Question

Find the value of the expression
$\left| \begin{matrix} {{a}^{2}}+{{\lambda }^{2}} & ab+c\lambda & ca-b\lambda \\\ ab-c\lambda & {{b}^{2}}+{{\lambda }^{2}} & bc+a\lambda \\\ ca+b\lambda & bc-a\lambda & {{c}^{2}}+{{\lambda }^{2}} \\\ \end{matrix} \right|\times \left| \begin{matrix} \lambda & c & -b \\\ -c & \lambda & a \\\ b & -a & \lambda \\\ \end{matrix} \right|$

Answer 1

Hint:Observe the first determinant with respect to cofactors of the second determinant in the product. Determinant made by cofactors of a determinant is square of that determinant i.e. $\Delta '={{\Delta }^{2}}$ , where $\Delta '$ is the determinant made by cofactors of all elements of $\Delta$ . And hence, simplify the determinant formed to get the answer.

Complete step-by-step answer:
Expression is
$\left| \begin{matrix} {{a}^{2}}+{{\lambda }^{2}} & ab+c\lambda & ca-b\lambda \\\ ab-c\lambda & {{b}^{2}}+{{\lambda }^{2}} & bc+a\lambda \\\ ca+b\lambda & bc-a\lambda & {{c}^{2}}+{{\lambda }^{2}} \\\ \end{matrix} \right|\times \left| \begin{matrix} \lambda & c & -b \\\ -c & \lambda & a \\\ b & -a & \lambda \\\ \end{matrix} \right|=?.........\left( i \right)$
Let us observe the determinant
$D=\left| \begin{matrix} {{a}^{2}}+{{\lambda }^{2}} & ab+c\lambda & ca-b\lambda \\\ ab-c\lambda & {{b}^{2}}+{{\lambda }^{2}} & bc+a\lambda \\\ ca+b\lambda & bc-a\lambda & {{c}^{2}}+{{\lambda }^{2}} \\\ \end{matrix} \right|......\left( ii \right)$
We know cofactor of any element can be determined by the following way:
Let we have

