## Linear Algebra and Its Applications, Review Exercise 1.18

Review exercise 1.18. Given the following matrix $A = \begin{bmatrix} 1&v_1&0&0 \\ 0&v_2&0&0 \\ 0&v_3&1&0 \\ 0&v_4&0&1 \end{bmatrix}$

(a) Factor $A$ into the form $A = LU$ (if $v_2 \ne 0$).

(b) Show that $A^{-1}$ exists and has the same form as $A$.

Answer: (a) We start elimination by subtracting $v_3/v_2$ times the second row from the third row ( $l_{32} = v_3/v_2$) and subtracting $v_4/v_2$ times the second row from the fourth row ( $l_{32} = v_4/v_2$) : $\begin{bmatrix} 1&v_1&0&0 \\ 0&v_2&0&0 \\ 0&v_3&1&0 \\ 0&v_4&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&v_1&0&0 \\ 0&v_2&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix}$

This completes elimination, and we have $L = \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&v_3/v_2&1&0 \\ 0&v_4/v_2&0&1 \end{bmatrix} \quad \rm and \quad U = \begin{bmatrix} 1&v_1&0&0 \\ 0&v_2&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix}$

so that $A = \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&v_3/v_2&1&0 \\ 0&v_4/v_2&0&1 \end{bmatrix} \begin{bmatrix} 1&v_1&0&0 \\ 0&v_2&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} = LU$

(b) We use Gauss-Jordan elimination, starting with the same elimination steps as above: $\begin{bmatrix} 1&v_1&0&0&\vline&1&0&0&0 \\ 0&v_2&0&0&\vline&0&1&0&0 \\ 0&v_3&1&0&\vline&0&0&1&0 \\ 0&v_4&0&1&\vline&0&0&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&v_1&0&0&\vline&1&0&0&0 \\ 0&v_2&0&0&\vline&0&1&0&0 \\ 0&0&1&0&\vline&0&-v_3/v_2&1&0 \\ 0&0&0&1&\vline&0&-v_4/v_2&0&1 \end{bmatrix}$

This completes forward elimination, and we start backward elimination by multiplying $v_1/v_2$ times the second row and subtracting it from the first: $\begin{bmatrix} 1&v_1&0&0&\vline&1&0&0&0 \\ 0&v_2&0&0&\vline&0&1&0&0 \\ 0&0&1&0&\vline&0&-v_3/v_2&1&0 \\ 0&0&0&1&\vline&0&-v_4/v_2&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&0&0&\vline&1&-v_1/v_2&0&0 \\ 0&v_2&0&0&\vline&0&1&0&0 \\ 0&0&1&0&\vline&0&-v_3/v_2&1&0 \\ 0&0&0&1&\vline&0&-v_4/v_2&0&1 \end{bmatrix}$

This completes backward elimination, and we divide the second row by $v_2$: $\begin{bmatrix} 1&0&0&0&\vline&1&-v_1/v_2&0&0 \\ 0&v_2&0&0&\vline&0&1&0&0 \\ 0&0&1&0&\vline&0&-v_3/v_2&1&0 \\ 0&0&0&1&\vline&0&-v_4/v_2&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&0&0&\vline&1&-v_1/v_2&0&0 \\ 0&1&0&0&\vline&0&1/v_2&0&0 \\ 0&0&1&0&\vline&0&-v_3/v_2&1&0 \\ 0&0&0&1&\vline&0&-v_4/v_2&0&1 \end{bmatrix}$

We thus have $A^{-1} = \begin{bmatrix} 1&-v_1/v_2&0&0 \\ 0&1/v_2&0&0 \\ 0&-v_3/v_2&1&0 \\ 0&-v_4/v_2&0&1 \end{bmatrix}$

with $A^{-1}$ having the same form as $A$.

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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