Q.If $A = \begin{bmatrix} 0 & -\tan\frac{\alpha}{2} \\ \tan\frac{\alpha}{2} & 0 \end{bmatrix}$, then show that $I + A = (I-A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$, where $I$ is the identity matrix of order 2. OR If $A = \begin{bmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{bmatrix}$, then find $A^{-1}$; and hence solve the system of equations $x+2y-3z=-4$, $2x+3y+2z=2$, $3x-3y-4z=11$.

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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