Question #176902

Consider the following system of linear equations:

-x1+2x2+x3=4

5x1-2x2+3x3=-28

2x1-x2+4x3=-23


Use Cramer’s rule to solve for x^2. (When working out determinants, indicate

with which row or column co-factor expansion is done.)


1
Expert's answer
2021-03-31T10:48:20-0400

The determinant of the coefficient omitting the constant,

121523214\begin{vmatrix} -1 & 2 & 1\\ 5 & -2 & 3\\ 2 & -1 & 4 \end{vmatrix}\\

Using the row 1 co-factor expansion

D0=1231425324+15221D0=1(11)2(14)+1(3)D0=20the determinant of the coefficient omitting,x2termD2=11453282423Row 1 expansionD2=1328323152842345324D2=1(43)1(3)4(14)D2=16x2=D2D0x2=1620x2=45D_0=-1\begin{vmatrix} -2 & -3 \\ -1 & 4 \end{vmatrix}-2\begin{vmatrix} 5 & 3 \\ 2 & 4 \end{vmatrix}+1\begin{vmatrix} 5 & -2 \\ 2 & -1 \end{vmatrix}\\ D_0= -1(-11)-2(14)+1(-3)\\ D_0=-20\\ \text{the determinant of the coefficient omitting},x^2 term\\ D_2= \begin{vmatrix} -1 & 1 & -4\\ 5 & 3 & 28\\ 2 & 4 & 23 \end{vmatrix}\\ \text{Row 1 expansion} D_2=-1\begin{vmatrix} 3 & 28 \\ 3 & 23 \end{vmatrix}-1\begin{vmatrix} 5 & 28\\ 4 & 23 \end{vmatrix}-4\begin{vmatrix} 5 & 3\\ 2 & 4 \end{vmatrix}\\ D_2= -1(-43)-1(3)-4(14)\\ D_2=-16\\ x^2=\frac{D_2}{D_0}\\ x^2=\frac{-16}{-20}\\ x^2=\frac{4}{5}


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