Use Cramer's rule to find the solution of the following systems of linear equation in terms of the parameter K.
a) 5x - ky = 6. b) 2x - 3y = k
- 2x + 2ky = -3 X + 2y = - 2
a)
{5x−ky=6−2x+2ky=−3Δ=∣5−k−22k∣=10k−2k=8kΔ1=∣6−k32∣=12k−3k=9kΔ2=∣56−2−3∣=−15+12=−3x=Δ1Δ=9k8k=98y=Δ2Δ=−38k\begin{cases} 5x-ky = 6 \\ -2x+2ky = -3 \end{cases} \\ \Delta =\begin{vmatrix} 5 & -k \\ -2 & 2k \end{vmatrix} =10k -2k = 8k \\ \Delta_1 = \begin{vmatrix} 6 &-k\\ 3&2 \end{vmatrix} = 12k-3k = 9k\\ \Delta_2 = \begin{vmatrix} 5&6\\ -2&-3 \end{vmatrix} = -15+12 = -3 \\ x = \cfrac{\Delta_1}{\Delta} = \cfrac{9k}{8k} = \cfrac{9}{8}\\ y = \cfrac{\Delta_2}{\Delta} =\cfrac{-3}{8k}{5x−ky=6−2x+2ky=−3Δ=∣∣5−2−k2k∣∣=10k−2k=8kΔ1=∣∣63−k2∣∣=12k−3k=9kΔ2=∣∣5−26−3∣∣=−15+12=−3x=ΔΔ1=8k9k=89y=ΔΔ2=8k−3
b)
{2x−3y=kx+2y=−2Δ=∣2−312∣=7Δ1=∣k−3−22∣=2k−6Δ2=∣2k1−2∣=−4−kx=Δ1Δ=2k−67y=Δ2Δ=−4+k7\begin{cases} 2x - 3y = k\\ x+2y = -2 \end{cases}\\ \Delta = \begin{vmatrix} 2 & -3 \\ 1 & 2 \end{vmatrix} = 7\\ \Delta_1 = \begin{vmatrix} k&-3\\ -2&2 \end{vmatrix} =2k-6\\ \Delta_2= \begin{vmatrix} 2&k\\ 1&-2 \end{vmatrix} = -4-k\\ x = \cfrac{\Delta_1}{\Delta} =\cfrac{2k-6}{7}\\ y = \cfrac{\Delta_2}{\Delta} = -\cfrac{4+k}{7}{2x−3y=kx+2y=−2Δ=∣∣21−32∣∣=7Δ1=∣∣k−2−32∣∣=2k−6Δ2=∣∣21k−2∣∣=−4−kx=ΔΔ1=72k−6y=ΔΔ2=−74+k
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