Exercise 4.4 · Q11: Inverse of a matrix with $\cos\alpha$ and $\sin\alpha$

Basics you need

Solution

1. Compute $|A|$

   Row 1 is $1, 0, 0$⁠, the row with the most zeros. Expand along it; its signs are $+, -, +$⁠:

   $$\begin{aligned} |A| &= 1\cdot M_{11} - 0\cdot M_{12} + 0\cdot M_{13}\\ &= M_{11} \end{aligned}$$ $$\begin{aligned} M_{11} &= \begin{vmatrix} \cos\alpha & \sin\alpha \\ \sin\alpha & -\cos\alpha \end{vmatrix}\\ &= \cos\alpha(-\cos\alpha) - \sin\alpha\sin\alpha\\ &= -\cos^2\alpha - \sin^2\alpha\\ &= -(\cos^2\alpha + \sin^2\alpha) = -1 \end{aligned}$$

   So $|A| = -1$⁠. This is non-zero for every $\alpha$⁠, so $A^{-1}$ always exists.

   The entries depend on $\alpha$⁠, so a priori $|A|$ might vanish for some angles and the answer would be "exists only when …". The identity $\cos^2\alpha+\sin^2\alpha = 1$ removes $\alpha$ completely, so no case analysis is needed. The minus sign in front of $\cos\alpha$ in position $(3,3)$ is what turns the product of the diagonal into $-\cos^2\alpha$⁠.
   Pythagorean identity. Basics: algebra the chapter leans on

2. Cofactors of row 1

   $$\begin{aligned} M_{11} &= -1 \quad (\text{Step 1})\\ A_{11} &= (-1)^{1+1}M_{11} = -1\\[2mm] M_{12} &= \begin{vmatrix} 0 & \sin\alpha \\ 0 & -\cos\alpha \end{vmatrix} = 0\\ A_{12} &= (-1)^{1+2}M_{12} = 0\\[2mm] M_{13} &= \begin{vmatrix} 0 & \cos\alpha \\ 0 & \sin\alpha \end{vmatrix} = 0\\ A_{13} &= (-1)^{1+3}M_{13} = 0 \end{aligned}$$
   Deleting column 2 or column 3 leaves part of column 1, whose lower entries are both $0$⁠. A $2\times2$ determinant with a whole column of zeros is $0\cdot d - b\cdot 0 = 0$⁠. In detail, $M_{12} = 0(-\cos\alpha) - (\sin\alpha)(0) = 0$ and $M_{13} = 0(\sin\alpha) - (\cos\alpha)(0) = 0$⁠.

3. Cofactors of row 2

   $$\begin{aligned} M_{21} &= \begin{vmatrix} 0 & 0 \\ \sin\alpha & -\cos\alpha \end{vmatrix} = 0\\ A_{21} &= (-1)^{2+1}M_{21} = 0\\[2mm] M_{22} &= \begin{vmatrix} 1 & 0 \\ 0 & -\cos\alpha \end{vmatrix} = -\cos\alpha\\ A_{22} &= (-1)^{2+2}M_{22} = -\cos\alpha\\[2mm] M_{23} &= \begin{vmatrix} 1 & 0 \\ 0 & \sin\alpha \end{vmatrix} = \sin\alpha\\ A_{23} &= (-1)^{2+3}M_{23} = -\sin\alpha \end{aligned}$$
   $M_{21}$ has a row of zeros (what is left of row 1), so it is $0$⁠. $M_{22}$ deletes row 2 and column 2, leaving rows 1, 3 and columns 1, 3: $1\cdot(-\cos\alpha) - 0\cdot0$⁠. $M_{23}$ deletes row 2 and column 3, leaving rows 1, 3 and columns 1, 2: $1\cdot\sin\alpha - 0\cdot0$⁠. Position $(2,3)$ has $2+3=5$⁠, odd, so its sign is $-$⁠.

