Modular arithmetic - re-arranging equation to get outcome

Question

Is it possible for me to show

$a=7\beta-7\alpha,\ b=3\alpha-2\beta$

If I know that

$\alpha=4a+b,\ \beta=19a+b$

in $mod\ 26$

Any help would be much appreciated, I'm new to modular arithmetic, so still figuring things out.

Answer 1

If you know that $\alpha=4a+b$ and $\beta=19a+b$, then you can show what you want as follows:

$\beta-\alpha=(19a+b)-(4a+b)=15a$, so $7\beta-7\alpha=7\times15a=105a=4\times26a+a\equiv a \bmod 26,$

and $3\alpha-2\beta=3(4a+b)-2(19a+b)=(12a+3b)-(38a+2b)=-26a+b\equiv b\bmod 26$.

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You could also say $\pmatrix{\alpha\\\beta}=\pmatrix{4&&1\\19&&1}\pmatrix{a\\b}$, so $\pmatrix{a\\b}=\pmatrix{4&&1\\19&&1}^{-1}\pmatrix{\alpha\\\beta}$,

and $\pmatrix{4&&1\\19&&1}^{-1}\equiv {\pmatrix{1&&-1\\7&&4}}(15)^{-1}\equiv{\pmatrix{1&&-1\\7&&4}}(19)={\pmatrix{-7&&7\\3&&-2}}$.

Answer 2

$4a + b \equiv \alpha \pmod{26}$

$19a + b\equiv \beta \pmod{26}$ so subtract

$15a \equiv \beta - \alpha \pmod{26}$ multiply but $7$

$105a \equiv 7(\beta -\alpha)\pmod {26}$

But $105a \equiv (4*26+1)a\equiv a \pmod {26}$ so $a \equiv 7\beta - 7\alpha$.

The other is not so straightforward byt the same idea.

$19(4a + b) \equiv 19\alpha\pmod{26}$

$4(19a + b) \equiv 4\beta\pmod{26}$ so subtract

$15b \equiv 19\alpha - 4\beta\pmod{26}$. Multiply by $7$

$7*15b \equiv 7*19\alpha - 7*4\beta \pmod{26}$

$105 b \equiv 133\alpha - 28\beta \pmod {26}$

$(4*26 + 1)b \equiv (5*26+3)\alpha - (26+2)\beta\pmod {26}$

$b \equiv 3\alpha - 2\beta\pmod {26}$.

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Alternatively if $\alpha \equiv 4a+b$ and $\beta \equiv 19a + b$ then

$7\beta - 7\alpha \equiv 7(19a + b)-7(4a+b)\equiv$

$133a -28a\equiv$

$105a\equiv a$.

And $3\alpha - 2\beta \equiv 3(4a+b)-2(19a+b)\equiv$

$(12a + 3b)-(38a +2b) \equiv$

$-26a + b\equiv b$.

Actually, I think maybe we were supposed to do it the second way rather than the first.