## Linear Algebra and Its Applications, Exercise 1.6.10

Exercise 1.6.10. Determine the inverses of the following matrices: $A_1 = \begin{bmatrix} 0&0&0&1 \\ 0&0&2&0 \\ 0&3&0&0 \\ 4&0&0&0 \end{bmatrix} \quad A_2 = \begin{bmatrix} 1&0&0&0 \\ -\frac{1}{2}&1&0&0 \\ 0&-\frac{2}{3}&1&0 \\ 0&0&-\frac{3}{4}&1 \end{bmatrix} \quad A_3 = \begin{bmatrix} a&b&0&0 \\ c&d&0&0 \\ 0&0&a&b \\ 0&0&c&d \end{bmatrix}$

Answer: The first matrix has the form of a diagonal matrix, only flipped horizontally, and it’s therefore worth seeing if its inverse has an analogous form to the inverse of a diagonal matrix. By analogy the inverse would also be a reverse-diagonal matrix. However a bit of trial and error reveals that each entry on the reverse-diagonal should not be the inverse of the corresponding entry in the original matrix, but rather should be the inverse of the corresponding entry in the original matrix’s transpose.

Forming this matrix B we have $A_1B = \begin{bmatrix} 0&0&0&1 \\ 0&0&2&0 \\ 0&3&0&0 \\ 4&0&0&0 \end{bmatrix} \begin{bmatrix} 0&0&0&\frac{1}{4} \\ 0&0&\frac{1}{3}&0 \\ 0&\frac{1}{2}&0&0 \\ 1&0&0&0 \end{bmatrix} = \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} = I$

and $BA_1 = \begin{bmatrix} 0&0&0&\frac{1}{4} \\ 0&0&\frac{1}{3}&0 \\ 0&\frac{1}{2}&0&0 \\ 1&0&0&0 \end{bmatrix} \begin{bmatrix} 0&0&0&1 \\ 0&0&2&0 \\ 0&3&0&0 \\ 4&0&0&0 \end{bmatrix} = \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} = I$

so that $A_1^{-1} = B = \begin{bmatrix} 0&0&0&\frac{1}{4} \\ 0&0&\frac{1}{3}&0 \\ 0&\frac{1}{2}&0&0 \\ 1&0&0&0 \end{bmatrix}$

For the second matrix we use Gauss-Jordan elimination: $\begin{bmatrix} 1&0&0&0&1&0&0&0 \\ -\frac{1}{2}&1&0&0&0&1&0&0 \\ 0&-\frac{2}{3}&1&0&0&0&1&0 \\ 0&0&-\frac{3}{4}&1&0&0&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&0&0&1&0&0&0 \\ 0&1&0&0&\frac{1}{2}&1&0&0 \\ 0&-\frac{2}{3}&1&0&0&0&1&0 \\ 0&0&-\frac{3}{4}&1&0&0&0&1 \end{bmatrix}$ $\rightarrow \begin{bmatrix} 1&0&0&0&1&0&0&0 \\ 0&1&0&0&\frac{1}{2}&1&0&0 \\ 0&0&1&0&\frac{1}{3}&\frac{2}{3}&1&0 \\ 0&0&-\frac{3}{4}&1&0&0&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&0&0&1&0&0&0 \\ 0&1&0&0&\frac{1}{2}&1&0&0 \\ 0&0&1&0&\frac{1}{3}&\frac{2}{3}&1&0 \\ 0&0&0&1&\frac{1}{4}&\frac{1}{2}&\frac{3}{4}&1 \end{bmatrix}$

At this point forward elimination is complete, and there is no need to do reverse elimination. We therefore have $A_2^{-1} = \begin{bmatrix} 1&0&0&0 \\ \frac{1}{2}&1&0&0 \\ \frac{1}{3}&\frac{2}{3}&1&0 \\ \frac{1}{4}&\frac{1}{2}&\frac{3}{4}&1 \end{bmatrix}$

The third 4 by 4 matrix is composed of two 2 by 2 nonzero submatrices (in the upper left and lower right) and two 2 by 2 zero submatrices (in the upper right and lower left). Consider a second 4 by 4 matrix B also containing two 2 by 2 nonzero submatrices and two 2 by 2 zero submatrices, in the same positions as the original matrix: $B = \begin{bmatrix} b_{11}&b_{12}&0&0 \\ b_{21}&b_{22}&0&0 \\ 0&0&b_{33}&b_{34} \\ 0&0&b_{43}&b_{44} \end{bmatrix}$

In order for $A_3B = \begin{bmatrix} a&b&0&0 \\ c&d&0&0 \\ 0&0&a&b \\ 0&0&c&d \end{bmatrix} \begin{bmatrix} b_{11}&b_{12}&0&0 \\ b_{21}&b_{22}&0&0 \\ 0&0&b_{33}&b_{34} \\ 0&0&b_{43}&b_{44} \end{bmatrix} = \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix}$

we must have $\begin{bmatrix} a&b \\ c&d \end{bmatrix} \begin{bmatrix} b_{11}&b_{12} \\ b_{21}&b_{22} \end{bmatrix} = \begin{bmatrix} a&b \\ c&d \end{bmatrix} \begin{bmatrix} b_{33}&b_{34} \\ b_{43}&b_{44} \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}$

That means that $B = 1/(ad - bc) \begin{bmatrix} d&-b&0&0 \\ -c&a&0&0 \\ 0&0&d&-b \\ 0&0&-c&a \end{bmatrix}$

assuming that (ad – bc) is nonzero. We then have $A_3B = 1/(ad - bc) \begin{bmatrix} a&b&0&0 \\ c&d&0&0 \\ 0&0&a&b \\ 0&0&c&d \end{bmatrix} \begin{bmatrix} d&-b&0&0 \\ -c&a&0&0 \\ 0&0&d&-b \\ 0&0&-c&a \end{bmatrix}$ $= 1/(ad - bc) \begin{bmatrix} ad - bc&-ab+ab&0&0 \\ cd-cd&ad-bc&0&0 \\ 0&0&ad-bc&-ab+ab \\ 0&0&cd-cd&ad-bc \end{bmatrix} = \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix}$

and $BA_3 = 1/(ad - bc) \begin{bmatrix} d&-b&0&0 \\ -c&a&0&0 \\ 0&0&d&-b \\ 0&0&-c&a \end{bmatrix} \begin{bmatrix} a&b&0&0 \\ c&d&0&0 \\ 0&0&a&b \\ 0&0&c&d \end{bmatrix}$ $= 1/(ad - bc) \begin{bmatrix} ad - bc&-ac+ac&0&0 \\ -bd+bd&ad-bc&0&0 \\ 0&0&ad-bc&bd-bd \\ 0&0&-ac+ac&ad-bc \end{bmatrix} = \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix}$

so that $A_3^{-1} = B = 1/(ad - bc) \begin{bmatrix} d&-b&0&0 \\ -c&a&0&0 \\ 0&0&d&-b \\ 0&0&-c&a \end{bmatrix}$

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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