Q.If $A = \begin{bmatrix} 0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0 \end{bmatrix}$ and $I$ is the identity matrix of order 2, show that $I + A = (I - A) \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$.

(a) verified using $\cos\alpha=\tfrac{1-t^2}{1+t^2},\sin\alpha=\tfrac{2t}{1+t^2}$ with $t=\tan\tfrac\alpha2$; (b) the determinant is $0$, so $x=y=0$. (OR: $x=2,y=1,z=3$.)

Main (a). Let $t=\tan\tfrac{\alpha}{2}$, so $A=\begin{bmatrix}0&-t\\t&0\end{bmatrix}$, $I-A=\begin{bmatrix}1&t\\-t&1\end{bmatrix}$, and $I+A=\begin{bmatrix}1&-t\\t&1\end{bmatrix}$. Use the half-angle forms $\cos\alpha=\dfrac{1-t^{2}}{1+t^{2}}$, $\sin\alpha=\dfrac{2t}{1+t^{2}}$. Compute

$(I-A)\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}=\begin{bmatrix}\cos\alpha+t\sin\alpha&-\sin\alpha+t\cos\alpha\\-t\cos\alpha+\sin\alpha&t\sin\alpha+\cos\alpha\end{bmatrix}.$

Entry $(1,1)$: $\dfrac{1-t^{2}}{1+t^{2}}+t\cdot\dfrac{2t}{1+t^{2}}=\dfrac{1+t^{2}}{1+t^{2}}=1$. Entry $(1,2)$: $-\dfrac{2t}{1+t^{2}}+t\cdot\dfrac{1-t^{2}}{1+t^{2}}=\dfrac{-t-t^{3}}{1+t^{2}}=-t$. Entry $(2,1)$: $-t\cdot\dfrac{1-t^{2}}{1+t^{2}}+\dfrac{2t}{1+t^{2}}=\dfrac{t+t^{3}}{1+t^{2}}=t$. Entry $(2,2)$: $t\cdot\dfrac{2t}{1+t^{2}}+\dfrac{1-t^{2}}{1+t^{2}}=1$. This is exactly $\begin{bmatrix}1&-t\\t&1\end{bmatrix}=I+A$. Proved.

Main (b). Expand along the first row:

$\begin{vmatrix}6i&-3i&1\\4&3i&-1\\20&3&i\end{vmatrix}=6i\begin{vmatrix}3i&-1\\3&i\end{vmatrix}+3i\begin{vmatrix}4&-1\\20&i\end{vmatrix}+1\begin{vmatrix}4&3i\\20&3\end{vmatrix}.$ …

Main: multiply out $(I-A)R$ where $R$ is the given rotation-like matrix, using $t=\tan\frac\alpha2$ half-angle formulas for $\cos\alpha,\sin\alpha$; every entry matches $I+A$. OR: compute $\det A$, the adjugate, hence $A^{-1}$, then solve $AX=B$ via $X=A^{-1}B$.

Main question. $A=\begin{bmatrix}0&-\tan\frac\alpha2\\\tan\frac\alpha2&0\end{bmatrix}$. Let $t=\tan\frac\alpha2$, so $A=\begin{bmatrix}0&-t\\t&0\end{bmatrix}$.

$I+A = \begin{bmatrix}1&-t\\t&1\end{bmatrix},\qquad I-A=\begin{bmatrix}1&t\\-t&1\end{bmatrix}.$

Recall the half-angle identities: $\cos\alpha=\dfrac{1-t^2}{1+t^2}$, $\sin\alpha=\dfrac{2t}{1+t^2}$.

Compute $(I-A)\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}$ entrywise (row of $I-A$ dotted with column of the rotation matrix):

• $(1,1)$: $\cos\alpha+t\sin\alpha = \dfrac{1-t^2}{1+t^2}+t\cdot\dfrac{2t}{1+t^2} = \dfrac{1-t^2+2t^2}{1+t^2}=\dfrac{1+t^2}{1+t^2}=1$. ✓ matches $(I+A)_{11}=1$.
• $(1,2)$: $-\sin\alpha+t\cos\alpha = \dfrac{-2t+t(1-t^2)}{1+t^2} = \dfrac{-t-t^3}{1+t^2} = -t$. ✓ matches $(I+A)_{12}=-t$.
• $(2,1)$: $-t\cos\alpha+\sin\alpha = \dfrac{-t(1-t^2)+2t}{1+t^2} = \dfrac{t+t^3}{1+t^2} = t$. ✓ matches $(I+A)_{21}=t$.
• $(2,2)$: $t\sin\alpha+\cos\alpha = \dfrac{2t^2+1-t^2}{1+t^2}=\dfrac{1+t^2}{1+t^2}=1$. ✓ matches $(I+A)_{22}=1$.

All four entries agree, so $(I-A)\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix} = I+A$, as required.

OR question. $A=\begin{bmatrix}1&2&-3\\2&3&2\\3&-3&-4\end{bmatrix}$.

Step 1 — determinant. Expanding along row 1:

$|A| = 1[3(-4)-2(-3)] - 2[2(-4)-2(3)] + (-3)[2(-3)-3(3)] = 1(-6)-2(-14)-3(-15) = -6+28+45 = 67.$

Step 2 — cofactors and adjugate.

$C_{11}=-6,\ C_{12}=14,\ C_{13}=-15,\quad C_{21}=17,\ C_{22}=5,\ C_{23}=9,\quad C_{31}=13,\ C_{32}=-8,\ C_{33}=-1.$

$\mathrm{adj}(A)$ is the transpose of the cofactor matrix:

$\mathrm{adj}(A)=\begin{bmatrix}-6&17&13\\14&5&-8\\-15&9&-1\end{bmatrix}.$

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