Question

If $$ A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & -2 \end{bmatrix} $$ then find the value of: $$ |\operatorname{adj}(\operatorname{adj} A)| $$ 
 

• A. \( 16 \)
• B. \( 256 \)
• C. \( 128 \)
• D. \( -256 \)
• E. \( -16 \)

Hint:

For a \( 3 \times 3 \) matrix, remember the shortcut \( |\adj A|=|A|^2 \). If adjoint appears twice, apply the same rule again carefully.

Answer 1

The Correct Option is B

Step 1: Recall the determinant of an upper triangular matrix.
The given matrix is $$ A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & -2 \end{bmatrix} $$ Since $A$ is an upper triangular matrix, its determinant is the product of diagonal entries: $$ |A| = 1 \cdot 2 \cdot (-2) = -4 $$
Step 2: Recall the formula for determinant of adjoint.
For an $n \times n$ matrix $A$, $$ |\operatorname{adj} A| = |A|^{\,n-1} $$ Here $A$ is a $3 \times 3$ matrix, so $n=3$. Thus, $$ |\operatorname{adj} A| = |A|^2 $$ Substituting $|A| = -4$, we get $$ |\operatorname{adj} A| = (-4)^2 = 16 $$
Step 3: Apply the same formula again to $\operatorname{adj} A$.
Now we need $$ |\operatorname{adj}(\operatorname{adj} A)| $$ Treat $\operatorname{adj} A$ as a new $3 \times 3$ matrix. Using the same formula: $$ |\operatorname{adj} B| = |B|^{\,n-1} $$ So, $$ |\operatorname{adj}(\operatorname{adj} A)| = |\operatorname{adj} A|^2 $$
Step 4: Substitute the value of $|\operatorname{adj} A|$.
$$ |\operatorname{adj} A| = 16 $$ Hence, $$ |\operatorname{adj}(\operatorname{adj} A)| = 16^2 = 256 $$
Step 5: Alternative identity check.
For an invertible $n \times n$ matrix: $$ \operatorname{adj}(\operatorname{adj} A) = |A|^{\,n-2} A $$ For $n=3$: $$ \operatorname{adj}(\operatorname{adj} A) = |A| A $$ Taking determinant: $$ |\operatorname{adj}(\operatorname{adj} A)| = ||A| A| $$ Using $|kA| = k^n |A|$ for $3 \times 3$ matrix: $$ |\operatorname{adj}(\operatorname{adj} A)| = |A|^3 \cdot |A| = |A|^4 $$ Since $|A| = -4$: $$ |\operatorname{adj}(\operatorname{adj} A)| = (-4)^4 = 256 $$
Step 6: Final conclusion.
$$ \boxed{256} $$ Hence, the correct option is: $$ \boxed{(2)\ 256} $$