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Chapter 3: Q29E (page 143)

Find a basis of the subspace of 鈩�3defined by the equation2x1+3x2+x3=0.

Short Answer

Expert verified

The basis of the subspace V of 鈩�3defined by the equation 2x1+3x2+x3=0is 鈭�320,鈭�102.

Step by step solution

01

Finding the basis of 2x1+3x2+x3=0

Let V be a subspace鈩�3 such that

V={(x1,x2,x3)鈭�鈩�3:2x1+3x2+x3=0}

鈬�V={(2x1,2x2,2x3)鈭�鈩�3:2x1+3x2+x3=0}

鈬�V={(鈭�3x2鈭�x3,2x2,2x3)鈭�鈩�3:2x1=鈭�3x2鈭�x3}

鈬�V=x2鈭�320+x3鈭�102

鈬�Basis of V=鈭�320,鈭�102

02

Final Answer

The basis of the subspace of 鈩�3defined by the equation 2x1+3x2+x3=0is

鈭�320,鈭�102

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Most popular questions from this chapter

In Exercises 21 through 25, find the reduced row-echelon form of the given matrix A. Then find a basis of the image of A and a basis of the kernel of A.

22.[248451793]

In Problem 46 through 55, Find all the cubics through the given points. You may use the results from Exercises 44 and 45 throughout. If there is a unique cubic, make a rough sketch of it. If there are infinitely many cubics, sketch two of them.

(0,0),(1,0),(2,0),(3,0),(4,0),(0,1),(0,2),(0,3),(1,1)

Give an example of a linear transformation whose kernel is the plane x+2y+3z=0in鈩�3.

Consider two subspaces VandW of 鈩�n, where Vis contained in W. Explain why dim(V)鈮�dim(W). (This statement seems intuitively rather obvious. Still, we cannot rely on our intuition when dealing with 鈩�n.)

Can you find a 3脳3 matrix Asuch that im(A)=ker(A)? Explain.
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