Cramers Rule

System of Equations	Coefficient Matrix	Variable Matrix	Constants Matrix
${a x + b y = e c x + d y = f$	$A = [a c b d]$	$x = [x y]$	$b = [e f]$

System of Equations

Coefficient Matrix

Variable Matrix

Constants Matrix

{a x + b y = e c x + d y = f

A = [a c b d]

x = [x y]

b = [e f]

System of Equations	Matrix Equation
$⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a_{1} x + b_{1} y + c_{1} z = d_{1} a_{2} x + b_{2} y + c_{2} z = d_{2} a_{3} x + b_{3} y + c_{3} z = d_{3}$	$⎣ ⎢ ⎡ a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} c_{1} c_{2} c_{3} ⎦ ⎥ ⎤ \cdot ⎣ ⎢ ⎡ x y z ⎦ ⎥ ⎤ = ⎣ ⎢ ⎡ d_{1} d_{2} d_{3} ⎦ ⎥ ⎤$
Solutions
$x = \frac{∣ ∣ ∣ ∣ ∣ ∣ ∣ d _{1} d _{2} d _{3} b _{1} b _{2} b _{3} c _{1} c _{2} c _{3} ∣ ∣ ∣ ∣ ∣ ∣ ∣}{∣ A ∣}; y = \frac{∣ ∣ ∣ ∣ ∣ ∣ ∣ a _{1} a _{2} a _{3} d _{1} d _{2} d _{3} c _{1} c _{2} c _{3} ∣ ∣ ∣ ∣ ∣ ∣ ∣}{∣ A ∣};$ $z = \frac{∣ ∣ ∣ ∣ ∣ ∣ ∣ a _{1} a _{2} a _{3} b _{1} b _{2} b _{3} d _{1} d _{2} d _{3} ∣ ∣ ∣ ∣ ∣ ∣ ∣}{∣ A ∣}$

System of Equations

Matrix Equation

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a_{1} x + b_{1} y + c_{1} z = d_{1} a_{2} x + b_{2} y + c_{2} z = d_{2} a_{3} x + b_{3} y + c_{3} z = d_{3}

⎣ ⎢ ⎡ a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} c_{1} c_{2} c_{3} ⎦ ⎥ ⎤ \cdot ⎣ ⎢ ⎡ x y z ⎦ ⎥ ⎤ = ⎣ ⎢ ⎡ d_{1} d_{2} d_{3} ⎦ ⎥ ⎤

Solutions

x = \frac{∣ ∣ ∣ ∣ ∣ ∣ ∣ d _{1} d _{2} d _{3} b _{1} b _{2} b _{3} c _{1} c _{2} c _{3} ∣ ∣ ∣ ∣ ∣ ∣ ∣}{∣ A ∣}; y = \frac{∣ ∣ ∣ ∣ ∣ ∣ ∣ a _{1} a _{2} a _{3} d _{1} d _{2} d _{3} c _{1} c _{2} c _{3} ∣ ∣ ∣ ∣ ∣ ∣ ∣}{∣ A ∣};

z = \frac{∣ ∣ ∣ ∣ ∣ ∣ ∣ a _{1} a _{2} a _{3} b _{1} b _{2} b _{3} d _{1} d _{2} d _{3} ∣ ∣ ∣ ∣ ∣ ∣ ∣}{∣ A ∣}

Consider a system of two equations. This system can be solved using the Elimination Method.

{a x + b y = e c x + d y = f (I) (II)

Start by eliminating the

y -

variable. To do this, multiply Equation (I) by

d

and Equation (II) by

b .

Then, subtract the equations.

{a x + b y = e c x + d y = f (I) (II)

▼

Solve by elimination

$(I):$ Multiply by $d$

{a d x + b d y = d e c x + d y = f (I) (II)

$(II):$ Multiply by $b$

{a d x + b d y = d e b c x + b d y = b f (I) (II)

$(I):$ Subtract $(II)$

{a d x - b c x = d e - b f b c x + b d y = b f

$(I):$ Factor out $x$

{x (a d - b c) = d e - b f b c x + b d y = b f

Notice that the factor

(a d - b c)

is equal to the determinant of matrix

A

and that the right-hand side expression is equal to the determinant of matrix

A_{1} .

∣ A ∣ ∣ A_{1} ∣ = ∣ ∣ ∣ ∣ ∣ a c b d ∣ ∣ ∣ ∣ ∣ = a d - b c = ∣ ∣ ∣ ∣ ∣ e f b d ∣ ∣ ∣ ∣ ∣ = d e - b f

Using these equations, Equation (I) can be rewritten in terms of determinants.

{x (a d - b c) = d e - b f b c x + b d y = b f

$(I):$ $a d - b c = ∣ A ∣$ , $d e - b f = ∣ A_{1} ∣$

{x \cdot ∣ A ∣ = ∣ A_{1} ∣ b c x + b d y = b f

Finally, since

∣ A ∣ \neq = 0,

both sides of Equation (I) can be divided by

∣ A ∣ .

This manipulation creates the first equation of Cramer's Rule.

x = \frac{∣ A _{1} ∣}{∣ A ∣} = \frac{∣ ∣ ∣ ∣ ∣ e f b d ∣ ∣ ∣ ∣ ∣}{∣ A ∣}

A similar process to eliminate the $x -$ variable gives the second equation.

y = \frac{∣ A _{2} ∣}{∣ A ∣} = \frac{∣ ∣ ∣ ∣ ∣ a c e f ∣ ∣ ∣ ∣ ∣}{∣ A ∣}

In conclusion, when $∣ A ∣ \neq = 0,$ the system has a unique solution $x$ whose components are the ones above.

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