## 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: