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Cramer's rule solves $2 x + y = 5$ and $x - y = 1$ as $(x, y) = (2, 1)$

Example

Cramer's rule solves $2 x + y = 5$ , $x - y = 1$ over $Q$ as $(x, y) = (2, 1)$ .

Facts & Assumptions

Given: $A = (\begin{array}{cc} 2 & 1 \\ 1 & - 1 \end{array})$ and $b = (\begin{array}{c} 5 \\ 1 \end{array})$ .

[F1]

$Q$ is a field (The rationals form a field).

[F2]

A $2 \times 2$ determinant is its two-term Leibniz sum (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[L1]

Over a field, a nonzero determinant gives the unique solution $x_{j} = \det (A_{j} (b)) / \det (A)$ (Over a field, for each right-hand side $b$ , $A x = b$ has a unique solution exactly when $\det (A)$ is nonzero, and then Cramer's quotient formula holds).

Verification

technique · direct

1.1

Direct calculation gives $\det (A) = 2 (- 1) - 1 = - 3 \neq 0$ .

F1F2algebra

1.2

Replacing the first and second columns gives $\det (\begin{array}{cc} 5 & 1 \\ 1 & - 1 \end{array}) = - 6, \det (\begin{array}{cc} 2 & 5 \\ 1 & 1 \end{array}) = - 3.$

F2algebra

2.1

By [L1], $x = (- 6) / (- 3) = 2$ and $y = (- 3) / (- 3) = 1$ .

step 1.1step 1.2L1

3.1

Substitution gives $2 (2) + 1 = 5$ and $2 - 1 = 1$ , verifying both equations and the claimed solution.

step 2.1algebra∎

Depends on

Used by

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Cramer's rule solves 2x+y=52x+y=5 and x−y=1x−y=1 as (x,y)=(2,1)(x,y)=(2,1)