# Finding the determinant of matrix with certain co-factor condition

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

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)}}.

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.

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

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