Question ID: #385
Let A be a $3\times3$ matrix such that $X^T AX=0$ for all nonzero $3\times1$ matrices $X$. If $A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}1\\4\\-5\end{bmatrix}$ and $A\begin{bmatrix}1\\2\\1\end{bmatrix}=\begin{bmatrix}0\\4\\-8\end{bmatrix}$, and $\det(\text{adj}(2(A+I))) = 2^\alpha 3^\beta 5^\gamma$, $\alpha, \beta, \gamma \in \mathbb{N}$, then $\alpha^2+\beta^2+\gamma^2$ is
Solution:
Condition $X^T A X = 0$ for all $X$ implies that $A$ is a Skew-Symmetric Matrix.
Let $A = \begin{bmatrix} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{bmatrix}$.
Given $A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}1\\4\\-5\end{bmatrix}$:
$\begin{bmatrix} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} x+y \\ -x+z \\ -y-z \end{bmatrix} = \begin{bmatrix} 1 \\ 4 \\ -5 \end{bmatrix}$
Equations:
1) $x+y=1$
2) $-x+z=4$
3) $-y-z=-5 \Rightarrow y+z=5$
Given $A\begin{bmatrix}1\\2\\1\end{bmatrix}=\begin{bmatrix}0\\4\\-8\end{bmatrix}$:
$\begin{bmatrix} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 2x+y \\ -x+z \\ -y-2z \end{bmatrix} = \begin{bmatrix} 0 \\ 4 \\ -8 \end{bmatrix}$
Equations:
4) $2x+y=0$
From (1) and (4):
$(2x+y) – (x+y) = 0 – 1 \Rightarrow x = -1$.
Substitute $x=-1$ in (1): $-1 + y = 1 \Rightarrow y = 2$.
Substitute $x=-1$ in (2): $-(-1) + z = 4 \Rightarrow 1+z=4 \Rightarrow z = 3$.
Check consistency with other equations: $y+z = 2+3=5$ (True).
So, matrix $A = \begin{bmatrix} 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 \end{bmatrix}$.
Calculate $B = 2(A+I) = 2 \left( \begin{bmatrix} 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \right) = 2 \begin{bmatrix} 1 & -1 & 2 \\ 1 & 1 & 3 \\ -2 & -3 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -2 & 4 \\ 2 & 2 & 6 \\ -4 & -6 & 2 \end{bmatrix}$.
Let’s find $\det(B) = |B|$.
$|B| = 2^3 \begin{vmatrix} 1 & -1 & 2 \\ 1 & 1 & 3 \\ -2 & -3 & 1 \end{vmatrix}$ (Taking 2 common from each row)
Determinant of inner matrix:
$1(1 – (-9)) – (-1)(1 – (-6)) + 2(-3 – (-2))$
$= 1(10) + 1(7) + 2(-1) = 10 + 7 – 2 = 15$.
So, $|B| = 8 \times 15 = 120$.
We need $\det(\text{adj}(B))$. For a matrix of order $n=3$, $|\text{adj}(B)| = |B|^{n-1} = |B|^2$.
$|\text{adj}(B)| = (120)^2 = 14400$.
However, the question format asks for $2^\alpha 3^\beta 5^\gamma$.
$120 = 2^3 \times 3^1 \times 5^1$.
$|B|^2 = (2^3 \cdot 3 \cdot 5)^2 = 2^6 \cdot 3^2 \cdot 5^2$.
So, $\alpha = 6, \beta = 2, \gamma = 2$.
Calculate $\alpha^2 + \beta^2 + \gamma^2$:
$= 6^2 + 2^2 + 2^2 = 36 + 4 + 4 = 44$.
Ans. (44)
Was this solution helpful?
YesNo