Question

Let ${\Delta }_{\text{o}}=$ $\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]$ and let ${\Delta }_1$ denote the determinant formed by the cofactors of elements of ${\Delta }_0$ and ${\Delta }_2$ denote the determinant formed by the cofactor of ${\Delta }_1,$ similarly ${\Delta }_n$ denotes the determinant formed by the cofactors of ${\Delta }_{n-1}$ then the determinant value of ${\Delta }_n$ is

• A. ${{\Delta }_0}^{2n}$
• B. ${{\Delta }_0}^{2^n}$
• C. ${{\Delta }_0}^{n^2}$
• D. ${{\Delta }^2}_0$

Answer 1

The correct option is B ${{\Delta }_0}^{2^n}$

Let $A$ be a matrix of order $m\times m$ and $C$ be its co-factor matrix

We know that $A.adj\left(A\right)=\left|A\right|.I$

$\Rightarrow \left|A\right|\left|adj\left(A\right)\right|={\left|A\right|}^m.\left|I\right|$

$\Rightarrow \left|A\right|\left|adj\left(A\right)\right|={\left|A\right|}^m$ ......$\left(1\right)$ since $\left|I\right|=1$

But $adj\left(A\right)=C^T$

$\therefore \left|adj\left(A\right)\right|=\left|C^T\right|=\left|C\right|$ .......$\left(2\right)$ since $\left|C\right|=\left|C^T\right|$

From $\left(1\right)$ and $\left(2\right)$ we have

$\left|A\right|\left|C\right|={\left|A\right|}^m$

$\Rightarrow \left|C\right|={\left|A\right|}^{m-1}$

Let ${\Delta }_i$ be the cofactor matrix then

${\Delta }_i={\left({\Delta }_{i-1}\right)}^{m-1}$ where $m=3$

$\Rightarrow {\Delta }_i={\left({\Delta }_{i-1}\right)}^{3-1}$

$\Rightarrow {\Delta }_i={\left({\Delta }_{i-1}\right)}^2$

$\Rightarrow {\Delta }_n={\left({\Delta }_{n-1}\right)}^2={\left({\left({\Delta }_{n-2}\right)}^2\right)}^2={\left({\left({\left({\Delta }_{n-3}\right)}^2\right)}^2\right)}^2$ and so on.

$\Rightarrow {\Delta }_{n-1}={\left({\Delta }_{n-2}\right)}^2$

$\Rightarrow {\Delta }_{n-2}={\left({\Delta }_{n-3}\right)}^2$

and so on.

Hence ${\Delta }_0={\left({\Delta }_0\right)}^{2^n}$