Question

1) given that A1=2i-j+k, A2=i+3j-2k, A3=3i+2j+5k, and A4=3i+2j+5k
find scalars a,b,c such that A4=aA1+bA2+cA3.

2) if a and b are non-collinear vectors and A=(x+y)a+(2x+y+1)b

3) given the scalar defined by phy(x,y,z)=3x^2-xy^2+5

Answer 1

Answer on Question #67830 – Math – Linear Algebra

Question

1) Given that

$$a_1 = 2i - j + k$$ $$a_2 = i + 3j - 2k$$ $$a_3 = 3i + 2j + 5k$$ $$a_4 = 3i + 2j + 5k$$

find scalars $a, b, c$ such that

$$a_4 = aa_1 + ba_2 + ca_3$$

Solution

If $a_4 = aa_1 + ba_2 + ca_3$, then

$$3i + 2j + 5k = a(2i - j + k) + b(i + 3j - 2k) + c(3i + 2j + 5k)$$ $$3i + 2j + 5k = (2a + b + 3c)i + (-a + 3b + 2c)j + (a - 2b + 5c)k$$ $$\left\{ \begin{array}{l} 2a + b + 3c = 3 \\ -a + 3b + 2c = 2 \\ a - 2b + 5c = 5 \end{array} \right.$$

It follows from the third equation that

$$a = 5 + 2b - 5c$$

Add the second and the third equations

$$b + 7c = 7,$$

hence

$$b = 7 - 7c$$

Substitute (2) into (1)

$$a = 5 + 2b - 5c = 5 + 2(7 - 7c) - 5c = 5 + 14 - 14c - 5c = 19 - 19c,$$

that is,

$$a = 19 - 19c$$

Substitute (2) and (3) into the first equation of the system

$$2a + b + 3c = 3$$ $$2(19 - 19c) + 7 - 7c + 3c = 3$$ $$38 - 38c + 7 - 7c + 3c = 3$$ $$-42c = -42$$

Hence

$$c = 1$$

Substitute (4) into (2) and (3)

$$b = 7 - 7c = 7 - 7 \cdot 1 = 7 - 7 = 0$$ $$a = 19 - 19 \cdot 1 = 19 - 19 = 0$$

Finally one gets

$$a = b = 0; \; c = 1$$

Answer: $a = b = 0$; $c = 1$

Question

2) If $a$ and $b$ are non-collinear vectors and $A = (x + y)a + (2x + y + 1)b$

Answer: the statement of the question is not complete and it is not known what one should calculate there.

Question

3) Given the scalar defined by $\phi(x, y, z) = 3x^2 - xy^2 + 5$

Answer: the statement of the question is not complete and it is not known what one should calculate there.

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