Let the determinant of a square matrix A of order $ m $ be $ m - n $, where $ m $ and $ n $ satisfy $ 4m + n = 22 $ and $ 17m + 4n = 93 $. If $ \text{det} (n \, \text{adj}(\text{adj}(mA))) = 3^a 5^b 6^c $, then $ a + b + c $ is equal to:

Question

The value of the determinant $\begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix}$ is:

Answer 1

To solve the problem, let's break it down step-by-step:

We are given that the determinant of matrix $ A $ is $ m-n $. The equations given are:

$4m + n = 22$

$17m + 4n = 93$

We can solve these two equations simultaneously to find the values of $ m $ and $ n $.

Multiply the first equation by 4 to eliminate $ n $:

$16m + 4n = 88$

Subtract this new equation from the second equation:

$(17m + 4n) - (16m + 4n) = 93 - 88$

$m = 5$

Substitute $ m = 5 $ back into the first equation:

$4(5) + n = 22$

$20 + n = 22$

$n = 2$

We have found $ m = 5 $ and $ n = 2 $. The determinant of matrix $ A $ is therefore $ 5 - 2 = 3 $.

Now, we need to find $\text{det}(n \, \text{adj}(\text{adj}(mA)))$:

The formula for the determinant of the adjugate of a matrix is given by:

$\text{det(adj}(A)) = (\text{det}(A))^{m-1}$

Therefore, for matrix $ mA $:

$\text{det}(mA) = m^m \cdot \text{det}(A)$

$\text{det}(mA) = 5^5 \cdot 3$

Since $\text{dim}(A) = m = 5$, $\text{det(adj}(mA)) = (\text{det}(mA))^{4}$.

$[ \text{det}(mA) ]^{4} = [ 5^{5} \cdot 3 ]^{4}$

$= 5^{20} \cdot 3^{4}$

Hence, $\text{det}(n \, \text{adj}(\text{adj}(mA))) = n^{5m-5} \cdot (\text{det}(mA))^{(5m-6)}$.

Evaluating for given values, we find:

Final calculation with powers:

$\text{det}(n \text{ adj}(\text{adj}(mA)) = [2^{20} \cdot 5^{20} \cdot 3^{4} \cdot 5]$

$= 2^{20} \cdot 5^{25} \cdot 3^5 \div 3$

After simplifying, the product is given by:

$= 3^4 \cdot 5^{25} \cdot 2^{20}$

Finally, for the given expression:

$a+b+c = 96$

Hence, the correct answer is 96.

Let the determinant of a square matrix A of order \( m \) be \( m - n \), where \( m \) and \( n \) satisfy \( 4m + n = 22 \) and \( 17m + 4n = 93 \). If \( \text{det} (n \, \text{adj}(\text{adj}(mA))) = 3^a 5^b 6^c \), then \( a + b + c \) is equal to: