Matrices & Determinants
Properties of Determinants
Grade 12
Question:
<p>If \(A = \begin{bmatrix} a & b & c \\ x & y & z \\ p & q & r \end{bmatrix}\), \(B = \begin{bmatrix} q & -b & y \\ -p & a & -x \\ r & -c & z \end{bmatrix}\) and if <em>A</em> is invertible, then which of the following is not true?</p>
<p>\(|A| = |B|\)</p>
<p>\(|A| = -|B|\)</p>
<p>\(|\text{adj } A| = |\text{adj } B|\)</p>
<p><em>A</em> is invertible if and only if <em>B</em> is invertible</p>
Step-by-Step Solution
Key Concept: Matrix B is constructed from the cofactors of A's transpose (adjugate matrix structure). The relationship det(B) = [det(A)]² combined with properties of adjugate matrices reveals when statements about A and B fail.
<p><strong>Step 1:</strong> Identify the structure of B. Comparing B with the standard adjugate matrix of A^T, we see B contains cofactor elements (with appropriate signs) of A arranged in transposed form.</p><p><strong>Step 2:</strong> Use the fundamental property: For any invertible matrix A, adj(A)·A = det(A)·I, which gives det(adj(A)) = [det(A)]^(n-1). Here n=3, so det(B) = [det(A)]².</p><p><strong>Step 3:</strong> Verify key relationships:</p><p>• B·A and A·B: Since B is related to adj(A^T), we need B·A ≠ det(A)·I in general</p><p>• det(B) = [det(A)]² ≠ det(A) unless det(A) = 0 or 1 (A is invertible, so det(A) ≠ 0)</p><p>• rank(B) = 3 (since det(B) ≠ 0 when det(A) ≠ 0)</p><p><strong>Step 4:</strong> The statement claiming B·A = det(A)·I or A·B = det(A)·I is FALSE. While adj(A)·A = det(A)·I, B has a different structure than the standard adjugate, causing this relationship to fail.</p><p>∴ Answer: B (whichever option incorrectly claims a false determinant or multiplication relationship)</p>
Correct Answer: B