Question

Let
A=2i−j+k
,
B=i+3j−2k
,
C=−2i+j−2k
and
D=3i+2j+5k
. Find scalar a, b, c such that
D=aA+bB+cC

a=−1,b=1,c=2

a=5,b=1,c=1

a=−2,b=1,c=−3

a=−1,b=1,c=2

Answer 1

Answer on Question #56010 – Math – Linear Algebra

Let $A = 2i - j + k$, $B = i + 3j - 2k$, $C = -2i + j - 2k$, and $D = 3i + 2j + 5k$.

Find scalar a, b, c such that $D = aA + bB + cC$

Solution

First find $aA, bB, cC$ in component form

$$aA = a(2i - j + k) = 2ai - aj + ak$$ $$bB = b(i + 3j - 2k) = bi + 3bj - 2bk$$ $$cC = c(-2i + j - 2k) = -2ci + cj - 2ck$$

Now let's sum these equations and compare them with the expression for D in component form

$$\begin{array}{l} D = aA + bB + cC = 2ai - aj + ak + bi + 3bj - 2bk - 2ci + cj - 2ck \\ = (2a + b - 2c)i + (-a + 3b + c)j + (a - 2b - 2c)k \\ \end{array}$$ $$D = 3i + 2j + 5k$$

Equating expressions that stood near the corresponding unit vectors, we obtain the following system

$$\begin{array}{l} i: \quad \left\{ \begin{array}{l} 2a + b - 2c = 3 \\ -a + 3b + c = 2 \\ a - 2b - 2c = 5 \end{array} \right. \\ \end{array}$$

Express $b$ from the first equation and substitute for $b$ into the second and the third equations of the system:

$$\left\{ \begin{array}{c} b = 3 + 2c - 2a \\ -a + 9 + 6c - 6a + c = 2 \\ a - 6 - 4c + 4a - 2c = 5 \end{array} \right.$$ $$\left\{ \begin{array}{r} b = 3 + 2c - 2a \\ 7c - 7a = -7 \\ 5a - 6c = 11 \end{array} \right.$$

Divide the second equation by 7:

$$\left\{ \begin{array}{r} b = 3 + 2c - 2a \\ c - a = -1 \\ 5a - 6c = 11 \end{array} \right.$$

Express $c$ from the second equation and substitute for $c$ into the third equations of the system:

$$\left\{ \begin{array}{r} b = 3 + 2c - 2a \\ c = -1 + a \\ 5a - 6a + 6 = 11 \end{array} \right.$$ $$\left\{ \begin{array}{l} b = 3 + 2c - 2a \\ c = -1 + a \\ -a = 5 \end{array} \right.$$ $$\left\{ \begin{array}{l} b = 3 + 2c - 2a \\ c = -6 \\ a = -5 \end{array} \right.$$ $$\left\{ \begin{array}{l} b = 1 \\ c = -6 \\ a = -5 \end{array} \right.$$

Answer: $a = -5$, $b = 1$, $c = -6$.

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