NCERT Class 12 Maths · Chapter 4 Determinants · Exercise 4.4 Question 16

Question

For A=$\left[\begin{matrix}2 & -1 & 1\-1 & 2 & -1\1 & -1 & 2\end{matrix}\right]$, show A³−6A²+9A−4I=0 and hence find A⁻¹.

Accepted Answer

Final Answer: $\left[\begin{matrix}\frac{3}{4} & \frac{1}{4} & - \frac{1}{4}\\frac{1}{4} & \frac{3}{4} & \frac{1}{4}\- \frac{1}{4} & \frac{1}{4} & \frac{3}{4}\end{matrix}\right]$ Step 1: Compute A²=$\left[\begin{matrix}6 & -5 & 5\-5 & 6 & -5\5 & -5 & 6\end{matrix}\right]$. Step 2: Also A³=$\left[\begin{matrix}22 & -21 & 21\-21 & 22 & -21\21 & -21 & 22\end{matrix}\right]$. Step 3: Substituting the powers into the stated polynomial gives $\left[\begin{matrix}0 & 0 & 0\0 & 0 & 0\0 & 0 & 0\end{matrix}\right]$. Step 4: The constant term is nonzero. Rearrange the identity to isolate I as A times a polynomial in A, divided by the negative constant term. Step 5: Explicitly, isolate the inverse as A⁻¹=(A²−6A+9I)/4. The resulting inverse is $\left[\begin{matrix}\frac{3}{4} & \frac{1}{4} & - \frac{1}{4}\\frac{1}{4} & \frac{3}{4} & \frac{1}{4}\- \frac{1}{4} & \frac{1}{4} & \frac{3}{4}\end{matrix}\right]$. Multiplying by A verifies the identity matrix.

NCERT Class 12 Maths · Chapter 4 Determinants · Exercise 4.4 Question 16

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