Class 12 Matrices · Long Answer3 Marksmatrix_polynomials

Class 12 Maths · Matrices · Long Answer #101

For A=\(\begin{bmatrix}2 & 0 & 1\\2 & 1 & 3\\1 & -1 & 0\end{bmatrix}\), find A²−3A+2I.

✅ Step-by-Step Solution

Final Answer: \(\begin{bmatrix}1 & -1 & -1\\3 & -3 & -4\\-3 & 2 & 0\end{bmatrix}\)

Compute A²=\(\begin{bmatrix}5 & -1 & 2\\9 & -2 & 5\\0 & -1 & -2\end{bmatrix}\).
Compute 3A=\(\begin{bmatrix}6 & 0 & 3\\6 & 3 & 9\\3 & -3 & 0\end{bmatrix}\) and 2I=\(\begin{bmatrix}2 & 0 & 0\\0 & 2 & 0\\0 & 0 & 2\end{bmatrix}\).
Subtract 3A from A² and then add 2I entrywise to get \(\begin{bmatrix}1 & -1 & -1\\3 & -3 & -4\\-3 & 2 & 0\end{bmatrix}\).
Therefore A²−3A+2I=\(\begin{bmatrix}1 & -1 & -1\\3 & -3 & -4\\-3 & 2 & 0\end{bmatrix}\).

[0.5 marks] Compute A²=\(\begin{bmatrix}5 & -1 & 2\\9 & -2 & 5\\0 & -1 & -2\end{bmatrix}\).
[1 mark] Compute 3A=\(\begin{bmatrix}6 & 0 & 3\\6 & 3 & 9\\3 & -3 & 0\end{bmatrix}\) and 2I=\(\begin{bmatrix}2 & 0 & 0\\0 & 2 & 0\\0 & 0 & 2\end{bmatrix}\).
[1 mark] Subtract 3A from A² and then add 2I entrywise to get \(\begin{bmatrix}1 & -1 & -1\\3 & -3 & -4\\-3 & 2 & 0\end{bmatrix}\).
[0.5 marks] Therefore A²−3A+2I=\(\begin{bmatrix}1 & -1 & -1\\3 & -3 & -4\\-3 & 2 & 0\end{bmatrix}\).

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