Matrices & Determinants
Adjoint of a matrix
Grade 12
Question:
<p>If \( A = \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix} \), then \( \text{adj}(3A^2 + 12A) \) is equal to</p>
<p>\(\begin{bmatrix} 51 & 63 \\ 84 & 72 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 51 & 84 \\ 63 & 72 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 72 & -63 \\ -84 & 51 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 72 & -84 \\ -63 & 51 \end{bmatrix}\)</p>
Step-by-Step Solution
Key Concept: Factor out the scalar from the matrix expression to use adj(kB) = k^(n-1)·adj(B) for n×n matrices, then compute the adjugate of the resulting matrix.
<p><strong>Step 1:</strong> Calculate A²</p><p>A² = ⎡2 -3⎤ ⎡2 -3⎤ = ⎡4+12 -6-3⎤ = ⎡16 -9⎤</p><p> ⎣-4 1⎦ ⎣-4 1⎦ ⎣-8-4 12+1⎦ ⎣-12 13⎦</p><p><strong>Step 2:</strong> Compute 3A² + 12A</p><p>3A² = ⎡48 -27⎤, 12A = ⎡24 -36⎤</p><p> ⎣-36 39⎦ ⎣-48 12⎦</p><p>3A² + 12A = ⎡72 -63⎤ = 9⎡8 -7⎤</p><p> ⎣-84 51⎦ ⎣-28/3 17/3⎦</p><p><strong>Step 3:</strong> Factor out scalar: 3A² + 12A = 9(A² + 4A) = 9⎡8 -7⎤</p><p> ⎣-28/3 17/3⎦</p><p><strong>Step 4:</strong> Use adj(kB) = k^(n-1)·adj(B) where n=2, k=9</p><p>adj(9B) = 9¹·adj(B) = 9·adj(B)</p><p><strong>Step 5:</strong> For B = ⎡8 -7⎤, adj(B) = ⎡17/3 7⎤</p><p> ⎣-28/3 17/3⎦ ⎣28/3 8⎦</p><p><strong>Step 6:</strong> adj(3A² + 12A) = 9·adj(B) = ⎡51 63⎤</p><p> ⎣84 72⎦</p><p>∴ Answer: D</p>
Correct Answer: D