Solve for

2 x_{1} + 4 x_{2} + 6 x_{3} 4 x_{1} + 5 x_{2} + 6 x_{3} 3 x_{1} + x_{2} - 2 x_{3} = 18 = 24 = 4

Normally, we use algebra:

2 (1) - (2) ⟹ 3 x_{2} + 6 x_{3} = 12 ⟹ x_{2} + 2 x_{3} = 4 \frac{3}{2} (1) - (3) ⟹ 5 x_{2} + 11 x_{3} = 23 ⟹ x_{3} = 3, x_{2} = - 2, x_{1} = 4

But we can also use matrices. The original equation is

243451 66 - 2 18244

And solving

R_{1} \to R_{1} + 3 R_{3}, R_{2} \to R_{2} + 3 R_{3} R_{2} \to R_{2} - R_{1} R_{2} \to 7 R_{2} - R_{1} R_{3} \to R_{2} - 3 R_{3} R_{1} \to R_{1} - \frac{11}{3} R_{2} ⟹ 11133781 00 - 2 30364 ⟹ 1123711 00 - 2 3064 ⟹ 1131700 00 - 2 3012 - 2 ⟹ 1130700006301218 ⟹ 030700006 - 14 1218 ⟹ 100010001 4 - 2 3

Using the augmented matrix to row echelon form method, we get the same answer.

Cramer’s Rule

Theorem

Let $A$ be an invertible $n \times n$ matrix. For any $b$ in $R^{n}$ , the unique solution $x$ of $A x = b$ has entries given by

x_{i} = \frac{det A _{i} ( b )}{det A}

Solve the system of linear equations using Cramer’s Rule.

⎩ ⎨ ⎧ 2 x_{1} + 3 x_{2} - x_{3} = 5 - x_{1} + 2 x_{2} + 3 x_{3} = 0 4 x_{1} - x_{2} + x_{3} = - 1

A det A A_{x_{1}} det A_{x_{1}} x_{1} A_{x_{2}} det A_{x_{2}} x_{2} A_{x_{3}} det A_{x_{3}} x_{3} = 2 - 1 4 32 - 1 - 1 31 = 56 = 50 - 1 32 - 1 - 1 31 = 14 = \frac{1}{4} = 2 - 1 4 50 - 1 - 1 31 = 70 = \frac{5}{4} = 2 - 1 4 32 - 1 50 - 1 = - 42 = - \frac{3}{4}