Question

solve the following systems of linear equations by cramers rule x+y+z=2,2x+4y+2z=4,x-2y-z=0​

Answer 1

$\large\underline{\sf{Solution-}}$

★ Given equations are

• x + y + z = 2

• 2x + 4y + 2z = 4

• x - 2y - z = 0

★ The matrix representation is

$\rm :\longmapsto\:A = \begin{gathered}\sf \left[\begin{array}{ccc}1&1&1\\2&4&2\\1&-2& - 1\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:\begin{gathered}\sf B=\left[\begin{array}{c}2\\4\\0\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:\begin{gathered}\sf X=\left[\begin{array}{c}x\\y\\z\end{array}\right]\end{gathered}$

Such that,

$\rm :\longmapsto\:AX=B$

Now,

Consider,

$\rm :\longmapsto\: |A| = \begin{gathered}\sf \left | \begin{array}{ccc}1& 1&1\\ 2&4&2\\1& - 2& - 1\end{array}\right |\end{gathered}$

$\: \: \rm= \: \:1( - 4 + 4) - 1( - 2 - 2) + 1( - 4 - 4)$

$\: \: \rm= \: \:1(0) - 1( -4) + 1( -8)$

$\: \: \rm= \: \:4 - 8$

$\: \: \rm= \: \: - 4$

$\rm :\implies\: |A| \: \ne \: 0$

★ It implies, System of equations is consistent having unique solution.

Consider,

$\rm :\longmapsto\: D_1= \begin{gathered}\sf \left | \begin{array}{ccc}2& 1&1\\ 4&4&2\\0& - 2& - 1\end{array}\right |\end{gathered}$

$\: \: \rm= \: \:2( - 4 + 4) - 1( - 4 - 0) + 1( - 8 - 0)$

$\: \: \rm= \: \:2(0) - 1( - 4) + 1( - 8)$

$\: \: \rm= \: \:4 - 8$

$\: \: \rm= \: \: - 4$

Consider,

$\rm :\longmapsto\: D_2 = \begin{gathered}\sf \left | \begin{array}{ccc}1& 2&1\\ 2&4&2\\1& 0& - 1\end{array}\right |\end{gathered}$

$\: \: \rm= \: \:1( - 4 - 0) - 2( - 2 - 2) + 1(0 - 4)$

$\: \: \rm= \: \:1( - 4) - 2( -4) + 1( - 4)$

$\: \: \rm= \: \: - 4 + 8 - 4$

$\: \: \rm= \: \: 0$

Consider,

$\rm :\longmapsto\: D_3= \begin{gathered}\sf \left | \begin{array}{ccc}1& 1&2\\ 2&4&4\\1& - 2& 0\end{array}\right |\end{gathered}$

$\: \: \rm= \: \:1(0 + 8) - 1(0 - 4) + 2( - 4 - 4)$

$\: \: \rm= \: \:1(8) - 1( - 4) + 2( -8)$

$\: \: \rm= \: \:8 + 4 - 16$

$\: \: \rm= \: \: - 4$

Therefore,

$\bf :\longmapsto\:x = \dfrac{D_1}{ |A| } = \dfrac{ - 4}{ - 4} = 1$

$\bf :\longmapsto\:y = \dfrac{D_2}{ |A| } = \dfrac{0}{ - 4} = 0$

$\bf :\longmapsto\:z = \dfrac{D_3}{ |A| } = \dfrac{ - 4}{ - 4} = 1$