_{V3} = \quad \begin{vmatrix} 1 & 1 & 9\\ 1 & -1 & 3\\ 1 & 1 & -1 \end{vmatrix}$
$V_{3} = \frac{D_{V3}}{D} = \frac{16}{4} = 4$
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