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已知向量组a1a2a3线性无关,向量组a1a2a3a4线性相关,向量组a1a2a3a4的秩为4,证明a1a2a3a5-a

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已知向量组a1a2a3线性无关,向量组a1a2a3a4线性相关,向量组a1a2a3a4的秩为4,证明a1a2a3a5-a4线性无关?
已知向量组a1a2a3线性无关,向量组a1a2a3a4线性相关,向量组a1a2a3a4的秩为4,证明a1a2a3a5-a
I suppose:"向量组a1a2a3a5的秩为4"
instead of:"向量组a1a2a3a4的秩为4"
向量组a1a2a3a5的秩为4 => a1,a2,a3,a5线性无关
a1a2a3a4线性相关
=> a4=m1a1+m2a2+m3a3
k1a1+k2a2+k3a3+k4(a5-a4)=0
k1a1+k2a2+k3a3+ k4a5 - k4(m1a1+m2a2+m3a3)=0
(k1-k4m1)a1+ (k2-k4m2)a2+(k3-k4m3)a3+ k4a5=0
=>
(k1-k4m1)=0 (1) and
(k2-k4m2)=0 (2) and
(k3-k4m3)=0 (3) and
k4=0 (4)
then k3=k2=k1=0
ie
a1,a2,a3,a5-a4线性无关