Determinant Matrix Problems

1. Find the value of x \(\left| \begin{matrix} 3x-8 & 3 & 3 \\ 3 & 3x-8 & 3 \\ 3 & 3 & 3x-8 \\\end{matrix} \right|=0\).

Solution: Given that \(\left| \begin{matrix} 3x-8 & 3 & 3 \\ 3 & 3x-8 & 3 \\ 3 & 3 & 3x-8 \\\end{matrix} \right|=0\),

C₁ → C₁ + C₂ + C₃

\(\left| \begin{matrix} 3x-2 & 3 & 3 \\ 3x-2 & 3x-8 & 3 \\ 3x-2 & 3 & 3x-8 \\\end{matrix} \right|=0\),

\((3x-2)\left| \begin{matrix} 1 & 3 & 3 \\ 1 & 3x-8 & 3 \\ 1 & 3 & 3x-8 \\\end{matrix} \right|=0\),

R₂ → R₂ – R₁, R₃ → R₃ – R₁

\((3x-2)\left| \begin{matrix} 1 & 3 & 3 \\ 0 & 3x-11 & 0 \\ 0 & 0 & 3x-11 \\\end{matrix} \right|=0\),

⇒ (3x – 2) (3x – 11)² = 0,

⇒ x = ⅔ (or) 11/3.

2. Find the x value if \(\left| \begin{matrix} 15-x & 11 & 10 \\ 11-3x & 17 & 16 \\ 7-x & 14 & 13 \\\end{matrix} \right|=0\).

Solution: Given that \(\left| \begin{matrix} 15-x & 11 & 10 \\ 11-3x & 17 & 16 \\ 7-x & 14 & 13 \\\end{matrix} \right|=0\),

R₂ → R₂ – 3R₁, R₃ → R₃ – R₁

\(\left| \begin{matrix} 15-x & 11 & 10 \\ -34 & -16 & -14 \\ -8 & 3 & 3 \\\end{matrix} \right|=0\),

C₂ → C₂ – C₃

\(\left| \begin{matrix} 15-x & 1 & 10 \\ -34 & -2 & -14 \\ -8 & 0 & 3 \\\end{matrix} \right|=0\),

(15 – x) (-6) – 1 ()-102 – 112) + 10 (0 – 16) = 0,

6x = 36,

x = 6.