Matrix Determinant and Adjugate Calculation
This solution provides the steps to calculate the determinant of the adjugate of matrix A, given information about matrix B and its determinant.
1. Calculate Determinant of A
First, find the determinant of matrix $A = \begin{bmatrix} \alpha & 1 & 2 \\ 2 & 3 & 0 \\ 0 & 4 & 5 \end{bmatrix}$.
Using cofactor expansion along the first row:
$ \det(A) = \alpha \begin{vmatrix} 3 & 0 \\ 4 & 5 \end{vmatrix} - 1 \begin{vmatrix} 2 & 0 \\ 0 & 5 \end{vmatrix} + 2 \begin{vmatrix} 2 & 3 \\ 0 & 4 \end{vmatrix} $
$ \det(A) = \alpha(15 - 0) - 1(10 - 0) + 2(8 - 0) $
$ \det(A) = 15\alpha - 10 + 16 $
$ \det(A) = 15\alpha + 6 $
2. Determinant of Adjugate Formula
The determinant of the adjugate matrix is related to the determinant of the original matrix by the formula:
$ \det(\text{adj}(A)) = (\det(A))^{n-1} $
For a $3 \times 3$ matrix ($n=3$), this simplifies to:
$ \det(\text{adj}(A)) = (\det(A))^{2} $
3. Construct and Evaluate Determinant of B
Matrix $B$ is defined as $B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -5\alpha & 0 \\ 0 & 4\alpha & -2\alpha \end{bmatrix} + \text{adj}(A)$.
The adjugate of A is $\text{adj}(A) = \begin{bmatrix} 15 & 3 & -6 \\ -10 & 5\alpha & 4 \\ 8 & -4\alpha & 3\alpha - 2 \end{bmatrix}$.
Adding the matrices to find B:
$ B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -5\alpha & 0 \\ 0 & 4\alpha & -2\alpha \end{bmatrix} + \begin{bmatrix} 15 & 3 & -6 \\ -10 & 5\alpha & 4 \\ 8 & -4\alpha & 3\alpha - 2 \end{bmatrix} = \begin{bmatrix} 16 & 3 & -6 \\ -10 & 0 & 4 \\ 8 & 0 & \alpha - 2 \end{bmatrix} $
Calculate the determinant of B:
$ \det(B) = 16 \begin{vmatrix} 0 & 4 \\ 0 & \alpha - 2 \end{vmatrix} - 3 \begin{vmatrix} -10 & 4 \\ 8 & \alpha - 2 \end{vmatrix} + (-6) \begin{vmatrix} -10 & 0 \\ 8 & 0 \end{vmatrix} $
$ \det(B) = 16(0) - 3((-10)(\alpha - 2) - 32) - 6(0) $
$ \det(B) = -3(-10\alpha + 20 - 32) = -3(-10\alpha - 12) $
$ \det(B) = 30\alpha + 36 $
4. Solve for Alpha
Given that $\det(B) = 66$, we set the expression equal to 66:
$ 30\alpha + 36 = 66 $
Solving for $\alpha$:
$ 30\alpha = 30 $
$ \alpha = 1 $
5. Final Calculation of Det(adj(A))
Substitute the value $\alpha = 1$ back into the expression for $\det(A)$:
$ \det(A) = 15(1) + 6 = 21 $
Now, apply the formula for the determinant of the adjugate matrix:
$ \det(\text{adj}(A)) = (\det(A))^2 = (21)^2 $
$ \det(\text{adj}(A)) = 441 $
Similar Questions
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
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Important Questions from Algebra
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
- Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
- If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
- Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
