Linear Algebra and Its Applications, Exercise 1.5.5

Exercise 1.5.5. Given the following system of linear equations

Ax = \begin{bmatrix} 2&3&3 \\ 0&5&7 \\ 6&9&8 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \\ 5 \end{bmatrix} = b

find the factors L and U of A and the vector c for which Ux = c.

Answer: The 2,1 position of A is already zero, so there’s no need for an elimination step to make it zero. We then subtract 3 times the first row from the third, which makes the first two entries in the third row zero and produces the upper triangular matrix U:

\begin{bmatrix} 2&3&3 \\ 0&5&7 \\ 6&9&8 \end{bmatrix} \Rightarrow \begin{bmatrix} 2&3&3 \\ 0&5&7 \\ 0&0&-1 \end{bmatrix} = U

We then have L as the identity matrix with zero in the 2,1 position, 3 in the 3,1 position, and zero in the 3,2 position:

L = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 3&0&1 \end{bmatrix}

so that

LU = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 3&0&1 \end{bmatrix} \begin{bmatrix} 2&3&3 \\ 0&5&7 \\ 0&0&-1 \end{bmatrix} = \begin{bmatrix} 2&3&3 \\ 0&5&7 \\ 6&9&8 \end{bmatrix} = A

We then apply the same elimination steps to the righthand vector b, multiplying the first element by 3 and subtracting it from the third element:

\begin{bmatrix} 2 \\ 2 \\ 5 \end{bmatrix} \Rightarrow \begin{bmatrix} 2 \\ 2 \\ -1 \end{bmatrix}

producing the upper triangular triangular system Ux = c:

\begin{bmatrix} 2&3&3 \\ 0&5&7 \\ 0&0&-1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \\ -1 \end{bmatrix}

NOTE: This continues 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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