## 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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