Vamos a calcular el rango
de la matriz A siguiente en función de los valores que tome su parámetro
x.
A=(
1
0
−2
3
1
2
−1
3
0
2
4
x
−1
6
4
)
MathType@MTEF@5@5@+=feaafeart1ev1aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8qrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeaabaWaaqaafaaakeaacaWGbbGaeyypa0ZaaeWaaeaafaqaceWafaaaaeaacaaIXaaabaGaaGimaaqaaiabgkHiTiaaikdaaeaacaaIZaaabaGaaGymaaqaaiaaikdaaeaacqGHsislcaaIXaaabaGaaG4maaqaaiaaicdaaeaacaaIYaaabaGaaGinaaqaaiaadIhaaeaacqGHsislcaaIXaaabaGaaGOnaaqaaiaaisdaaaaacaGLOaGaayzkaaaaaa@4A26@
Elegimos
un menor no nulo de orden 2:
|
3
1
0
2
|=6
MathType@MTEF@5@5@+=feaafeart1ev1aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8qrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeaabaWaaqaafaaakeaadaabdaqaauaabiqaciaaaeaacaaIZaaabaGaaGymaaqaaiaaicdaaeaacaaIYaaaaaGaay5bSlaawIa7aiabg2da9iaaiAdaaaa@408A@
. Al orlarlo obtenemos tres posibilidades:
|
−2
3
1
3
0
2
−1
6
4
|=0, |
0
3
1
−1
0
2
x
6
4
|=6+6x, |
1
3
1
2
0
2
4
6
4
|=0
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8qrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@68B3@
Vemos
que hay un menor no nulo de orden 3 si 6+6x es distinto de 0, es decir, si
x≠−1
MathType@MTEF@5@5@+=feaafeart1ev1aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8qrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeaabaWaaqaafaaakeaacaWG4bGaeyiyIKRaeyOeI0IaaGymaaaa@3D0F@
.
Por tanto:
Rango(A)=
{
2, si x=−1
3, si x≠−1
MathType@MTEF@5@5@+=feaafeart1ev1aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8qrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeaabaWaaqaafaaakeaadaGabaqaauaabeqaceaaaeaacaaIYaGaaiilaiaaywW7caWGZbGaamyAaiaaysW7caWG4bGaeyypa0JaeyOeI0IaaGymaaqaaiaaiodacaGGSaGaaGzbVlaadohacaWGPbGaaGjbVlaadIhacqGHGjsUcqGHsislcaaIXaaaaaGaay5Eaaaaaa@4EBC@
|