Question

If $A = \begin{pmatrix} 1 & 5 & 2 \\ 4 & 1 & 3 \\ 2 & 6 & 3 \end{pmatrix}$, then $|(\text{Adj A})^{-1}| =$

• A. -1
• B. 1
• C. 4
• D. -4

Hint:

This problem is solved efficiently by knowing determinant properties. Direct calculation of the adjoint, its inverse, and then its determinant would be extremely time-consuming. Always look for ways to use properties like $|M^{-1}|=1/|M|$ and $|\text{Adj A}|=|A|^{n-1}$.

Answer 1

The Correct Option is B

To find the value of \( |(\text{Adj A})^{-1}| \), we first need to understand the formula relating to the determinant and adjugate of a matrix \( A \).

The adjugate matrix, \(\text{Adj A}\), is related to the determinant and the inverse by the formula:

\(A^{-1} = \frac{1}{\det(A)} \cdot \text{Adj } A\)

Consequently, the determinant of the inverse of the adjugate of matrix \( A \) is:

\(|(\text{Adj A})^{-1}| = \frac{1}{|\text{Adj A}|}\)

For an \(n \times n\) matrix, the determinant of the adjugate is related to the determinant of the matrix itself:

\(|\text{Adj A}| = (\det(A))^{n-1}\)

For a \(3 \times 3\) matrix like \( A \), this becomes:

\(|\text{Adj A}| = (\det(A))^{2}\)

Thus:

\(|(\text{Adj A})^{-1}| = \frac{1}{(\det(A))^{2}}\)

Now, let's calculate \(\det(A)\) for matrix \( A = \begin{pmatrix} 1 & 5 & 2 \\ 4 & 1 & 3 \\ 2 & 6 & 3 \end{pmatrix}\):

Using the determinant formula for a \(3 \times 3\) matrix:

\(\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)\)

Substitute the values from matrix \( A \):

\(\det(A) = 1(1 \cdot 3 - 3 \cdot 6) - 5(4 \cdot 3 - 3 \cdot 2) + 2(4 \cdot 6 - 1 \cdot 2)\)

\(= 1(3 - 18) - 5(12 - 6) + 2(24 - 2)\)

\(= 1(-15) - 5(6) + 2(22)\)

\(= -15 - 30 + 44\)

\(\det(A) = -1\)

Now, calculate \(|(\text{Adj A})^{-1}|\):

\(|(\text{Adj A})^{-1}| = \frac{1}{(\det(A))^{2}} = \frac{1}{(-1)^{2}} = \frac{1}{1} = 1\)

Therefore, the correct answer is 1.