Question

Let $A=\left[\begin{matrix}1& 1& 1\\ 2& 2& 3\\ x& y& 2\end{matrix}\right]$ and let $V=\{(x,y,z)\in R^{3}:\det(A)=0\}$. Then the dimension of V equals}

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

A single linear constraint in $\mathbb{R}^n$ always defines a subspace of dimension $n-1$.

Answer 1

The Correct Option is C

Step 1: Concept
The determinant of a $3\times3$ matrix is calculated using the expansion method. The condition $\det(A)=0$ defines a subspace in $\mathbb{R}^3$.

Step 2: Meaning
Calculate $\det(A)$: $1(2(2)-3(y)) - 1(2(2)-3(x)) + 1(2y-2x) = 0$. $4-3y - (4-3x) + 2y-2x = 0$. $4-3y-4+3x+2y-2x = 0$. $x-y=0$.

Step 3: Analysis
The equation $x-y=0$ (or $x=y$) represents a plane in $\mathbb{R}^3$. Note that the variable $z$ is free, so any point $(x,x,z)$ satisfies the condition.

Step 4: Conclusion
A plane in $\mathbb{R}^3$ is a 2-dimensional subspace. Therefore, the dimension of $V$ equals 2. Final Answer: (C)