PLS HELP WILL MARK BRAINLIEST HAVE TO INCLUDE SOLUTION!!
1 answer:
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We see that in these 3 terms, 6 is a common factor. So, let's factor out a 6:
We can set this equal to 0 and factor by using the quadratic formula which is:
So, let's do just that:
Note that the 6 goes away if you divide both sides by 6. In this case, a = 1, b = -1, and c = -2. Let's plug that into the quadratic equation:
So, we can write this as:
Notice that the 6 comes back because it was only temporarily mad. And that the roots have opposite signs in the parentheses because to find the roots, you need to set each parentheses equal to 0 and solve for n. With that in mind, your final answer is 6(n-1)(n+2). Hope I could help you!
Answer:
A. {u1,u2,u3,u4} is a linearly independent set of vectors unless one of {u1,u2,u3} is the zero vector.
Step-by-step explanation:
Given that u4 is not a linear combination of {u1,u2,u3}
This means there is no possibility to write u4 = au1+bu2+cu3 for three scalars a,b,and c.
This gives that
This implies that these four vectors are not linearly dependent but linearly independent.
Hence option a is right.
A. {u1,u2,u3,u4} is a linearly independent set of vectors unless one of {u1,u2,u3} is the zero vector.