If $a^2-b^2=2$ then what is the least possible value of: \begin{vmatrix} 1+a^2-b^2 & 2ab &-2b\\ 2ab & 1-a^2+b^2&2a\\2b&-2a&1-a^2-b^2 \end{vmatrix}

I tried to express the determinant as a product of two determinants but could not do so. Seeing no way out, I tried expanding it but that took too long and was difficult to evaluate. Please help me with this one, thanks.

  • You would be surprised by the simplicity of the determinant. – Claude Leibovici Sep 21 '16 at 6:28
  • @ Claude Leibovici please let me know,I looked it several times but found it difficult to evaluate(not at all simple) – Navin Sep 21 '16 at 6:38
up vote 0 down vote accepted

\begin{align*} A = \begin{vmatrix} 1+a^2-b^2 & 2ab & -2b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{vmatrix} &= (1+a^2+b^2)^3 \end{align*} To get $1+a^2+b^2$ in the first row, we can apply $R_1 \rightarrow R_1+bR_3$. This will give \begin{align*} A = \begin{vmatrix} 1+a^2+b^2 & 0 & -b(1+a^2+b^2) \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{vmatrix} \end{align*} Taking $1+a^2+b^2$ out from the first row, we get \begin{align*} A = (1+a^2+b^2)\begin{vmatrix} 1 & 0 & -b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{vmatrix} \end{align*} We need one more zero in the first row. This can be obtained by taking $C_3 \rightarrow C_3+bC_1$. This gives \begin{align*} A = (1+a^2+b^2)\begin{vmatrix} 1 & 0 & 0 \\ 2ab & 1-a^2+b^2 & 2a(1+b^2) \\ 2b & -2a & 1-a^2+b^2 \end{vmatrix} \end{align*} Now, expanding by the first row, we obtain \begin{align*} A &= (1+a^2+b^2)[(1-a^2+b^2)^2 + 4a^2(1+b^2)] \\ &= (1+a^2+b^2)[(1+b^2-a^2)^2 + 4a^2(1+b^2)] \\ &= (1+a^2+b^2)[(1+b^2)^2 + a^4 -2a^2(1+b^2)+ 4a^2(1+b^2)] \\ &\qquad \qquad \qquad \qquad \text{expanding } ((1+b^2) - a^2)^2 \\ &= (1+a^2+b^2[(1+b^2)^2 + a^4 +2a^2(1+b^2)] \\ &= (1+a^2+b^2)(1+a^2+b^2)^2 \\ &= (1+a^2+b^2)^3 \end{align*}
Since $a^2-b^2 = 2$, the determinant can be written as $$ (1+ b^2 + 2 + b^2)^3 = (3+2b^2)^3 $$ This is least when $b^2$ is least. The least value of $b^2$ is 0 and hence the least value is 27.

Try to express the determinant using the so-called Rule of Sarrus. This will yield a function in two variables, namely $a$ and $b$. Then find the minimum of this function using differentiation in several variables and your second condition $a^2 - b^2 = 2$. Can you take it from here?

  • This rule, which I know as the diagonals rule, doesn't seem to be very effective in this case, as one would get pretty long expressions in most products... – DonAntonio Sep 21 '16 at 7:10
  • Of course you'd have to use appropriate transformations (binomial formulas, etc. and the additional condition $a^2 - b^2 = 2$) to get a nice and relatively short expression... – ComplexFlo Sep 21 '16 at 11:57

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