Class 12 Matrices · Long Answer3 Marksmatrix_polynomials_and_inverse
Class 12 Maths · Matrices · Long Answer #70
For A=\(\begin{bmatrix}1 & 2 & 3\\3 & -2 & 1\\4 & 2 & 1\end{bmatrix}\), prove A³−23A−40I=O.
✅ Step-by-Step Solution
Final Answer: A³−23A−40I=O
- Compute A²=\(\begin{bmatrix}19 & 4 & 8\\1 & 12 & 8\\14 & 6 & 15\end{bmatrix}\).
- Multiply once more to obtain A³=\(\begin{bmatrix}63 & 46 & 69\\69 & -6 & 23\\92 & 46 & 63\end{bmatrix}\).
- Independently, 23A+40I=\(\begin{bmatrix}63 & 46 & 69\\69 & -6 & 23\\92 & 46 & 63\end{bmatrix}\), which equals the calculated A³ entry by entry.
- Therefore A³−23A−40I is the zero 3×3 matrix.
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] Compute A²=\(\begin{bmatrix}19 & 4 & 8\\1 & 12 & 8\\14 & 6 & 15\end{bmatrix}\).
- [1 mark] Multiply once more to obtain A³=\(\begin{bmatrix}63 & 46 & 69\\69 & -6 & 23\\92 & 46 & 63\end{bmatrix}\).
- [1 mark] Independently, 23A+40I=\(\begin{bmatrix}63 & 46 & 69\\69 & -6 & 23\\92 & 46 & 63\end{bmatrix}\), which equals the calculated A³ entry by entry.
- [0.5 marks] Therefore A³−23A−40I is the zero 3×3 matrix.
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