4. Cofactors of row 3

   $$\begin{aligned} M_{31} &= \begin{vmatrix} 0 & 0 \\ \cos\alpha & \sin\alpha \end{vmatrix} = 0\\ A_{31} &= (-1)^{3+1}M_{31} = 0\\[2mm] M_{32} &= \begin{vmatrix} 1 & 0 \\ 0 & \sin\alpha \end{vmatrix} = \sin\alpha\\ A_{32} &= (-1)^{3+2}M_{32} = -\sin\alpha\\[2mm] M_{33} &= \begin{vmatrix} 1 & 0 \\ 0 & \cos\alpha \end{vmatrix} = \cos\alpha\\ A_{33} &= (-1)^{3+3}M_{33} = \cos\alpha \end{aligned}$$
   $M_{32}$ deletes row 3 and column 2, leaving rows 1, 2 and columns 1, 3, that is $1, 0$ over $0, \sin\alpha$⁠. $M_{33}$ keeps the top-left $2\times2$ block. Test: row 2 of $A$ times its own cofactors is $0\cdot0 + \cos\alpha(-\cos\alpha) + \sin\alpha(-\sin\alpha)$⁠, which is $-(\cos^2\alpha+\sin^2\alpha) = -1 = |A|$⁠, as it must be.
   Row times its own cofactors gives $|A|$ (Remark, Section 4.4). Lesson: minors and cofactors

5. Matrix of cofactors and $\operatorname{adj}A$

   $$[A_{ij}] = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha \end{bmatrix}$$

   Transposing swaps entry $(i,j)$ with entry $(j,i)$⁠. Here the only off-diagonal pair that is not zero is $A_{23} = A_{32} = -\sin\alpha$⁠, so the matrix is symmetric and the transpose changes nothing:

   $$\operatorname{adj}A = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha \end{bmatrix}$$
   Even when the transpose turns out to change nothing, it must be checked: it is part of the definition of $\operatorname{adj}A$⁠. Compare with $A$⁠: every entry of $\operatorname{adj}A$ is the negative of the matching entry of $A$⁠, so $\operatorname{adj}A = -A$⁠.
   A matrix equal to its own transpose is symmetric. Basics: transpose and identity · Lesson: the adjoint

6. Divide by $|A| = -1$

   $$A^{-1} = \frac{1}{-1}\operatorname{adj}A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}$$

   This is $A$ itself: $A^{-1} = A$.

   Theorem 4 gives $A^{-1} = \frac{1}{|A|}\operatorname{adj}A$⁠. Since $|A| = -1$ and $\operatorname{adj}A = -A$⁠, the two minus signs cancel. A matrix that is its own inverse satisfies $A\cdot A = I$⁠, which we check next.
   Lesson: inverse of a matrix

7. Check: $A\,A^{-1} = A\cdot A = I$

   Both factors are $3\times3$⁠. Write $c$ for $\cos\alpha$ and $s$ for $\sin\alpha$ to keep the rows short. Row 1 of $A$ is $1, 0, 0$⁠, so row 1 of the product is row 1 of the second factor, $1, 0, 0$⁠. Likewise column 1 of the second factor is $1, 0, 0$⁠, so $c_{21} = c_{31} = 0$⁠. The remaining four entries:

   $$\begin{aligned} c_{22} &= 0\cdot0 + c\cdot c + s\cdot s = 1\\ c_{23} &= 0\cdot0 + c\cdot s + s(-c) = 0\\ c_{32} &= 0\cdot0 + s\cdot c + (-c)s = 0\\ c_{33} &= 0\cdot0 + s\cdot s + (-c)(-c) = 1 \end{aligned}$$

   (using $c^2 + s^2 = 1$ in $c_{22}$ and $c_{33}$⁠). So

   $$A\cdot A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$$
   The product confirms the inverse for every $\alpha$⁠, because it only uses $c^2+s^2 = 1$ and $cs = sc$⁠, which hold for all angles. The figure gives the geometric reason: $A$ is a mirror reflection (in the last two coordinates), and doing a reflection twice returns every point to where it started. To see that the $2\times2$ block really is a reflection, write $h = \alpha/2$⁠, so $\alpha - h = h$⁠. By the formulas for $\cos(\alpha - h)$ and $\sin(\alpha - h)$⁠, the block sends $(\cos h, \sin h)$ to $(\cos(\alpha - h), \sin(\alpha - h))$⁠, which is $(\cos h, \sin h)$ itself, and sends $(-\sin h, \cos h)$ to $(\sin h, -\cos h)$⁠, its negative. So it fixes every point of the violet line and reverses the direction perpendicular to it, which is exactly what a reflection in that line does.
   Row times column, then the Pythagorean identity. Basics: matrix facts the chapter uses

Every fact used, and where it comes from