Question

Let $\Delta_0$ = $\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{32} & a_{32} & a_{33} \end{vmatrix}$ 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 at $\Delta_1$ similarly $\Delta_n$ denotes the determinant formed by the cofactors at $\Delta_{n-1}$ then the determinant value of $\Delta_n$ is

• A. $\Delta_0^{2n}$
• B. $\Delta_0^{2^n}$
• C. $\Delta_0^{n^2}$
• D. $\Delta_0^2$

Answer 1

Answer: (B)

$\Delta_0^{2^n}$

Answer 2

For a 3×3 matrix $A$ with determinant $\Delta_0$, the cofactor matrix is the adjugate $\text{adj}(A)$. A known result is:

$$\det(\text{adj}(A)) = (\det A)^{3-1} = (\Delta_0)^2.$$

Thus, if we denote by $\Delta_1$ the determinant of the matrix of cofactors of $A$, then:

$$\Delta_1 = (\Delta_0)^2.$$

Now, the process is repeated. For an invertible 3×3 matrix $B$, the determinant of its cofactor matrix is also $(\det B)^2$. Therefore:

$$\Delta_2 = (\Delta_1)^2 = \left[(\Delta_0)^2\right]^2 = (\Delta_0)^4.$$

Continuing in this manner, we see that:

$$\Delta_n = (\Delta_{n-1})^2 = (\Delta_0)^{2^n}.$$

Therefore, the correct answer is $\Delta_0^{2^n}$.