Question:
If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, then $A^2 - 5A - 2I =$
If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, then $A^2 - 5A - 2I =$
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Do not waste time manually multiplying $A \times A$! The trace ($\text{tr}(A)$) is $5$ and the determinant is $-2$. The identity $A^2 - (\text{tr }A)A + (\det A)I = O$ always holds.
Updated On: Jun 3, 2026
- $O$
- $I$
- $A$
- $2A$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
According to the Cayley-Hamilton Theorem, every square matrix satisfies its own characteristic equation, $\det(A - \lambda I) = 0$.
Step 2: Meaning
For a $2 \times 2$ matrix $A$, the characteristic equation is given by: \[ \lambda^2 - \text{tr}(A)\lambda + \det(A) = 0 \] Substituting matrix $A$ in place of $\lambda$ yields $A^2 - \text{tr}(A)A + \det(A)I = O$.
Step 3: Analysis
Given $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$:
• Trace, $\text{tr}(A) = 1 + 4 = 5$
• Determinant, $\det(A) = (1)(4) - (2)(3) = 4 - 6 = -2$
Substituting these values into the characteristic equation: \[ A^2 - 5A - 2I = O \]
Step 4: Conclusion
Thus, the expression $A^2 - 5A - 2I$ evaluates directly to the null matrix $O$.
Final Answer: (A)
According to the Cayley-Hamilton Theorem, every square matrix satisfies its own characteristic equation, $\det(A - \lambda I) = 0$.
Step 2: Meaning
For a $2 \times 2$ matrix $A$, the characteristic equation is given by: \[ \lambda^2 - \text{tr}(A)\lambda + \det(A) = 0 \] Substituting matrix $A$ in place of $\lambda$ yields $A^2 - \text{tr}(A)A + \det(A)I = O$.
Step 3: Analysis
Given $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$:
• Trace, $\text{tr}(A) = 1 + 4 = 5$
• Determinant, $\det(A) = (1)(4) - (2)(3) = 4 - 6 = -2$
Substituting these values into the characteristic equation: \[ A^2 - 5A - 2I = O \]
Step 4: Conclusion
Thus, the expression $A^2 - 5A - 2I$ evaluates directly to the null matrix $O$.
Final Answer: (A)
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