Question

If $A = \begin{bmatrix} 2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2 \end{bmatrix}$ and $P = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 2 \\ 0 & 1 & 5 \end{bmatrix}$, then $\left|P^{-1}AP - 2I\right| =$

• A. $55$
• B. $69$
• C. $96$
• D. $38$

Hint:

For any scalar $k$ and similarity transformation, $\left|P^{-1}AP - kI\right| = \left|A - kI\right|$. You never need to calculate $P^{-1}$ or perform full matrix multiplication in such problems!

Answer 1

The Correct Option is B

Concept: The problem asks for the determinant of the matrix expression $P^{-1}AP - 2I$. We can simplify this expression using properties of matrix multiplication and determinants. Specifically, the identity matrix $I$ commutes with any matrix, so we can write $2I = P^{-1}(2I)P$. This allows us to factor out $P^{-1}$ and $P$: \[ P^{-1}AP - 2I = P^{-1}AP - P^{-1}(2I)P = P^{-1}(A - 2I)P \] Taking the determinant on both sides and using the distributive property of determinants over multiplication ($\left|XYZ\right| = \left|X\right|\left|Y\right|\left|Z\right|$), we get: \[ \left|P^{-1}AP - 2I\right| = \left|P^{-1}(A - 2I)P\right| = \left|P^{-1}\right| \cdot \left|A - 2I\right| \cdot \left|P\right| \] Since $\left|P^{-1}\right| = \frac{1}{\left|P\right|}$, the terms $\left|P^{-1}\right|$ and $\left|P\right|$ cancel out, leaving: \[ \left|P^{-1}AP - 2I\right| = \left|A - 2I\right| \] Thus, we only need to compute the determinant of the matrix $A - 2I$.

Step 1: Compute the matrix $A - 2I$.
Given the matrix $A$: \[ A = \begin{bmatrix} 2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2 \end{bmatrix} \] The scalar multiple $2I$ of the $3 \times 3$ identity matrix is: \[ 2I = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} \] Subtracting $2I$ from $A$ entry-wise: \[ A - 2I = \begin{bmatrix} 2-2 & 1-0 & 2-0 \\ 6-0 & 2-2 & 11-0 \\ 3-0 & 3-0 & 2-2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0 \end{bmatrix} \]

Step 2: Calculate the determinant $\left|A - 2I\right|$.
We expand the determinant along the first row: \[ \left|A - 2I\right| = \begin{vmatrix} 0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0 \end{vmatrix} \] \[ = 0 \cdot \begin{vmatrix} 0 & 11 \\ 3 & 0 \end{vmatrix} - 1 \cdot \begin{vmatrix} 6 & 11 \\ 3 & 0 \end{vmatrix} + 2 \cdot \begin{vmatrix} 6 & 0 \\ 3 & 3 \end{vmatrix} \] Evaluating each $2 \times 2$ determinant: \[ \begin{vmatrix} 6 & 11 \\ 3 & 0 \end{vmatrix} = (6 \cdot 0) - (11 \cdot 3) = 0 - 33 = -33 \] \[ \begin{vmatrix} 6 & 0 \\ 3 & 3 \end{vmatrix} = (6 \cdot 3) - (0 \cdot 3) = 18 - 0 = 18 \] Substituting these values back into our expansion: \[ \left|A - 2I\right| = 0 - 1(-33) + 2(18) \] \[ = 33 + 36 = 69 \] Therefore, $\left|P^{-1}AP - 2I\right| = 69$.