## Linear Algebra and Its Applications, exercise 1.2.2

Exercise 1.2.2. Given the system of equations

$\setlength\arraycolsep{0.2em}\begin{array}{rrrrrrr} u&+&v&+&w&=&b_1 \\ &&v&+&w&=&b_2 \\ &&&&w&=&b_3 \end{array}$

solve for u, v, and w, and then use the solution to find a linear combination of column vectors that equals the vector $(b_1, b_2, b_3)$.

Answer: First, subtract the third equation from the other first two to get

$\setlength\arraycolsep{0.2em}\begin{array}{rrrrrrl} u&+&v&&&=&b_1 - b_3 \\ &&v&&&=&b_2 - b_3 \\ &&&&w&=&b_3 \end{array}$

Then subtract the second equation from the first to get

$\setlength\arraycolsep{0.2em}\begin{array}{rcl} u&=&b_1 - b_2 \\ v&=&b_2 - b_3 \\ w&=&b_3 \end{array}$

We can restate the original set of equations as

$u\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix} + v\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} + w\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix} = \begin{bmatrix}b_1 \\ b_2 \\ b_3\end{bmatrix}$

and then substitute in the solutions for u, v, and w:

$(b_1 - b_2)\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix} + (b_2 - b_3)\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} + b_3\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix} = \begin{bmatrix}b_1 \\ b_2 \\ b_3\end{bmatrix}$

This expresses the right-hand side as a linear combination of the vectors (1, 0, 0), (1, 1, 0), and (1, 1, 1).

NOTE: This begins a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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### 2 Responses to Linear Algebra and Its Applications, exercise 1.2.2

1. d says:

u+v+w=b1 (x)
u-v-w=b1 (o)

• hecker says:

I’m not sure what you’re asking. Could you please clarify your question?