## Linear Algebra and Its Applications, Exercise 1.5.19

Exercise 1.5.19. For the following two matrices, specify the values of a, b, and c for which elimination requires row exchanges, and the values for which the matrices in question are singular.

$A = \begin{bmatrix} 1&2&0 \\ a&8&3 \\ 0&b&5 \end{bmatrix} \quad A = \begin{bmatrix} c&2 \\ 6&4 \end{bmatrix}$

Answer: For the first matrix a row exchange would be necessary if the first step of elimination caused the second entry of the second row, i.e., in the (2,2) position, to become 0. This would occur if we multiplied the first row by 4 and subtracted it from the second row, as we would then have $8 - 4 \cdot 2 = 0$ in the (2,2) position. Since the entry in the (1,1) position is 1, multiplying the first row by 4 would be necessary if a were 4:

$\begin{bmatrix} 1&2&0 \\ 4&8&3 \\ 0&b&5 \end{bmatrix} \rightarrow \begin{bmatrix} 1&2&0 \\ 0&0&3 \\ 0&b&5 \end{bmatrix} \rightarrow \begin{bmatrix} 1&2&0 \\ 0&b&5 \\ 0&0&3 \end{bmatrix}$

If we had both a = 4 and b = 0 then the matrix would be singular:

$\begin{bmatrix} 1&2&0 \\ 4&8&3 \\ 0&0&5 \end{bmatrix} \rightarrow \begin{bmatrix} 1&2&0 \\ 0&0&3 \\ 0&0&5 \end{bmatrix}$

For the second matrix a row exchange would be necessary if c = 0:

$\begin{bmatrix} 0&2 \\ 6&4 \end{bmatrix} \rightarrow \begin{bmatrix} 6&4 \\ 0&2 \end{bmatrix}$

On the other hand, if c = 3 then the first elimination produces zeroes in both entries of the second row, and the matrix is singular:

$\begin{bmatrix} 3&2 \\ 6&4 \end{bmatrix} \rightarrow \begin{bmatrix} 3&2 \\ 0&0 \end{bmatrix}$

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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