# Given a linear system of equations below. The matrix equation of the linear system is given

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.
$x+3y+4z=3$
$2z+6y+9z=5$
$3x+y-2z=7$

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casincal

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, $A\left[\left(1,3,4\right),\left(2,6,9\right),\left(3,1,-2\right)\right]$
given, $|A|=8$
Find the value of x by using cramer's rule
$x=\frac{|{D}_{x}|}{|D|}$
here, ${D}_{x}=\left[\left(3,3,4\right),\left(5,6,9\right),\left(7,1,-2\right)\right]$
and, $|D|$ is the determinant of the matrix A
Since,when system of equation as,
${a}_{1}x+{b}_{1}y+{c}_{1}z={d}_{1}$
${a}_{2}x+{b}_{2}y+{c}_{2}z={d}_{2}$
${a}_{3}+{b}_{3}+{c}_{3}z={d}_{3}$
Then, ${D}_{x}=\left[\left({d}_{1},{b}_{1},{c}_{1}\right),\left({d}_{2},{b}_{2},{c}_{2}\right),\left({d}_{3},{b}_{3},{c}_{3}\right)\right]$
Now, the value of x find as,
$x=\frac{|{D}_{x}|}{|D|}$
$=|\left(3,3,4\right),\left(5,6,9\right),\frac{7,1,-2}{|A|}$
$=\frac{\left(-36+20+189\right)-\left(168+27-30\right)}{8}$
$=\frac{173-165}{8}$
$=\frac{8}{8}$
$=1$
Thus, $x=1$