Use Cramer's Rule to solve (if possible) the system of linear equations. (If not possible, enter IMPOSSIBLE.) 4x − 2y + 3z = −9 2x + 2y + 5z = −1 8x − 5y − 2z = 15

We start by calculating what it's called the determinant of the system. Represent by the letter D. This determinant is formed by the coefficients on the three equations of the system.

[tex]4x -2y + 3z =-9 (eq.1)\\2x +2y +5z=-1 (eq.2)\\8x-5y-2z=15(eq.3)\\[/tex]

unknow quantities: x, y, z

x y z

D=[tex]\left[\begin{array}{ccc}4&-2&3\\2&2&5\\8&-5&-2\end{array}\right][/tex]

For solve this determinant of 3x3 we use the Rule of Sarrus. This rule consist in repeating line 1 and 2 under the line 3.

[tex]D= \left[\begin{array}{ccccc}4&-2&3\\2&2&5\\8&-5&-2\\4&-2&3\\2&2&5\end{array}\right][/tex]

Then we multiply the main diagonals and then we add them.

[tex](4*2*-2)= -16\\(2*-5*3)= -30\\(8*-2*5)= -80\\\\-16-30-80=-126[/tex]

After that we multiply the secondary diagonals and we subtract that value from the result of adding the multiplication of main diagonals.

[tex]3*2*8=48\\5*-5*4=-100\\-2*-2*2=8\\\\48 -100 + 8= -44[/tex]

[tex]D= \left[\begin{array}{ccccc}4&-2&3\\2&2&5\\8&-5&-2\\4&-2&3\\2&2&5\end{array}\right]= -126 - (-44) = -82[/tex]

So the system's determinant is

[tex]D= -82[/tex]

We continue now with the x's determinant.

In the original x column we substitute the independent values, those are after the equal.

[tex]Dx=\left[\begin{array}{ccc}-9&-2&3\\-1&2&5\\15&-5&-2\end{array}\right][/tex]

Then we repeat the use of Sarru's Rule. Which is repeating line 1 and 2 under the line 3. After that we multiply the main diagonals and add the result. Then we multiply the the secondary diagonals and we subtract that value from the result of adding the multiplication of main diagonals.

1)

[tex]Dx= \left[\begin{array}{ccccc}-9&-2&3\\-1&2&5\\15&-5&-2\\-9&-2&3\\-1&2&5\end{array}\right][/tex]

2) Multiplication of main diagonals

[tex](-9*2*-2) = 36 \\\\(-1*-5*3)= 15\\\\(15*-2*5)= -150\\\\36+15-150= -99[/tex]

3) Multiplicational of secondary diagonals

[tex](3*2*15)=90\\(5*-5*-9)= 225\\(-2*-2*-1)= -4\\90+225-4=311[/tex]

4)

[tex]Dx=-90- (311)= -401[/tex]

So the x's determinant is

[tex]Dx= -401[/tex]

We continue now with the determinant of y and z columns. Following the previous 4 steps. In the original y and z column we substitute the independent values, those are after the equal.

[tex]Dy =\left[\begin{array}{ccc}4&-9&3\\2&-1&5\\8&15&-2\end{array}\right][/tex] [tex]Dz =\left[\begin{array}{ccc}4&-2&-9\\2&-2&-1\\8&-5&15\end{array}\right][/tex]

Then we repeat the use of Sarru's Rule. Which is repeating line 1 and 2 under the line 3. After that we multiply the main diagonals and add the result. Then we multiply the the secondary diagonals and we subtract that value from the result of adding the multiplication of main diagonals.

1)

[tex]Dy =\left[\begin{array}{ccccc}4&-9&3\\2&-1&5\\8&15&-2\\4&-9&3\\2&-1&5\end{array}\right][/tex] [tex]Dz =\left[\begin{array}{ccc}4&-2&-9\\2&-2&-1\\8&-5&15\\4&-2&-9\\2&-2&-1\end{array}\right][/tex]

2) Multiplication of main diagonals

For Dy:

[tex](4*-1*-2) = 8 \\(2*15*3)= 90\\(8*-9*5)= -360\\\\8+90-360= -262[/tex]

For Dz:

[tex](4*2*15) = 120 \\(2*-5*9)= -90\\(8*-2*-1)= 16\\\\120-90+16= 46[/tex]

3) Multiplicational of secondary diagonals

For Dy:

[tex](3*-1*8)=-24\\(5*15*4)= 300\\(-2*-9*2)= 36\\\\-24+300+36=312[/tex]

For Dz:

[tex](-9*2*8)=-144\\(-1*-5*4)= 20\\(15*-2*2)= -60\\\\-144+20-60= -184[/tex]

4)

[tex]Dy=-262- (312)= 50\\Dz= 46-(-184)= 230[/tex]

So the y's determinant is

[tex]Dy= 50[/tex]

And the z's determinant is

[tex]Dz= 230[/tex]

Now for the final phase, we have to find the values of x, y and z as shown bellow:

[tex]x=\frac{Dx}{D} ; y=\frac{Dy}{D} ; z=\frac{Dz}{D}[/tex]

[tex]D=-82;Dx=-401;Dy=50;Dz=230[/tex]

[tex]x=\frac{-401}{-82}=4.89\\ \\y=\frac{50}{-82}=-0.609\\ \\z=\frac{230}{-82} =-2.804[/tex]

[tex]x=4.89\\y= -0.609\\z=-2.804[/tex]