سوال ۲۶ کنکور ارشد علوم کامپیوتر ۱۴۰۲

Use the Cayley-Hamilton theorem: every matrix satisfies its own characteristic polynomial. The fastest route to $A^{-1}$ is to build the characteristic polynomial from the three invariants of $A$.

Step 1 - Invariants of $A$. For a $3\times 3$ matrix the characteristic polynomial is $p(\lambda) = \lambda^{3} - (\operatorname{tr} A)\,\lambda^{2} + (m_{2})\,\lambda - \det A,$ where $m_{2}$ is the sum of the three principal $2\times 2$ minors.

• $\operatorname{tr} A = 2 + 0 + 4 = 6.$
• $m_{2} = \det\begin{bmatrix}0 & 1\\-1 & 4\end{bmatrix} + \det\begin{bmatrix}2 & 1\\3 & 4\end{bmatrix} + \det\begin{bmatrix}2 & -1\\1 & 0\end{bmatrix} = 1 + 5 + 1 = 7.$
• $\det A = 2(0\cdot 4 - 1\cdot(-1)) - (-1)(1\cdot 4 - 1\cdot 3) + 1(1\cdot(-1) - 0\cdot 3) = 2 + 1 - 1 = 2.$

So $p(\lambda) = \lambda^{3} - 6\lambda^{2} + 7\lambda - 2.$

Step 2 - Apply Cayley-Hamilton. Substituting $A$ for $\lambda$: $A^{3} - 6A^{2} + 7A - 2I = 0.$

Step 3 - Solve for $A^{-1}$. Since $\det A = 2 \ne 0$, $A$ is invertible. Multiply the identity on the left by $A^{-1}$: $A^{2} - 6A + 7I - 2A^{-1} = 0 \quad\Longrightarrow\quad 2A^{-1} = A^{2} - 6A + 7I.$ $\boxed{\,A^{-1} = \tfrac{1}{2}\left(A^{2} - 6A + 7I\right)\,}$

This is option 1. (A direct check gives $A^{-1} = \begin{bmatrix} \tfrac12 & \tfrac32 & -\tfrac12 \\ -\tfrac12 & \tfrac52 & -\tfrac12 \\ -\tfrac12 & -\tfrac12 & \tfrac12 \end{bmatrix}$, which agrees.)

Exam trick: The constant term of the characteristic polynomial is $-\det A$, and the last term before it carries $A^{1}$, so $A^{-1}$ is always $\frac{1}{\det A}$ times a polynomial in $A$ of degree $n-1$. For $n=3$, just lift the coefficients $1,\,-\operatorname{tr}A,\,+m_{2}$ from $p(\lambda)$ and divide by $\det A$ - no row reduction needed.