Question

Given a linear system of equations below. The matrix equation of the linear system is given by: (see image)
Given a linear system of equations below.The matrix equation of the linear system is given by:Ax=b.The determinant of A is 8.Using Cramers's rule find the value for x.
\(x+3y+4z=3\)
\(2z+6y+9z=5\)
\(3x+y-2z=7\)

Answer 1

Consider the provided system of equation
\(x+3y+4z=3\)
\(2z+6y+9z=5\)
\(3x+y-2z=7\)
The matrix equation of the linear system is given by,
AX=b
Where, \(A[(1,3,4),(2,6,9),(3,1,-2)]\)
given, \(|A|=8\)
Find the value of x by using cramer's rule
\(x=\frac{|D_x|}{|D|}\)
here, \(D_x=[(3,3,4),(5,6,9),(7,1,-2)]\)
and, \(|D|\) is the determinant of the matrix A
Since,when system of equation as,
\(a_1x+b_1y+c_1z=d_1\)
\(a_2x+b_2y+c_2z=d_2\)
\(a_3+b_3+c_3z=d_3\)
Then, \(D_x=[(d_1,b_1,c_1),(d_2,b_2,c_2),(d_3,b_3,c_3)]\)
Now, the value of x find as,
\(x=\frac{|D_x|}{|D|}\)
\(=|(3,3,4),(5,6,9),\frac{7,1,-2}{|A|}\)
\(=\frac{(-36+20+189)-(168+27-30)}{8}\)
\(=\frac{173-165}{8}\)
\(=\frac{8}{8}\)
=1
Thus,x=1