AMC 8 1998 Problem 2

AMC8 1998 年第 2 题

表达式与运算★☆☆☆☆
If (abcd)=ad−bc\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc, then (2113)+(3214)=\begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix} + \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix} =
若 (abcd)=ad−bc\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc,则 (2113)+(3214)=\begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix} + \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix} =
  1. A.-2
  2. B.-1
  3. C.0
  4. D.1
  5. E.2

正确答案 · Correct Answer

E

解析 · Solution

Solution 1

Plugging in values for aa, bb, cc, and dd, we get
a=3, b=4, c=1, d=2,
a×d=3×2=6a\times d=3×2=6, b×c=4×1=4b\times c=4×1=4, 6-4=2, EE

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