Matrices & Determinants
System of Linear Equations
Grade 12
Question:
<p>Let \(A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 0 & 5 \\ 0 & 2 & 1 \end{bmatrix}\) and \(B = \begin{bmatrix} 0 \\ -3 \\ 1 \end{bmatrix}\). Which of the following is true?</p>
<p>\(AX = B\) has a unique solution</p>
<p>\(AX = B\) has exactly three solutions</p>
<p>\(AX = B\) has infinitely many solutions</p>
<p>\(AX = B\) is inconsistent</p>
Step-by-Step Solution
Key Concept: To solve AX = B, you must find A⁻¹ and compute X = A⁻¹B. This requires calculating det(A) and the adjugate matrix using cofactor expansion.
<p><strong>Step 1:</strong> Calculate det(A) using cofactor expansion along Row 1:</p><p>det(A) = 1(0·1 - 5·2) - 2(2·1 - 5·0) + 3(2·2 - 0·0)</p><p>= 1(-10) - 2(2) + 3(4) = -10 - 4 + 12 = -2</p><p><strong>Step 2:</strong> Since det(A) ≠ 0, A is invertible. Calculate the cofactor matrix C:</p><p>C₁₁ = -10, C₁₂ = -2, C₁₃ = 4</p><p>C₂₁ = 4, C₂₂ = 1, C₂₃ = -2</p><p>C₃₁ = 10, C₃₂ = 1, C₃₃ = -4</p><p><strong>Step 3:</strong> adj(A) = C^T = <begin>bmatrix</begin> -10 & 4 & 10 \\ -2 & 1 & 1 \\ 4 & -2 & -4 \end{bmatrix}</p><p><strong>Step 4:</strong> A⁻¹ = (1/-2)adj(A) = <begin>bmatrix</begin> 5 & -2 & -5 \\ 1 & -1/2 & -1/2 \\ -2 & 1 & 2 \end{bmatrix}</p><p><strong>Step 5:</strong> X = A⁻¹B = <begin>bmatrix</begin> 5 & -2 & -5 \\ 1 & -1/2 & -1/2 \\ -2 & 1 & 2 \end{bmatrix} <begin>bmatrix</begin> 0 \\ -3 \\ 1 \end{bmatrix} = <begin>bmatrix</begin> 6-5 \\ 3/2-1/2 \\ -3+2 \end{bmatrix} = <begin>bmatrix</begin> 1 \\ 1 \\ -1 \end{bmatrix}</p><p>∴ Answer: A</p>
Correct Answer: A