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

We need to find the value of $|(\text{Adj A})^{-1}|$.
We use the property that for any invertible matrix $M$, $|M^{-1}| = \frac{1}{|M|}$.
So, $|(\text{Adj A})^{-1}| = \frac{1}{|\text{Adj A}|}$.
We also use the property that for a square matrix $A$ of order $n$, $|\text{Adj A}| = |A|^{n-1}$.
Here, the order is $n=3$, so $|\text{Adj A}| = |A|^{3-1} = |A|^2$.
Combining these properties, we get $|(\text{Adj A})^{-1}| = \frac{1}{|A|^2}$.
Now, we need to calculate the determinant of matrix A.
$|A| = \begin{vmatrix} 1 & 5 & 2 \\ 4 & 1 & 3 \\ 2 & 6 & 3 \end{vmatrix} = 1(1 \cdot 3 - 3 \cdot 6) - 5(4 \cdot 3 - 3 \cdot 2) + 2(4 \cdot 6 - 1 \cdot 2)$
$|A| = 1(3-18) - 5(12-6) + 2(24-2) = 1(-15) - 5(6) + 2(22) = -15 - 30 + 44 = -1$.
Finally, we substitute the value of $|A|$ into our expression.
$|(\text{Adj A})^{-1}| = \frac{1}{(-1)^2} = \frac{1}{1} = 1$.