IMC / 2009 / Problems / Day 1, P2

IMC 2009 · Day 1 · P2

easy

Let $A$ , $B$ and $C$ be real square matrices of the same size, and suppose that $A$ is invertible. Prove that if $(A - B) C = B A^{-1}$ , then $C (A - B) = A^{-1} B$ .

Solution (official)

A straightforward calculation shows that $(A - B) C = B A^{-1}$ is equivalent to $AC - BC - BA^{-1} + AA^{-1} = I$ , where $I$ denotes the identity matrix. This is equivalent to $(A - B)(C + A^{-1}) = I$ . Hence, $(A - B)^{-1} = C + A^{-1}$ , meaning that $(C + A^{-1})(A - B) = I$ also holds. Expansion yields the desired result.

How the field did

contestants scored

334

average (of 10)

6.41

solved (≥ 80%)

61.4%

near-0 (≤ 10%)

29.0%

discrimination

0.45

Score distribution (field cohort)

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

IMC 2009 · Day 1 · P2

How the field did

Score distribution (field cohort)

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