NCERT Class 12 Maths · Chapter 4 Determinants · Exercise 4.4 Question 15

Question

For A=$\left[\begin{matrix}1 & 1 & 1\1 & 2 & -3\2 & -1 & 3\end{matrix}\right]$, show A³−6A²+5A+11I=0 and hence find A⁻¹.

Accepted Answer

Final Answer: $\left[\begin{matrix}- \frac{3}{11} & \frac{4}{11} & \frac{5}{11}\\frac{9}{11} & - \frac{1}{11} & - \frac{4}{11}\\frac{5}{11} & - \frac{3}{11} & - \frac{1}{11}\end{matrix}\right]$ Step 1: Compute A²=$\left[\begin{matrix}4 & 2 & 1\-3 & 8 & -14\7 & -3 & 14\end{matrix}\right]$. Step 2: Also A³=$\left[\begin{matrix}8 & 7 & 1\-23 & 27 & -69\32 & -13 & 58\end{matrix}\right]$. 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−5I)/11. The resulting inverse is $\left[\begin{matrix}- \frac{3}{11} & \frac{4}{11} & \frac{5}{11}\\frac{9}{11} & - \frac{1}{11} & - \frac{4}{11}\\frac{5}{11} & - \frac{3}{11} & - \frac{1}{11}\end{matrix}\right]$. Multiplying by A verifies the identity matrix.

NCERT Class 12 Maths · Chapter 4 Determinants · Exercise 4.4 Question 15

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