Matrix Power and Identity

Question

If \[ A = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} \] find \[ (i)\ A^2 – 5A – 14I,\quad (ii)\ A^2 – 5A + 14I,\quad (iii)\ A^3 \]


Solution

Step 1: Compute \(A^2\)

\[ A^2 = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} \]

Step 2: Compute \(A^2 – 5A\)

\[ A^2 – 5A = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} – \begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix} = \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix} = 14I \]

Step 3: Find \(A^2 – 5A – 14I\)

\[ = 14I – 14I = O \]

Step 4: Find \(A^2 – 5A + 14I\)

\[ = 14I + 14I = 28I \]

Step 5: Find \(A^3\)

\[ A^2 = 5A + 14I \] \[ A^3 = A \cdot A^2 = A(5A + 14I) \] \[ = 5A^2 + 14A \] Substitute \(A^2\): \[ = 5(5A + 14I) + 14A = 25A + 70I + 14A = 39A + 70I \]

Final Answers

\[ A^2 – 5A – 14I = O \] \[ A^2 – 5A + 14I = 28I \] \[ A^3 = 39A + 70I \]

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