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Linear Algebra With Applications

Otto Bretscher

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 442 pages

ISBN: 9780321796974

Chapter 6: 35 E (page 265)

Consider two distinct points $[\begin{array}{l} a_{1} \\ a_{2} \end{array}]$ and $[\begin{array}{l} b_{1} \\ b_{2} \end{array}]$ in the plane. Explain why the solutions $[\begin{array}{l} x_{1} \\ x_{2} \end{array}]$ of the equation $d e t [\begin{matrix} 1 & 1 & 1 \\ x_{1} & a_{1} & b_{1} \\ x_{2} & a_{2} & b_{2} \end{matrix}] = 0$ form a line and why this line goes through the two points $[\begin{array}{l} a_{1} \\ a_{2} \end{array}]$ and $[\begin{array}{l} b_{1} \\ b_{2} \end{array}]$ .

Short Answer

Expert verified

Therefore, $[\begin{array}{l} x_{1} \\ x_{2} \end{array}] = [\begin{array}{l} a_{1} \\ a_{2} \end{array}]$ and $[\begin{array}{l} x_{1} \\ x_{2} \end{array}] = [\begin{array}{l} b_{1} \\ b_{2} \end{array}]$ are a solution.

Step by step solution

01

Given

Consider two distinct points are,

$[\begin{array}{l} a_{1} \\ a_{2} \end{array}]$

$[\begin{array}{l} b_{1} \\ b_{2} \end{array}]$

02

To solve equations

We solve,

$\det [\begin{matrix} 1 & 1 & 1 \\ x_{1} & a_{1} & b_{1} \\ x_{2} & a_{2} & b_{2} \end{matrix}] = 0$

$a_{1} b_{2} + b_{1} x_{2} + a_{2} x_{1} 鈭� a_{1} x_{2} 鈭� a_{2} b_{1} 鈭� b_{2} x_{1} = 0$

$(a_{2} 鈭� b_{2}) x_{1} + (鈭� a_{1} + b_{1}) x_{2} + (a_{1} b_{2} 鈭� a_{2} b_{1}) = 0$

This is a linear equation. By computing, we can see that:

$[\begin{array}{l} x_{1} \\ x_{2} \end{array}] = [\begin{array}{l} a_{1} \\ a_{2} \end{array}]$

And

$[\begin{array}{l} x_{1} \\ x_{2} \end{array}] = [\begin{array}{l} b_{1} \\ b_{2} \end{array}]$ Satisfy the equation.

Therefore, $[\begin{array}{l} x_{1} \\ x_{2} \end{array}] = [\begin{array}{l} a_{1} \\ a_{2} \end{array}]$ and $[\begin{array}{l} x_{1} \\ x_{2} \end{array}] = [\begin{array}{l} b_{1} \\ b_{2} \end{array}]$ are a solution.

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