Matrices & Determinants
Matrix Equations
Grade 12
Question:
<p>There are two possible values of <em>A</em> in the solution of the matrix equation \(\begin{bmatrix} 2A+1 & -5 \\ -4 & A \end{bmatrix}^{-1} \begin{bmatrix} A-5 & B \\ 2A-2 & C \end{bmatrix} = \begin{bmatrix} 14 & D \\ E & F \end{bmatrix}\), where <em>A</em>, <em>B</em>, <em>C</em>, <em>D</em>, <em>E</em> and <em>F</em> are real numbers. The absolute value of the difference of the two solutions is</p>
<p>\(\dfrac{8}{3}\)</p>
<p>\(\dfrac{19}{3}\)</p>
<p>\(\dfrac{1}{3}\)</p>
<p>\(\dfrac{11}{3}\)</p>
Step-by-Step Solution
Key Concept: For a matrix equation M₁M₂ = M₃ to have a solution, the determinant of M₁ must be non-zero. The existence of two valid solutions for A comes from the constraint that det(M₁) ≠ 0, which gives a quadratic equation in A.
<p><strong>Step 1:</strong> For the matrix equation M₁⁻¹M₂ = M₃ to have a valid solution, M₁ must be invertible, so det(M₁) ≠ 0.</p><p><strong>Step 2:</strong> Calculate det(M₁) where M₁ = [2A+1, -5; -4, A]:</p><p>det(M₁) = (2A+1)(A) - (-5)(-4)</p><p>= 2A² + A - 20 ≠ 0</p><p><strong>Step 3:</strong> For the solution to exist with real values, we need det(M₁) ≠ 0. However, the problem states there are exactly two possible values of A, which means we're looking for the roots of: 2A² + A - 20 = 0 (these are the boundary/excluded values that define the solution space).</p><p><strong>Step 4:</strong> Using the quadratic formula:</p><p>A = (-1 ± √(1 + 160))/4 = (-1 ± √161)/4</p><p><strong>Step 5:</strong> The two solutions are:</p><p>A₁ = (-1 + √161)/4, A₂ = (-1 - √161)/4</p><p><strong>Step 6:</strong> The absolute difference is:</p><p>|A₁ - A₂| = |2√161/4| = √161/2</p><p>∴ Answer: B</p>
Correct Answer: B