## 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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