Question

If \( A = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 1 & 6 \end{pmatrix} \) and \( |adj(adj(A))|(adj A)^{-1} = kA \), then k =

• A. 1296
• B. 216
• C. 36
• D. 432

Hint:

Key identities: \(|adj(adj(A))| = |A|^{(n-1)^2}\) and \((adj A)^{-1} = \frac{A}{|A|}\). For \(n=3\), this leads to \(k = |A|^3\). Calculate \(|A|\) for the given matrix. \(|A| = 1(13) - 2(-4) + 3(-5) = 13+8-15 = 6\). Thus \(k = 6^3 = 216\).

Answer 1

The Correct Option is B

Let \(A\) be an \(n \times n\) non-singular matrix.
We use the properties: 1.
\(|adj(adj(A))| = |A|^{(n-1)^2}\) 2.
\((adj A)^{-1} = \frac{1}{|adj A|} adj(adj A)\).
Also, \((adj A)^{-1} = \frac{A}{|A|}\).
The given equation is \( |adj(adj(A))|(adj A)^{-1} = kA \).
For \(n=3\), \(|adj(adj(A))| = |A|^{(3-1)^2} = |A|^4\).
Substitute the properties into the equation: \( |A|^4 \cdot \left(\frac{A}{|A|}\right) = kA \) \( |A|^{4-1} A = kA \) \( |A|^3 A = kA \) Since \(A\) is given and is not a null matrix, we can equate the scalar parts (assuming \(|A| \neq 0\)).
\( k = |A|^3 \).
Now calculate \(|A|\) for \( A = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 1 & 6 \end{pmatrix} \): \( |A| = 1(3 \cdot 6 - 5 \cdot 1) - 2(1 \cdot 6 - 5 \cdot 2) + 3(1 \cdot 1 - 3 \cdot 2) \) \( |A| = 1(18 - 5) - 2(6 - 10) + 3(1 - 6) \) \( |A| = 1(13) - 2(-4) + 3(-5) \) \( |A| = 13 + 8 - 15 \) \( |A| = 21 - 15 = 6 \).
Since \(|A|=6 \neq 0\), \(A\) is non-singular.
Then, \( k = |A|^3 = 6^3 = 216 \).
\[ \boxed{216} \]