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}

However, these are not the only values of a and b for which the matrix is singular. Suppose we proceed with elimination. The first step would be to multiply the first row by l_1 = a and subtract the result from the second row:

\begin{bmatrix} 1&2&0 \\ a&8&3 \\ 0&b&5 \end{bmatrix} \rightarrow \begin{bmatrix} 1&2&0 \\ 0&8-2a&3 \\ 0&b&5 \end{bmatrix}

Assuming that a \ne 4 then the next step of elimination would multiply the second row of the matrix by l_2 = b/(8-2a) and subtract the result from the third row:

\begin{bmatrix} 1&2&0 \\ 0&8-2a&3 \\ 0&b&5 \end{bmatrix} \rightarrow \begin{bmatrix} 1&2&0 \\ 0&8-2a&3 \\ 0&0&5-3b/(8-2a) \end{bmatrix}

If the (3,3) entry of the resulting matrix above is zero then the original matrix is singular. In this case we have 3b/(8-2a) = 5. Multiplying both sides by (8-2a) we have 3b = 5 \cdot (8-2a) = 40 - 10a or 10a + 3b = 40.

So the first matrix is singular for all values of a and b for which 10a+3b=40. This includes the case a = 4, b=0 mentioned above, the case a=0, b = 40/3, the case a=5, b=-10/3, and an infinity of others on the line b=-\frac{10}{3}a + \frac{40}{3}.

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}

UPDATE: Prompted by Theodore’s comment, added a section deriving the complete set of a and b for which the first matrix is singular.

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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2 Responses to Linear Algebra and Its Applications, Exercise 1.5.19

  1. Theodore says:

    I also found a = 5 and b = -10/3 to also cause a singular matrix. My question is how do we find all values that can cause singularity

    • hecker says:

      Thank you for finding this problem. You are correct that there are multiple (actually, an infinity) of values of a and b for which the first matrix is singular. I added a new section to derive the correct answer.

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