Linear Algebra and Its Applications, Exercise 1.6.19

Exercise 1.6.19. Given the matrix A where

A = \begin{bmatrix} a&b&c \\ d&e&0 \\ f&0&0 \end{bmatrix} \quad \rm or \quad A = \begin{bmatrix} a&b&0 \\ c&d&0 \\ 0&0&e \end{bmatrix}

what values must the entries of A have in order for A to be invertible?

Answer: We know that a (square) matrix A is invertible if and only if it is nonsingular. We therefore perform Gaussian elimination on the first matrix A to see under which conditions it is nonsingular. We start by noting that f must be nonzero; otherwise there will be no pivot in the last row and elimination will fail.

Since we know for certain that f is nonzero, we can exchange the first and third rows to obtain the following matrix:

\begin{bmatrix} f&0&0 \\ d&e&0 \\ a&b&c \end{bmatrix}

For the next elimination step we multiply the first row by d/f; recall that we know from above that f is nonzero, so dividing d by f is possible. We then subtract the result from the second row, obtaining the following matrix:

\begin{bmatrix} f&0&0 \\ 0&e&0 \\ a&b&c \end{bmatrix}

We next multiple the first row by a/f and subtract the result from the third row, obtaining the following matrix:

\begin{bmatrix} f&0&0 \\ 0&e&0 \\ 0&b&c \end{bmatrix}

We note at this point that e must also be nonzero, for the same reason that f must be nonzero: If e were zero then we would have at least one row without a pivot. Since e is nonzero we can divide b by e, multiply the second row by b/e, and then subtract the result from the third row. This elimination step produces the following matrix:

\begin{bmatrix} f&0&0 \\ 0&e&0 \\ 0&0&c \end{bmatrix}

Finally, we note that c must also be nonzero, or else its row will not have a pivot.

Combining this with our previous conclusions, we see that the first matrix A will be nonsingular, and thus invertible, if and only if c, e, and f are all nonzero.

We now consider the second matrix A. Note that this is a block diagonal matrix consisting of two submatrices: a 2 by 2 matrix with elements a, b, c, and d, and a 1 by 1 submatrix with the single element e:

\begin{bmatrix} a&b&\vline&0 \\ c&d&\vline&0 \\ \hline 0&0&\vline&e \end{bmatrix}

As I discussed in the previous post, a block diagonal matrix is invertible if and only if all of its diagonal submatrices are invertible. From the formula for the inverse of a 2 by 2 matrix we know that the first submatrix is invertible only if ad - bc \ne 0, and since the inverse of the 1 by 1 matrix is simply the reciprocal of its (sole) entry, the second submatrix is invertible only if e \ne 0. Combining these results, we must have ad - bc \ne 0 and e \ne 0 for this matrix A to be invertible.

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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