Evaluate \(3A^2 – 2B + I\)

Solution:

Step 1: Compute \(A^2\)

\[ A^2 = \begin{bmatrix}2 & -1 \\ 3 & 2\end{bmatrix} \begin{bmatrix}2 & -1 \\ 3 & 2\end{bmatrix} = \begin{bmatrix} 2(2)+(-1)(3) & 2(-1)+(-1)(2) \\ 3(2)+2(3) & 3(-1)+2(2) \end{bmatrix} \] \[ = \begin{bmatrix} 4-3 & -2-2 \\ 6+6 & -3+4 \end{bmatrix} = \begin{bmatrix} 1 & -4 \\ 12 & 1 \end{bmatrix} \]

Step 2: Compute \(3A^2\)

\[ 3A^2 = \begin{bmatrix} 3 & -12 \\ 36 & 3 \end{bmatrix} \]

Step 3: Compute \(2B\)

\[ 2B = \begin{bmatrix} 0 & 8 \\ -2 & 14 \end{bmatrix} \]

Step 4: Compute \(3A^2 – 2B\)

\[ = \begin{bmatrix} 3-0 & -12-8 \\ 36-(-2) & 3-14 \end{bmatrix} = \begin{bmatrix} 3 & -20 \\ 38 & -11 \end{bmatrix} \]

Step 5: Add Identity Matrix \(I\)

\[ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] \[ 3A^2 – 2B + I = \begin{bmatrix} 3+1 & -20+0 \\ 38+0 & -11+1 \end{bmatrix} = \begin{bmatrix} 4 & -20 \\ 38 & -10 \end{bmatrix} \]

Final Answer:

\[ \boxed{ \begin{bmatrix} 4 & -20 \\ 38 & -10 \end{bmatrix} } \]