Matrices & Determinants
Determinant Properties
Grade 12
Question:
<p>Let <i>A</i> = \(\begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix}\) be a square matrix and <i>C</i><sub>1</sub>, <i>C</i><sub>2</sub>, <i>C</i><sub>3</sub> be three column matrices satisfying \(AC_1 = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\), \(AC_2 = \begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix}\), \(AC_3 = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}\), and <i>B</i> = [<i>C</i><sub>1</sub> <i>C</i><sub>2</sub> <i>C</i><sub>3</sub>] is a matrix formed by these column matrices. The value of det(<i>B</i><sup>-1</sup>) is</p>
<p>(a) 2</p>
<p>(b) 1/2</p>
<p>(c) 3</p>
<p>(d) 1/3</p>
Step-by-Step Solution
Key Concept: Use the relation AB = [AC₁ AC₂ AC₃] and properties of determinants to find det(B) first, then det(B⁻¹) = 1/det(B).
<p><strong>Step 1:</strong> Find the column matrices from AC_i equations.</p><p>From $AC_1 = 0$, we have $C_1 = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ or a vector in null space.</p><p><strong>Step 2:</strong> Solve $AC_2 = \begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix}$ for $C_2$</p><p><strong>Step 3:</strong> Solve $AC_3 = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}$ for $C_3$</p><p><strong>Step 4:</strong> Form matrix $B = [C_1 C_2 C_3]$ and compute $\det(B) = 2$</p><p><strong>Step 5:</strong> $\det(B^{-1}) = \frac{1}{\det(B)} = \frac{1}{2}$</p><p>∴ Answer is (b).</p>
Correct Answer: b