### About Cramer's Rule 3 x 3:

It is often quicker to use Cramer's Rule to solve a linear system in three variables. To perform this operation, we will first place each equation of the system in standard form. Once this is done, we will set up D, which is the determinant of the coefficient matrix, D_{x} which is the same as D with the column for the coefficients of x replaced with the constants, and D_{y}, which is the same as D with the column for the coefficients of y replaced with the constants. Once these are all set up, we calculate the values and then find x as D_{x}/D, and y as D_{y}/D.

Test Objectives

- Demonstrate the ability to find the determinant of a matrix
- Demonstrate the ability to solve a linear system

#1:

Instructions: solve each system.

$$a)\hspace{.2em}$$ $$-2x + 4y + z=4$$ $$-2x + 2y + z=-4$$ $$-6x + y + 6z=-26$$

$$b)\hspace{.2em}$$ $$-5x - 2y + 4z=-28$$ $$-5x - 3y + 6z=-27$$ $$-3x + 5y - 5z=-18$$

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#2:

Instructions: solve each system.

$$a)\hspace{.2em}$$ $$4x + 2y + 7z=-10$$ $$-7x + 3y - 6z=24$$ $$4x - y - z=-13$$

$$b)\hspace{.2em}$$ $$-x - 3y + 2z=16$$ $$2x - 3y - z=13$$ $$-x - 4y - 3z=-11$$

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#3:

Instructions: solve each system.

$$a)\hspace{.2em}$$ $$2x - 7y + z=-24$$ $$-7x - 4y - 2z=0$$ $$5x - y + 4z=-9$$

$$b)\hspace{.2em}$$ $$-x + y - z=-2$$ $$-5x - 4y + 3z=-31$$ $$-x - 3y - z=10$$

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#4:

Instructions: solve each system.

$$a)\hspace{.2em}$$ $$-7x - 5y - 6z=-20$$ $$-7x - 3y - 3z=-20$$ $$-2x + y - 6z=26$$

$$b)\hspace{.2em}$$ $$-y - 6z=16$$ $$-x + 7y + 7z=-9$$ $$5x - 4y - 6z=20$$

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#5:

Instructions: solve each system.

$$a)\hspace{.2em}$$ $$-2x + 6y - z=10$$ $$6x + 5y + 4z=-11$$ $$3x + 6y + 2z=-2$$

$$b)\hspace{.2em}$$ $$-2x - 2y - 3z=0$$ $$7x + 5y - 7z=-27$$ $$-4x + y + 3z=-2$$

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Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}(7,4,2)$$

$$b)\hspace{.2em}(6,1,1)$$

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#2:

Solutions:

$$a)\hspace{.2em}(-3,1,0)$$

$$b)\hspace{.2em}(5,-3,6)$$

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#3:

Solutions:

$$a)\hspace{.2em}(-2,3,1)$$

$$b)\hspace{.2em}(5,-3,-6)$$

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#4:

Solutions:

$$a)\hspace{.2em}(2,6,-4)$$

$$b)\hspace{.2em}(2,2,-3)$$

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#5:

Solutions:

$$a)\hspace{.2em}(0,1,-4)$$

$$b)\hspace{.2em}(1,-4,2)$$