Hello, bryan-duong!
(1) ax - by = c
(2) bx + ay = d
a,b,c,d represent constants
Solve the system and give the conditions that the system has a unique solution.
I assume you know the Elimination method.
Multiply (1) by a:
.a<sup>2</sup>x - aby
. =
.ac
Multiply (2) by b:
.b<sup>2</sup>x + aby
.=
.bd
Add:
.a<sup>2</sup>x + b<sup>2</sup>x
.=
.ac + bd
. . . . (This eliminates the y-terms)
Factor:
.(a<sup>2</sup> + b<sup>2</sup>)x
.=
.ac + bd
. . . . . . . . . . .ac + bd
Then:
. x
.=
.-----------
. . . . . . . . . . .a<sup>2</sup> + b<sup>2</sup>
Now use the same procedure to eliminate the x-terms
. . . . . . . . . . . . . . .ad - bc
. . and get:
. y
.=
.----------
. . . . . . . . . . . . . . .a<sup>2</sup> + b<sup>2</sup>
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
The system will not have a unique solution if the denominators equal zero.
. . This means:
.a<sup>2</sup> + b<sup>2</sup> = 0
Clearly this is impossible.
**
The system will <u>always</u> have a unique solution.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Well, okay . . . it
is possible if a = 0 and b = 0
. . but then we'd have:
.0 = c
.and
.0 = d
. . and we wouldn't have a system, would we?