21. Elective 4-2: Matrices and Transformations

This problem mainly tests basic knowledge of matrices and inverse matrices, tests computational ability, and tests transformation and conversion ideas. Full marks: 7 points.

Solution: (1) Since $|A| = 2 \times 1 - (-1) \times 4 = 2$,

we have $A^{-1} = \begin{pmatrix} \frac{3}{2} & -\frac{1}{2} \\ -2 & 1 \end{pmatrix}$.

(2) From $AC = B$ we get $(A^{-1}A)C = A^{-1}B$,

thus $C = A^{-1}B = \begin{pmatrix} \frac{3}{2} & -\frac{1}{2} \\ -2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} \frac{3}{2} & 2 \\ -2 & -3 \end{pmatrix}$.

Elective 4-4: Coordinate Systems and Parametric Equations

This problem mainly tests basic knowledge of conversion between polar and rectangular coordinates and parametric equations of circles, tests computational ability, and tests transformation and conversion ideas. Full marks: 7 points.

Solution: (1) Eliminating the parameter $t$, we obtain the ordinary equation of the circle: $(x-1)^2 + (y+2)^2 = 9$.

From $\sqrt{2}r\sin\left(q - \frac{p}{4}\right) = m$, we get $r\sin q - r\cos q - m = 0$,

so the rectangular coordinate equation of line $l$ is $x - y - m = 0$.

(2) According to the problem, the distance from center $C$ to line $l$ equals 2, i.e.,

$$\frac{|1 - (-2) - m|}{\sqrt{2}} = 2,$$

solving we get $m = -3 \pm 2\sqrt{2}$.

Elective 4-5: Inequalities

This problem mainly tests basic knowledge of absolute value inequalities and Cauchy inequality, tests reasoning and proof ability, and tests transformation and conversion ideas. Full marks: 7 points.

Solution: (1) Since $f(x) = |x+a| + |x+b| + c \geq |(