NCERT Class 12 Maths · Chapter 4 Determinants · Exercise 4.4 Question 13

Question

For A=$\left[\begin{matrix}3 & 1\-1 & 2\end{matrix}\right]$, show A²−5A+7I=0 and hence find A⁻¹.

Accepted Answer

Final Answer: $\left[\begin{matrix}\frac{2}{7} & - \frac{1}{7}\\frac{1}{7} & \frac{3}{7}\end{matrix}\right]$ Step 1: Compute A²=$\left[\begin{matrix}8 & 5\-5 & 3\end{matrix}\right]$. Step 2: Substituting the powers into the stated polynomial gives $\left[\begin{matrix}0 & 0\0 & 0\end{matrix}\right]$. Step 3: 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 4: Explicitly, isolate the inverse as A⁻¹=(5I−A)/7. The resulting inverse is $\left[\begin{matrix}\frac{2}{7} & - \frac{1}{7}\\frac{1}{7} & \frac{3}{7}\end{matrix}\right]$. Multiplying by A verifies the identity matrix.

NCERT Class 12 Maths · Chapter 4 Determinants · Exercise 4.4 Question 13

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