Question

If \( A \) is a square matrix of order \(3\) such that \( |2(\operatorname{adj}A)|=288 \), then the value of \( |A| \) is:

• A. \( 144 \)
• B. \( 36 \)
• C. \( \pm 12 \)
• D. \( \pm 6 \)

Hint:

Always remember these two very important determinant properties: \[ |kA|=k^n|A| \] where \(n\) is the order of the matrix. Also, \[ |\operatorname{adj}(A)|=|A|^{n-1} \] For a \(3\times3\) matrix specifically: \[ |\operatorname{adj}(A)|=|A|^2 \] These formulas are extremely common in determinant-based objective questions.

Answer 1

The Correct Option is D

Concept: This problem uses important determinant properties involving scalar multiplication and adjoint matrices. The two standard results required are:
• If \(A\) is a square matrix of order \(n\), then: \[ |kA|=k^n|A| \] where \(k\) is a scalar.
• For a square matrix \(A\) of order \(n\): \[ |\operatorname{adj}(A)|=|A|^{n-1} \] Since the matrix is of order \(3\), we will substitute \(n=3\) carefully into these formulas.

Step 1: Using the determinant property for scalar multiplication We are given: \[ |2(\operatorname{adj}A)|=288 \] Since \(A\) is a matrix of order \(3\), the adjoint matrix \(\operatorname{adj}(A)\) is also of order \(3\). Using: \[ |kB|=k^n|B| \] for an \(n\times n\) matrix \(B\), we get: \[ |2(\operatorname{adj}A)| = 2^3|\operatorname{adj}A| \] because the order is \(3\). Therefore, \[ 8|\operatorname{adj}A|=288 \]

Step 2: Finding the determinant of the adjoint matrix Divide both sides by \(8\): \[ |\operatorname{adj}A| = \frac{288}{8} \] \[ =36 \] Thus, \[ |\operatorname{adj}A|=36 \]

Step 3: Using the adjoint determinant property For a matrix of order \(n\): \[ |\operatorname{adj}A|=|A|^{n-1} \] Since the order is \(3\): \[ |\operatorname{adj}A|=|A|^{3-1} \] \[ |\operatorname{adj}A|=|A|^2 \] But we already found: \[ |\operatorname{adj}A|=36 \] Hence, \[ |A|^2=36 \]

Step 4: Finding the value of \( |A| \) Taking square root on both sides: \[ |A|=\pm \sqrt{36} \] \[ |A|=\pm 6 \] Final Answer: \[ \boxed{|A|=\pm 6} \] Therefore, the correct option is: \[ \boxed{(D)} \]