Question

If $A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \\ 1 & 3 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 3 & 4 \\ 3 & 2 & 2 \\ 2 & 4 & 2 \end{pmatrix}$, then $\sqrt{|\text{Adj}(AB)|} =$

• A. 176
• B. 208
• C. 198
• D. 234

Hint:

Remember the key properties of determinants and adjoints: $|\text{Adj}(M)| = |M|^{n-1}$ and $|AB| = |A||B|$. These properties save a significant amount of time compared to calculating the matrix product AB and then its adjoint.

Answer 1

The Correct Option is C

We need to find the value of $\sqrt{|\text{Adj}(AB)|}$.
We use the property that for a square matrix $M$ of order $n$, $|\text{Adj}(M)| = |M|^{n-1}$.
In this case, $M = AB$ and the order is $n=3$. So, $|\text{Adj}(AB)| = |AB|^{3-1} = |AB|^2$.
Therefore, $\sqrt{|\text{Adj}(AB)|} = \sqrt{|AB|^2} = |AB|$.
We also use the property that $|AB| = |A| |B|$.
First, let's calculate the determinant of matrix A.
$|A| = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \\ 1 & 3 & 1 \end{vmatrix} = 1(1 \cdot 1 - 1 \cdot 3) - 2(2 \cdot 1 - 1 \cdot 1) + 3(2 \cdot 3 - 1 \cdot 1)$
$|A| = 1(1-3) - 2(2-1) + 3(6-1) = 1(-2) - 2(1) + 3(5) = -2 - 2 + 15 = 11$.
Next, let's calculate the determinant of matrix B.
$|B| = \begin{vmatrix} 2 & 3 & 4 \\ 3 & 2 & 2 \\ 2 & 4 & 2 \end{vmatrix} = 2(2 \cdot 2 - 2 \cdot 4) - 3(3 \cdot 2 - 2 \cdot 2) + 4(3 \cdot 4 - 2 \cdot 2)$
$|B| = 2(4-8) - 3(6-4) + 4(12-4) = 2(-4) - 3(2) + 4(8) = -8 - 6 + 32 = 18$.
Now, we can find $|AB|$.
$|AB| = |A| |B| = 11 \times 18 = 198$.
So, $\sqrt{|\text{Adj}(AB)|} = |AB| = 198$.