Answer to Question #196275 in Linear Algebra for anuj

Question

Use Cramer’s rule to solve for y without solving for x, z and w in the system 2w + x + y + z = 3 −8w − 7x − 3y + 5z = −3 w + 4x + y + z = 6 w + 3x + 7y − z = 1 

Answer 1

$\begin{matrix} 2w + x + y + z = 3 \\ −8w − 7x − 3y + 5z = −3 \\ w + 4x + y + z = 6 \\ w + 3x + 7y − z = 1 \end{matrix}$

Write the matrix for the system and check that the determinant is not equal zero.

$A=\begin{bmatrix} 2 & 1 & 1 & 1 \\ -8 & -7 & -3 & 5 \\ 1 & 4 & 1 & 1 \\ 1 & 3 & 7 & -1 \\ \end{bmatrix}$

$\det A=\begin{vmatrix} 2 & 1 & 1 & 1 \\ -8 & -7 & -3 & 5 \\ 1 & 4 & 1 & 1 \\ 1 & 3 & 7 & -1 \\ \end{vmatrix}$

$=2\begin{vmatrix} -7 & -3 & 5 \\ 4 & 1 & 1 \\ 3 & 7 & -1 \\ \end{vmatrix}-\begin{vmatrix} -8 & -3 & 5 \\ 1 & 1 & 1 \\ 1 & 7 & -1 \\ \end{vmatrix}$

$+\begin{vmatrix} -8 & -7 & 5 \\ 1 & 4 & 1 \\ 1 & 3 & -1 \\ \end{vmatrix}-\begin{vmatrix} -8 & -7 & -3 \\ 1 & 4 & 1 \\ 1 & 3 & 7 \\ \end{vmatrix}$

$\begin{vmatrix} -7 & -3 & 5 \\ 4 & 1 & 1 \\ 3 & 7 & -1 \\ \end{vmatrix}=-4\begin{vmatrix} -3 & 5 \\ 7 & -1 \\ \end{vmatrix}+\begin{vmatrix} -7 & 5 \\ 3 & -1 \\ \end{vmatrix}$

$-\begin{vmatrix} -7 & -3 \\ 3 &7 \\ \end{vmatrix}=-4(-32)+(-8)-(-40)=160$

$\begin{vmatrix} -8 & -3 & 5 \\ 1 & 1 & 1 \\ 1 & 7 & -1 \\ \end{vmatrix}=-\begin{vmatrix} -3 & 5 \\ 7 & -1 \\ \end{vmatrix}+\begin{vmatrix} -8 & 5 \\ 1 & -1 \\ \end{vmatrix}$

$-\begin{vmatrix} -8 & -3 \\ 1 &7 \\ \end{vmatrix}=-(-32)+(3)-(-53)=88$

$\begin{vmatrix} -8 & -7 & 5 \\ 1 & 4 & 1 \\ 1 & 3 & -1 \\ \end{vmatrix}=-8\begin{vmatrix} 4 & 1 \\ 3 & -1 \\ \end{vmatrix}-\begin{vmatrix} -7 & 5 \\ 3 & -1\\ \end{vmatrix}$

$+\begin{vmatrix} -7 & 5 \\ 4 &1 \\ \end{vmatrix}=-8(-7)-(-8)+(-27)=37$

$\begin{vmatrix} -8 & -7 & -3 \\ 1 & 4 & 1 \\ 1 & 3 & 7 \\ \end{vmatrix}=-8\begin{vmatrix} 4 & 1 \\ 3 & 7 \\ \end{vmatrix}-\begin{vmatrix} -7 & -3 \\ 3 & 7 \\ \end{vmatrix}$

$+\begin{vmatrix} -7 & -3\\ 4 & 1 \\ \end{vmatrix}=-8(25)-(-40)+(5)=-155$ $\det A=2(160)-88+37-(-155)$

$\det A=424\not=0$

$B=\begin{bmatrix} 2 & 1 & 3 & 1 \\ -8 & -7 & -3 & 5 \\ 1 & 4 & 6 & 1 \\ 1 & 3 & 1 & -1 \\ \end{bmatrix}$

$\det B=\begin{vmatrix} 2 & 1 & 3 & 1 \\ -8 & -7 & -3 & 5 \\ 1 & 4 & 6 & 1 \\ 1 & 3 & 1 & -1 \\ \end{vmatrix}$

$=2\begin{vmatrix} -7 & -3 & 5 \\ 4 & 6 & 1 \\ 3 & 1 & -1 \\ \end{vmatrix}-\begin{vmatrix} -8 & -3 & 5 \\ 1 & 6 & 1 \\ 1 & 1 & -1 \\ \end{vmatrix}$

$+3\begin{vmatrix} -8 & -7 & 5 \\ 1 & 4 & 1 \\ 1 & 3 & -1 \\ \end{vmatrix}-\begin{vmatrix} -8 & -7 & -3 \\ 1 & 4 & 6 \\ 1 & 3 & 1 \\ \end{vmatrix}$

$\begin{vmatrix} -7 & -3 & 5 \\ 4 & 6 & 1 \\ 3 & 1 & -1 \\ \end{vmatrix}=-4\begin{vmatrix} -3 & 5 \\ 1 & -1 \\ \end{vmatrix}+6\begin{vmatrix} -7 & 5 \\ 3 & -1 \\ \end{vmatrix}$

$-\begin{vmatrix} -7 & -3 \\ 3 &1 \\ \end{vmatrix}=-4(-2)+6(-8)-(2)=-42$

$\begin{vmatrix} -8 & -3 & 5 \\ 1 & 6 & 1 \\ 1 & 1 & -1 \\ \end{vmatrix}=\begin{vmatrix} -3 & 5 \\ 6 & 1 \\ \end{vmatrix}-\begin{vmatrix} -8 & 5 \\ 1 & 1 \\ \end{vmatrix}$

$-\begin{vmatrix} -8 & -3 \\ 1 & 6 \\ \end{vmatrix}=-33-(-13)-(-45)=25$

$\begin{vmatrix} -8 & -7 & 5 \\ 1 & 4 & 1 \\ 1 & 3 & -1 \\ \end{vmatrix}=-8\begin{vmatrix} 4 & 1 \\ 3 & -1 \\ \end{vmatrix}-\begin{vmatrix} -7 & 5 \\ 3 & -1\\ \end{vmatrix}$

$+\begin{vmatrix} -7 & 5 \\ 4 &1 \\ \end{vmatrix}=-8(-7)-(-8)+(-27)=37$

$\begin{vmatrix} -8 & -7 & -3 \\ 1 & 4 & 6 \\ 1 & 3 & 1 \\ \end{vmatrix}=-8\begin{vmatrix} 4 & 6 \\ 3 & 1 \\ \end{vmatrix}-\begin{vmatrix} -7 & -3 \\ 3 & 1 \\ \end{vmatrix}$ $+\begin{vmatrix} -7 & -3\\ 4 & 6 \\ \end{vmatrix}=-8(-14)-(2)-(30)=80$

$\det B=2(-42)-25+3(37)-80=-78$

$y=\dfrac{\det B}{\det A}=\dfrac{-78}{424}=-\dfrac{39}{212}$

$y=-\dfrac{39}{212}$