IMC / 2003 / Problems / Day 2, P7

IMC 2003 · Day 2 · P7

easy

Let $A$ and $B$ be $n \times n$ real matrices such that $AB + A + B = 0$ . Prove that $AB = BA$ .

Solution (official)

Since $(A + I)(B + I) = AB + A + B + I = I$ ( $I$ is the identity matrix), matrices $A + I$ and $B + I$ are inverses of each other. Then $(A + I)(B + I) = (B + I)(A + I)$ and $AB + BA$ .

How the field did

contestants scored

185

average (of 20)

15.86

solved (≥ 80%)

75.7%

near-0 (≤ 10%)

11.9%

discrimination

0.44

Score distribution (field cohort)

Computed on contestants with a meaningful total (field cohort); discrimination is the corrected item–total correlation.

IMC 2003 · Day 2 · P7

How the field did

Score distribution (field cohort)

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