Matrices & Determinants
Matrices And Determinants
nta_abhyas_2025
Grade 12
Question:
If $\alpha$ and $\beta$ are the roots of the equation $|\begin{matrix} 1 & 5 \\ -4 & 7 \end{matrix}| \begin{matrix} \frac{x}{19} & -\frac{3}{19} \\ \frac{1}{19} & \frac{5}{19} \end{matrix} = \begin{matrix} 1 & 3 \\ -4 & 7 \end{matrix} \begin{matrix} \frac{x^2 - 5x + 6}{-3} \end{matrix} = -4|$, then the value of $(2 - \alpha)(2 - \beta)$ is
Step-by-Step Solution
Key Concept: The inverse of a $2 \times 2$ matrix uses the formula $A^{-1} = \frac{1}{\det(A)}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
For the matrix $A = \begin{bmatrix} 1 & 3 \\ -4 & 7 \end{bmatrix}$, we compute $\det(A) = 1(7) - 3(-4) = 7 + 12 = 19$. Then $A^{-1} = \frac{1}{19}\begin{bmatrix} 7 & -3 \\ 4 & 1 \end{bmatrix}$. Computing $AA^{-1} = \begin{bmatrix} 1 & 3 \\ -4 & 7 \end{bmatrix}\begin{bmatrix} \frac{7}{19} & -\frac{3}{19} \\ \frac{4}{19} & \frac{1}{19} \end{bmatrix} = I$. From $(2-\omega)(2-\bar{\omega}) = 4 - \omega - \bar{\omega}(2) + \omega\bar{\omega} = 4 - 5 - (-4) \times 3 = -12$.
Correct Answer: -12