{{A}_{11}} & {{A}_{12}} & {{A}_{13}} \\\ {{A}_{21}} & {{A}_{22}} & {{A}_{23}} \\\ {{A}_{31}} & {{A}_{32}} & {{A}_{33}} \\\ \end{matrix} \right|$$ So, cofactor of ${{A}_{11}}$ can be given as ${{C}_{11}}={{\left( -1 \right)}^{1+1}}\left| \begin{matrix} {{A}_{22}} & {{A}_{23}} \\\ {{A}_{32}} & {{A}_{33}} \\\ \end{matrix} \right|$ And cofactor of ${{A}_{12}}$ can be given as ${{C}_{12}}={{\left( -1 \right)}^{1+2}}\left| \begin{matrix} {{A}_{21}} & {{A}_{23}} \\\ {{A}_{31}} & {{A}_{33}} \\\ \end{matrix} \right|$ Similarly, cofactor of other elements can be calculated by the same way as calculated above. The determinants formed for the calculation of cofactor of any element is determined by leaving all the elements belonging to the element row and column. It means the determinant for cofactor of all elements will be of 2 x 2. And we need to use the formed determinant by ${{\left( -1 \right)}^{m+n}},$ where m and n are the row number and column number of the element of which we are going to find the cofactor Now, we know the determinant formed by the cofactors of elements of the determinant $\Delta $ is given as $$\Delta '=\left| \begin{matrix} {{C}_{11}} & {{C}_{12}} & {{C}_{13}} \\\ {{C}_{21}} & {{C}_{22}} & {{C}_{23}} \\\ {{C}_{31}} & {{C}_{32}} & {{C}_{33}} \\\ \end{matrix} \right|$$ Now, if we want to solve the $$\Delta '$$ and $$\Delta $$ , we can give relation between both the determinant as $$\Delta '={{\Delta }^{2}}.....\left( iii \right)$$ Now coming to the question, we supposed value of D as $D=\left| \begin{matrix} \lambda & c & -b \\\ -c & \lambda & a \\\ b & -a & \lambda \\\ \end{matrix} \right|$ Now, let us find out the cofactors of all the elements of determinant D, we get that Cofactor of $\begin{aligned} & {{A}_{11\text{ }}}i.e\,\text{ }\lambda \text{ =}{{\left( -1 \right)}^{1+1}}\left| \begin{matrix} \lambda & a \\\ -a & \lambda \\\ \end{matrix} \right| \\\ & ={{\lambda }^{2}}-\left( -{{a}^{2}} \right) \\\ & ={{\lambda }^{2}}+{{a}^{2}} \end{aligned}$ Cofactor of $\begin{aligned} & {{A}_{12\text{ }}}i.e\text{ }\lambda \text{ =}{{\left( -1 \right)}^{2+1}}\left| \begin{matrix} -c & a \\\ b & \lambda \\\ \end{matrix} \right| \\\ & =-\left( -c\lambda -ab \right) \\\ & =ab+c\lambda \end{aligned}$ Cofactor of ${{A}_{13}}$ i.e. $'b'=ac-b\lambda $ Similarly, we can calculate the cofactors of all the elements and hence, we can get a determinant formed by the cofactors of them. Hence, we can get determinant of cofactors formed as $D'=\left| \begin{matrix} {{a}^{2}}+{{\lambda }^{2}} & ab+c\lambda & ca-b\lambda \\\ ab-c\lambda & {{b}^{2}}+{{\lambda }^{2}} & bc+a\lambda \\\ ca+b\lambda & bc-a\lambda & {{c}^{2}}+{{\lambda }^{2}} \\\ \end{matrix} \right|$ Now we can get the value of D’ by calculating ${{D}^{2}}$ as per the property given in the equation (iii). So, we get $D'={{D}^{2}}={{\left| \begin{matrix} \lambda & c & -b \\\ -c & \lambda & a \\\ b & -a & \lambda \\\ \end{matrix} \right|}^{2}}$ Now, we can compare determinant D’ with the equation (i) and hence, we get the equation (i) as ${{\left| \begin{matrix} \lambda & c & -b \\\ -c & \lambda & a \\\ b & -a & \lambda \\\ \end{matrix} \right|}^{2}}\left| \begin{matrix} \lambda & c & -b \\\ -c & \lambda & a \\\ b & -a & \lambda \\\ \end{matrix} \right|=?$ Hence, we need to find the value of the expression $={{\left| \begin{matrix} \lambda & c & -b \\\ -c & \lambda & a \\\ b & -a & \lambda \\\ \end{matrix} \right|}^{3}}$ Now we can expand the above determinant along row 1 as $$\begin{aligned} & ={{\left[ \lambda \left( {{\lambda }^{2}}-\left( -{{a}^{2}} \right) \right)-c\left( -c\lambda -ab \right)-b\left( ac-b\lambda \right) \right]}^{3}} \\\ & ={{\left[ {{\lambda }^{3}}+{{a}^{2}}\lambda +{{c}^{2}}\lambda +abc-abc+{{b}^{2}}\lambda \right]}^{3}} \\\ & ={{\left[ {{\lambda }^{3}}+{{a}^{2}}\lambda +{{c}^{2}}\lambda +{{b}^{2}}\lambda \right]}^{3}} \\\ & ={{\left[ \lambda \left( {{\lambda }^{2}}+{{a}^{2}}+{{c}^{2}}+{{b}^{2}} \right) \right]}^{3}} \\\ & ={{\lambda }^{3}}{{\left( {{\lambda }^{2}}+{{a}^{2}}+{{c}^{2}}+{{b}^{2}} \right)}^{3}} \\\ \end{aligned}$$ Hence the value of the given determinant is $${{\lambda }^{3}}\left( {{\lambda }^{2}}+{{a}^{2}}+{{c}^{2}}+{{b}^{2}} \right)^{3}$$ Note: Observing the first determinant involved with the given equation, as a determinant of cofactors of second determinant given in the equation is the key point of the problem. One may try to apply some properties for solving or simplifying the problem, but it will be a complex and much longer approach of the question clear with the cofactors and minors’ concept and their properties with the determinant. Minors can also be written in the same way but here we do not need to multiply ${{\left( -1 \right)}^{m+n}}$ with the determinant formed by leaving the row and column of element of which we are calculated minor/cofactor. It is different from cofactor in this way only. So, take care of it with these kinds of questions.