Matrices & Determinants
Determinants
Grade 12
Question:
<p><strong>For Problems 12 and 13</strong><br>Let for \(A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}\), there be three row matrices \(R_1, R_2\) and \(R_3\), satisfying the relations, \(R_1 A = [1\ 0\ 0]\), \(R_2 A = [2\ 3\ 0]\) and \(R_3 A = [2\ 3\ 1]\). If \(B\) is square matrix of order 3 with rows \(R_1, R_2\) and \(R_3\) in order, then<br><br>The value of det.(B) is</p>
<p>\(-2\)</p>
<p>\(-1\)</p>
<p>\(2\)</p>
<p>\(3\)</p>
Step-by-Step Solution
Key Concept: Since BA = I (where rows of B satisfy RᵢA = eᵢ), matrix B is the left inverse of A, meaning B = A⁻¹. Therefore det(B) = 1/det(A).
<p><strong>Step 1: Recognize the matrix equation structure</strong></p><p>The three row conditions combine into:</p><p>$$\begin{bmatrix} R_1 \\ R_2 \\ R_3 \end{bmatrix} A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 3 & 0 \\ 2 & 3 & 1 \end{bmatrix}$$</p><p>This means: <strong>BA = I</strong> (where B is the matrix with rows R₁, R₂, R₃)</p><p><strong>Step 2: Apply determinant property</strong></p><p>Taking determinant of both sides:</p><p>$$\det(B) \cdot \det(A) = \det(I) = 1$$</p><p>Therefore: $$\det(B) = \frac{1}{\det(A)}$$</p><p><strong>Step 3: Calculate det(A)</strong></p><p>For lower triangular matrix A:</p><p>$$\det(A) = 1 \times 1 \times 1 = 1$$</p><p>Therefore: $$\det(B) = \frac{1}{1} = 1$$</p><p>∴ Answer: <strong>det(B) = 1</strong></p>
Correct Answer: C