# Problem on Determinant.

Q.

$$\text{If } \Delta = \left|\begin{array}{ccc} a & b & c \\ c & a & b \\ b & c & a \end{array}\right|,$$ $$\text{ then the value of }$$ $$\left|\begin{array}{ccc} a^2 - bc & b^2 - ca & c^2 - ab \\ c^2 - ab & a^2 - bc & b^2 - ca \\ b^2 - ca & c^2 - ab & a^2 - bc \end{array}\right| \text{ is:}$$

Express the Answer in terms of $\Delta$.

My Attempt - I tried Rearranging the Second determinant (by adding or subtracting particular rows or Columns) such that after this, It'll be easy for me write the second determinant as the product of two or more other Determinants that might be same as $\Delta$. But I Found that rearranging the Second Det. Wasn't a good choice as it seems to never simplify itself!! So, I tried Expressing the Second Det. directly as the product of two or more Det. ! Still It was no use. I finally concluded that the Second Det. Can't be expressed as a product without Rearrangement ! But Rearranging it makes it more creepier!

Could you please Guide me how to Rearrange it so that it could be easily expressed as a product? Or is their Another Way for this type of Problem? Please Let me know!

Any help would be Appreciated.

• Do you know how to write matrices? It's exactly the same, except that you use \vmatrix instead of \bmatrix or \pmatrix. – saulspatz Aug 14 '18 at 21:09
• Sorry, Not them too ! I'll get to know soon!! – Creep Anonymous Aug 14 '18 at 21:10
• Each entry of the second matrix is the cofactor from the first matrix. – Ng Chung Tak Aug 14 '18 at 21:10
• @NgChungTak I got that Earlier but How would it help? – Creep Anonymous Aug 14 '18 at 21:12
• Actually, @heropup did it in a way I wasn't familiar with. Take a look. – saulspatz Aug 14 '18 at 21:12

Notice that

$$\begin{bmatrix} a & b & c \\ c & a & b \\ b & c & a% \end{bmatrix}% ^{-1}=\frac{1}{\Delta }% \begin{bmatrix} a^{2}-bc & c^{2}-ab & b^{2}-ac \\ b^{2}-ac & a^{2}-bc & c^{2}-ab \\ c^{2}-ab & b^{2}-ac & a^{2}-bc% \end{bmatrix}.$$ Then, since $\det(A^{-1})=\det(A)^{-1}$ and $\det(cA)=c^{n}\det(A)$ for a matrix $A$ that is $n\times n$, taking determinants of both sides we get $$\frac{1}{\Delta}=\frac{1}{\Delta ^3}% \det\left(\begin{bmatrix} a^{2}-bc & c^{2}-ab & b^{2}-ac \\ b^{2}-ac & a^{2}-bc & c^{2}-ab \\ c^{2}-ab & b^{2}-ac & a^{2}-bc% \end{bmatrix}\right),$$ so $$\det\left(\begin{bmatrix} a^{2}-bc & c^{2}-ab & b^{2}-ac \\ b^{2}-ac & a^{2}-bc & c^{2}-ab \\ c^{2}-ab & b^{2}-ac & a^{2}-bc% \end{bmatrix}\right)=\Delta^2.$$

Finally, notice that $\det(A)=\det(A^T)$.

• Got It!! You're correct! – Creep Anonymous Aug 14 '18 at 21:19
• In the first equation, shouldn't you take the transpose of the matrix on the right-hand side? Of course, $\det A^T=\det A$ so it just a detail. – saulspatz Aug 14 '18 at 21:21
• @saulspatz I don't think so, but at the end I had to. Thank you. – mzp Aug 14 '18 at 21:24

Take $A$ the matrix inside the determinant so that $\Delta=\det(A)$. The transpose of the new matrix is the so called classical adjoint matrix of $A$ for which we have: $$\mathrm{adj}(A)A=\det(A) I.$$ Hence the determinant is equal to $\Delta^{k-1}$ where $k$ is the matrix dimension and is equal to $3$ here.

• Isn't Adjoint Matrix would be the Transpose of the Second one?? – Creep Anonymous Aug 14 '18 at 21:15
• @CreepAnonymous, Yes, you are right! slightly edited! – Arash Aug 14 '18 at 21:28
• But That wouldn't have mattered for at last we had to take the determinant!! – Creep Anonymous Aug 14 '18 at 21:30
• @CreepAnonymous, exactly! – Arash Aug 14 '18 at 21:30

To complete the answer given by @mzp, note that the matrix $$\left(\begin{array}{ccc} a^2 - bc & b^2 - ca & c^2 - ab \\ c^2 - ab & a^2 - bc & b^2 - ca \\ b^2 - ca & c^2 - ab & a^2 - bc \end{array}\right)$$ is $$\pmatrix{1&0&0\\0&0&1\\0&1&0}\left( \matrix{a^{2}-bc & c^{2}-ab & b^{2}-ac \\ b^{2}-ac & a^{2}-bc & c^{2}-ab \\ c^{2}-ab & b^{2}-ac & a^{2}-bc} \right)\pmatrix{1&0&0\\0&0&1\\0&1&0}$$ and that the determinant of $\pmatrix{1&0&0\\0&0&1\\0&1&0}$ is $-1$.

• Isn't it just the transpose? I added a sentence at the end. – mzp Aug 14 '18 at 21:28
• @mzp Yes, But he Mark Fischler proved that too! – Creep Anonymous Aug 14 '18 at 21:30