Question

Let \( A \) be a square matrix of order 3 such that \( \text{det}(A) = -2 \) and \( \text{det}(3 \cdot \text{adj}(-6 \cdot \text{adj}(3A))) = 2^{m+n} \cdot 3^{mn} \), where \( m > n \). Then \( 4m + 2n \) is equal to:

Hint:

For matrix determinant properties, remember that the determinant of a product is the product of determinants, and the properties of adjugates and scalar multiplication can simplify calculations.

Answer 1

Correct Answer: 34

We are given: \[ |A| = -2 \] \[ \text{det}(3 \cdot \text{adj}(-6 \cdot \text{adj}(3A))) = 3^3 \cdot \text{det}(\text{adj}(-\text{adj}(3A))) \] Simplifying this: \[ \text{det}(3 \cdot \text{adj}(-6 \cdot \text{adj}(3A))) = 3^3 \cdot 6^6 \cdot \text{det}(3A)^4 \] We know that: \[ 3^{21} \cdot 2^{10} = 3^7 \cdot 2^3 \] Thus, solving for \( m \) and \( n \), we find: \[ m + n = 10, \quad mn = 21 \] Solving for \( m \) and \( n \), we get: \[ m = 7, \quad n = 3 \] Therefore, \( 4m + 2n = 4 \times 7 + 2 \times 3 = 28 + 6 = 34 \).

Answer 2

Use the facts for $3\times3$ matrices:

\(\det(kM)=k^{3}\det(M)\) for scalar \(k\).

\(\det(\operatorname{adj}M)=\det(M)^{2}\) (since \(\det(\operatorname{adj}M)=\det(M)^{n-1}\) and \(n=3\)).

\(\operatorname{adj}(kM)=k^{2}\operatorname{adj}M\) for \(n=3\) (because adj scales by \(k^{n-1}\)).

Start from the innermost expression: \(3A\). \[ \det(3A)=3^{3}\det(A)=27\cdot(-2)=-54. \]

Next compute \(\operatorname{adj}(3A)\). We only need its determinant: \[ \det\big(\operatorname{adj}(3A)\big)=\det(3A)^{2}=(-54)^{2}=54^{2}=2916. \]

Form \(D=-6\cdot\operatorname{adj}(3A)\). Its determinant is \[ \det(D)=(-6)^{3}\det\big(\operatorname{adj}(3A)\big)=(-216)\cdot 2916. \] Instead of multiplying numerically, factor powers of \(2\) and \(3\): \[ -216 = -2^{3}3^{3},\qquad 2916=2^{2}3^{6}. \] Thus \[ \det(D)=-2^{3+2}\,3^{3+6}=-2^{5}3^{9}. \]

Now take \(\operatorname{adj}(D)\). Its determinant is \[ \det\big(\operatorname{adj}(D)\big)=\det(D)^{2}=(2^{5}3^{9})^{2}=2^{10}3^{18}. \]

Finally consider \(3\cdot\operatorname{adj}(D)\). Its determinant is \[ \det\big(3\cdot\operatorname{adj}(D)\big)=3^{3}\det\big(\operatorname{adj}(D)\big) =3^{3}\cdot 2^{10}3^{18}=2^{10}3^{21}. \]

Compare with the given form \(2^{m+n}3^{mn}\). Hence \[ m+n=10,\qquad mn=21. \] The pair of positive integers with product \(21\) and sum \(10\) is \((m,n)=(7,3)\) (since \(7>3\)).

Therefore \[ 4m+2n=4\cdot7+2\cdot3=28+6=\boxed{34}. \]

Answer

$4m+2n=34$