0
$\begingroup$

Let $\Delta = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}}\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right|$ , where ${D_1},{D_2}\& {D_3}$ are co-factor of ${c_1},{c_2}\& {c_3}$ respectively such that ${D_1}^2 + {D_2}^2 + {D_3}^2 = 16$ and ${c_1}^2 + {c_2}^2 + {c_3}^2 = 4$ then the maximum value of $\Delta $ is _______

My approach is as follow

$\Delta = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}}\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right|$

${D_1} = \left| {\begin{array}{*{20}{c}} {{a_2}}&{{a_3}}\\ {{b_2}}&{{b_3}} \end{array}} \right|;{D_2} = - \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_3}}\\ {{b_1}}&{{b_3}} \end{array}} \right|;{D_3} = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}\\ {{b_1}}&{{b_2}} \end{array}} \right|$

$\Delta = {c_1}\left| {\begin{array}{*{20}{c}} {{a_2}}&{{a_3}}\\ {{b_2}}&{{b_3}} \end{array}} \right| - {c_2}\left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_3}}\\ {{b_1}}&{{b_3}} \end{array}} \right| + {c_3}\left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}\\ {{b_1}}&{{b_2}} \end{array}} \right|$

How do I proceed from here

$\endgroup$

3 Answers 3

2
$\begingroup$

Let $x=(D_1,D_2,D_3,c_1,c_2,c_3)$. We wish to maximise $$f(x)=c_1D_1-c_2D_2+c_3D_3$$ with constraints $$ g_1(x)=D_1^2+D_2^2+D^3-16 =0, \quad g_2(x) = c_1^2+c_2^2+c^3-4 =0.\tag{$\ast$}$$ By the method of Lagrange multipliers (with multiple constraints) there exists $\lambda,\mu \in \mathbb R$ such that $$\nabla f(x) = \lambda \nabla g_1 + \mu \nabla g_2 . \tag{$\ast\ast$}$$ We have $$ \nabla f (x) = (c_1,-c_2,c_3,D_1,-D_2,D_3) \\ \nabla g_1(x)= 2(D_1,D_2,D_3,0,0,0) \\\nabla g_2(x) = 2 (0,0,0,c_1,-c_2,c_3).$$ Hence, $(\ast\ast)$ becomes $$c_1= 2 \lambda D_2 ,\quad c_2 = - 2 \lambda D_2, \quad c_3 = 2 \lambda D_3 \\ D_1 = 2 \mu c_1, \quad D_2 = -2 \mu c_2, \quad D_3=2 \mu c_3.$$ Substituting the first line of this into the constraints $(\ast)$ we obtain $$ 4 = c_1^2+c_2^2+c^3 = 4 \lambda^2(D_1^2+D_2^2+D^3)=32 \lambda^2. $$ Hence, $\lambda = \pm 1 /4$. Similarly, we obtain $\mu = \pm 1$. It follows that $4 \lambda \mu = \pm 1$ depending on the values of $\lambda$ and $\mu$. Note that we also have $$c_i = 4\lambda \mu c_i $$ for each $i=1,2,3$. Hence, if $4\lambda \mu = -1$ then $c_i=0$ for each $i$ which contradicts $(\ast)$. Thus, $4\lambda \mu = 1$. In this case $c_1,c_2,c_3$ are free variables provided they satisfy the constraint. Finally, we have $$f(2\mu c_1, -2 \mu c_2, 2 \mu c_3 ,c_1,c_2,c_3) = 2 \mu (c_1^2+c_2^2+c_3^2) = 8 \mu. $$ Thus, the maximum of $f$ is 8 which corresponds to $\mu =1$ and the minimum is $-8$ which corresponds to $\mu =-1$.


Edit: I've confirmed this answer via Mathematica: Entering

Maximize[{c1 D1 - c2 D2 + c3 D3, D1^2 + D2^2 + D3^2 == 16 , c1^2 + c2^2 + c3^2 == 4}, {c1, c2, c3, D1, D2, D3}]

returns

{8, {c1 -> -(7/8), c2 -> 1, c3 -> -(Sqrt[143]/8), D1 -> -(7/4), D2 -> -2, D3 -> -(Sqrt[143]/4)}}.

$\endgroup$
0
$\begingroup$

Use Cauchy-Schwarz inequality to write $\Delta=x_1y_1+x_2y_2+x_3y_3\leq \sqrt{x_1^2+x_2^2+x_3^2}\sqrt{y_1^2+y_2^2+y_3^2}$. Then, you can find the answer.

$\endgroup$
2
  • $\begingroup$ This only gives an upper bound for $\Delta$. How can you guarantee that this is the maximum? $\endgroup$
    – JackT
    Sep 12, 2021 at 4:51
  • $\begingroup$ You are right, this is just an upper bound. I don't know whether this is indeed the maximum. $\endgroup$
    – sankhya
    Sep 12, 2021 at 4:54
0
$\begingroup$

We can solve this using Lagrange's method.

We have $2$ constraints,

$g_1 = D_1^2+D_2^2+D_3^2 -16=0$
$g_2 = c_1^2+c_2^2+c_3^2 -4=0$

and the Determinant is $f(c_1,c_2,c_3,D_1,D_2,D_3) \equiv f= c_1D_1+c_2D_2+c_3D_3$

Now let, $F(c_1,c_2,c_3,D_1,D_2,D_3) \equiv F = f +kg_1+hg_2$

Differentiating partially,

$\begin{align}F_{c_1} = D_1+2kc_1 = 0 &&F_{D_1} = c_1+2hD_1 = 0\\ F_{c_2} = D_2+2kc_2 = 0 &&F_{D_2} = c_2+2hD_2 = 0\\ F_{c_3} = D_3+2kc_3 = 0 &&F_{D_3} = c_3+2hD_3 = 0\end{align}$

On solving, we get $\dfrac{c_1}{D_1}=\dfrac{c_2}{D_2}=\dfrac{c_3}{D_3} = -2h = -\dfrac{1}{2k}$

Using the above equalities in $g_2$,

$(-2h)^2(D_1^2+D_2^2+D_3^2) =4 \Rightarrow 64h^2 =4 \Rightarrow \boxed{h =\pm\frac{1}{4}}$

For maximum we need, $h =-\dfrac{1}{4}$ or $c_1 = \dfrac{D_1}{2},c_2 = \dfrac{D_2}{2},c_3 = \dfrac{D_3}{2}$

So, $f_{\max} = \dfrac{1}{2}(D_1^2+D_2^2+D_3^2) = 8$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .