Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be the linear transformation which reflects every vector in $\mathbb{R}^3$ through a two-dimensional subspace of $\mathbb{R}^3$. Let $P \in \mathbb{R}^{3 \times 3}$ be the matrix representation of $T$ using the basis
\[ \left\{ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right\}. \]
Then the value of $2 \times \text{trace}(P) - 3 \times \text{determinant}(P)$ is equal to ______. (answer in integer)

Question

Let $A$ be a square matrix. If $A^2 = A$, then the matrix $A$ is called:

Answer 1

Instead of arguing through eigenvalues in general, we can just pick one concrete two-dimensional subspace, write down the actual matrix of $T$ for it, and compute trace and determinant directly. Since the answer does not depend on which plane is chosen, any valid choice will do.

Choose a plane: Take the two-dimensional subspace to be the $xy$-plane, that is, all vectors of the form $(x, y, 0)$.
Write down what $T$ does: Reflecting through the $xy$-plane keeps the $x$ and $y$ coordinates the same and flips the sign of the $z$ coordinate, since $z$ is measured perpendicular to this plane. So $T(x,y,z) = (x, y, -z)$.
Write the matrix $P$: Using the standard basis $(1,0,0), (0,1,0), (0,0,1)$, this transformation has the matrix $$P = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}.$$
Compute trace and determinant: The trace is the sum of the diagonal entries, $1+1+(-1)=1$. The determinant of this diagonal matrix is the product of the diagonal entries, $1 \times 1 \times (-1) = -1$.

Now substitute these into the required expression: $2 \times \text{trace}(P) - 3 \times \text{determinant}(P) = 2(1) - 3(-1) = 2+3 = 5$.

Let's summarize:

Reflection through any plane in $\mathbb{R}^3$ always has trace 1 and determinant $-1$, since two directions stay fixed and one direction flips sign.
The specific plane chosen does not matter; only the type of transformation matters.

So the required value is 5 for any choice of plane.

$$\boxed{5}$$

Answer 2

Instead of arguing through eigenvalues in general, we can just pick one concrete two-dimensional subspace, write down the actual matrix of $T$ for it, and compute trace and determinant directly. Since the answer does not depend on which plane is chosen, any valid choice will do.

Choose a plane: Take the two-dimensional subspace to be the $xy$-plane, that is, all vectors of the form $(x, y, 0)$.
Write down what $T$ does: Reflecting through the $xy$-plane keeps the $x$ and $y$ coordinates the same and flips the sign of the $z$ coordinate, since $z$ is measured perpendicular to this plane. So $T(x,y,z) = (x, y, -z)$.
Write the matrix $P$: Using the standard basis $(1,0,0), (0,1,0), (0,0,1)$, this transformation has the matrix $$P = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}.$$
Compute trace and determinant: The trace is the sum of the diagonal entries, $1+1+(-1)=1$. The determinant of this diagonal matrix is the product of the diagonal entries, $1 \times 1 \times (-1) = -1$.

Now substitute these into the required expression: $2 \times \text{trace}(P) - 3 \times \text{determinant}(P) = 2(1) - 3(-1) = 2+3 = 5$.

Let's summarize:

Reflection through any plane in $\mathbb{R}^3$ always has trace 1 and determinant $-1$, since two directions stay fixed and one direction flips sign.
The specific plane chosen does not matter; only the type of transformation matters.

So the required value is 5 for any choice of plane.

$$\boxed{5}$$

Show Hint

Correct Answer: 5

Solution and Explanation

Questions Asked in GATE MA exam