Number Theory and Linear Algebra!

Algebra Level 4

Let $a,b,c,d$ be four integers and $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$

such that $A$ is an invertible matrix and $A^{-1}$ also have four integer entries.

Find all possible value(s) of $ad-bc$ and input the product of the possible value(s).

2 -1 1 -2

1 solution

Kelvin Hong
Sep 22, 2018

Since $A^{-1}=\dfrac1{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$

If all of the entries of $A^{-1}$ are integers, then we can let

$\begin{cases}a=k_1(ad-bc)\\b=k_2(ad-bc)\\c=k_3(ad-bc)\\d=k_4(ad-bc)\end{cases}$

Now we have the relation: $\begin{aligned}ad-bc&=k_1(ad-bc)k_4(ad-bc)-k_2(ad-bc)k_3(ad-bc)\\1&=(ad-bc)(k_1k_4-k_2k_3)\end{aligned}$

This can happen only when $ad-bc=\pm 1$ , since $k_1k_4-k_2k_3$ is also an integer. Therefore the answer is $1\times(-1)=\boxed{-1}$ .

Example for $ad-bc=1$ : $A=\begin{pmatrix}3&5\\1&2\end{pmatrix}, A^{-1}=\begin{pmatrix}2&-5\\-1&3\end{pmatrix}$ .

Example for $ad-bc=-1$ : $A=\begin{pmatrix}2&5\\1&2\end{pmatrix}, A^{-1}=\begin{pmatrix}-2&5\\1&-2\end{pmatrix}$ .

You are using the key fact that $\mathrm{det}\,(AB) = \mathrm{det}\,A \times \mathrm{det}\,B$ for any square matrices $A,B$ . Thus $\mathrm{det}\,A \times \mathrm{det}\,A^{-1} = 1$ . If both $A,A^{-1}$ have integer entries, their determinants are also integers, and hence both determinants are $1$ or both are $-1$ .

Mark Hennings - 2 years, 8 months ago

I don't think he's using that fact. This seems like a different solution!

Jordan Cahn - 2 years, 8 months ago

Yes he is. The inverse matrix is $\left( \begin{array}{cc} k_4 & -k_2 \\ -k_3 & k_1 \end{array} \right)$ and so the determinant of the inverse is $k_1k_4 - k_2k_3$ , so the key identity $(ad-bc)(k_1k_4 - k_2k_3) = 1$ is just the property of determinants.

Mark Hennings - 2 years, 8 months ago

@Mark Hennings – While that is correct, and one way to arrive at that identity, I don't believe that's how Kelvin arrived at it. Look at the line immediately above $1 = (ad-bc)(k_1k_4-k_2k_3)$ : $ad-bc = k_1(ad-bc)k_4(ad-bc)-k_2(ad-bc)k_3(ad-bc)$ Simply factor out $(ad-bc)^2$ and divide both sides by $(ad-bc)$ (which must be nonzero, as $A$ is invertible). This gives the desired identity without using $\det A\times \det B = \det(AB)$ .

Jordan Cahn - 2 years, 8 months ago

@Jordan Cahn – Sigh. Without realising it, maybe, this is just a proof of the determinant identity for $2\times2$ matrices.

Mark Hennings - 2 years, 8 months ago

@Mark Hennings – Hah. I believe it is.

Jordan Cahn - 2 years, 8 months ago

Yes, after I wrote this long solution I find your solution by my friend...

Kelvin Hong - 2 years, 8 months ago

Number Theory and Linear Algebra!

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