## Linear Algebra and Its Applications, Exercise 1.5.9

Exercise 1.5.9. (a) Assume that A is the product of three matrices as follows: $A = \begin{bmatrix} 1&0&0 \\ -1&1&0 \\ 0&-1&1 \end{bmatrix} \begin{bmatrix} d_1&& \\ &d_2& \\ &&d_3 \end{bmatrix} \begin{bmatrix} 1&-1&0 \\ 0&1&-1 \\ 0&0&1 \end{bmatrix}$

What must be true for A to be nonsingular?

(b) Given A above, assume we have a system of linear equations Ax = b with corresponding Lc = b as follows: $\begin{bmatrix} 1&0&0 \\ -1&1&0 \\ 0&-1&1 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = b$

Answer: (a) From the above we have A = LDU where L is a lower triangular matrix, D is a diagonal matrix, and U is an upper triangular matrix. Per p. 36 the diagonal entries of D are the pivots, and in order for A to be nonsingular all pivots must be nonzero. So A is nonsingular if and only if $d_1$, $d_2$, and $d_3$ are not zero.

(b) From above we have $b = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1&0&0 \\ -1&1&0 \\ 0&-1&1 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2-c_1 \\ c_3-c_2 \end{bmatrix} \Rightarrow c_1 = 0, c_2 = 0, c_3 = 1$

Since A = LDU and Ax = b we have LDUx = b. But we also have Lc = b which implies DUx = c. Substituting from above this gives us $\begin{bmatrix} d_1&& \\ &d_2& \\ &&d_3 \end{bmatrix} \begin{bmatrix} 1&-1&0 \\ 0&1&-1 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$

Multiplying on the left gives us $\begin{bmatrix} d_1&-d_1&0 \\ 0&d_2&-d_2 \\ 0&0&d_3 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \Rightarrow d_1u-d_1v=0, d_2v-d_2w=0, d_3w=1$

Assuming that none of $d_1$, $d_2$, and $d_3$ is zero, we then have $d_3w=1 \Rightarrow w = 1/d_3$ $d_2v-d_2w=0 \Rightarrow d_2v = d_2w \Rightarrow v = w = 1/d_3$ $d_1u-d_1v=0 \Rightarrow d_1u = d_1v \Rightarrow u = v = 1/d_3$

So if A is nonsingular then Ax = b has the unique solution $x = \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 1/d_3 \\ 1/d_3 \\ 1/d_3 \end{bmatrix}$

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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