Algebra

Let the three original numbers be $x,\;2x,\;3x$.

After adding $2,\;4,\;11$ respectively, the numbers become

$$x+2,\quad 2x+4,\quad 3x+11.$$

Since they are in G.P.,

$$(2x+4)^2=(x+2)(3x+11).$$

Expanding:

$$4x^2+16x+16=3x^2+17x+22$$ $$x^2-x-6=0$$ $$(x-3)(x+2)=0$$

So,

$$x=3 \quad \text{or} \quad x=-2.$$

If $x=-2$, the first two resulting numbers become $0,0$, which do not form a proper G.P. Hence $x=-2$ is rejected.

Therefore,

$$x=3,\quad 2x=6,\quad 3x=9.$$ $$\boxed{\text{The original three numbers are }3,\;6,\;9.}$$

Check: adding $2,4,11$ gives $5,10,20$, which are in G.P. with common ratio $2$.

Let the two geometric means be $G_1$ and $G_2$. Then

$9,\ G_1,\ G_2,\ 243$

are in GP. If the common ratio is $r$, then

$243=9r^3 \quad\Rightarrow\quad r^3=27 \quad\Rightarrow\quad r=3$

So,

$G_1=9\times 3=27,\qquad G_2=27\times 3=81$

The two geometric means are 27 and 81.

The system is $x+y+z=6,\qquad 2x+3y-z=5,\qquad 3x+4y+5z=26.$

Determinant of coefficients: $D=\begin{vmatrix}1&1&1\\2&3&-1\\3&4&5\end{vmatrix}=1(15+4)-1(10+3)+1(8-9)=19-13-1=5\ne0$

Replacing the columns by the constants: $D_x=\begin{vmatrix}6&1&1\\5&3&-1\\26&4&5\end{vmatrix}=6(19)-1(25+26)+1(20-78)=114-51-58=5$

$D_y=\begin{vmatrix}1&6&1\\2&5&-1\\3&26&5\end{vmatrix}=1(25+26)-6(10+3)+1(52-15)=51-78+37=10$

$D_z=\begin{vmatrix}1&1&6\\2&3&5\\3&4&26\end{vmatrix}=1(78-20)-1(52-15)+6(8-9)=58-37-6=15$

By Cramer's rule: $x=\frac{D_x}{D}=\frac55=1,\qquad y=\frac{D_y}{D}=\frac{10}{5}=2,\qquad z=\frac{D_z}{D}=\frac{15}{5}=3$

Check: $1+2+3=6$, $\;2+6-3=5$, $\;3+8+15=26$.

$\boxed{x=1,\quad y=2,\quad z=3}$

The true relation is $A^T=-A$.

$A^T=\begin{pmatrix}0&-3\\3&0\end{pmatrix}=-A$

So $A$ is a skew-symmetric matrix.

(a) 0

Expanding along the first row: $1(0\cdot0-1\cdot1)-(-1)((-1)\cdot0-1\cdot0)+(-1)((-1)\cdot1-0\cdot0)$ $=1(-1)+1(0)-1(-1)=-1+0+1=0$

Matrix form $AX=B$:

$$\begin{bmatrix} 3 & 3 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 6 \\ 2 \end{bmatrix}$$

So,

$$A=\begin{bmatrix}3&3\\1&1\end{bmatrix},\quad X=\begin{bmatrix}x\\y\end{bmatrix},\quad B=\begin{bmatrix}6\\2\end{bmatrix}$$

(ii) Determinant of $A$:

$$|A| = \begin{vmatrix}3&3\\1&1\end{vmatrix} = 3(1)-3(1)=0$$

(iii) Since $|A|=0$, the inverse of $A$ does not exist.

(iv) Since $|A|=0$, the system does not have a unique solution. The two equations are dependent, so there are infinitely many solutions.

(v) Using Gauss elimination, the augmented matrix is:

$$[A|B]=\begin{bmatrix}3&3&|&6\\1&1&|&2\end{bmatrix}$$

Apply row operations:

$$R_1 \to \frac{R_1}{3} \Rightarrow \begin{bmatrix}1&1&|&2\\1&1&|&2\end{bmatrix}$$ $$R_2 \to R_2-R_1 \Rightarrow \begin{bmatrix}1&1&|&2\\0&0&|&0\end{bmatrix}$$

Here,

$$\text{Rank}(A)=1,\quad \text{Rank}(A|B)=1,\quad n=2$$

Since

$$\text{Rank}(A)=\text{Rank}(A|B)<n$$

the system is consistent and has infinitely many solutions.

Hence, the system is dependent and has no unique solution.

Given, $A=\begin{pmatrix}1&2&-3\\2&-1&4\\3&1&3\end{pmatrix}$

Determinant of A: $|A|=1\begin{vmatrix}-1&4\\1&3\end{vmatrix}-2\begin{vmatrix}2&4\\3&3\end{vmatrix}+(-3)\begin{vmatrix}2&-1\\3&1\end{vmatrix}$ $=1(-3-4)-2(6-12)-3(2+3)$ $=-7+12-15=-10\ne0$ So, $A^{-1}$ exists.

Cofactors: $C_{11}=-7,\quad C_{12}=6,\quad C_{13}=5$ $C_{21}=-9,\quad C_{22}=12,\quad C_{23}=5$ $C_{31}=5,\quad C_{32}=-10,\quad C_{33}=-5$

Hence, the cofactor matrix is $C=\begin{pmatrix}-7&6&5\\-9&12&5\\5&-10&-5\end{pmatrix}$

Adjoint of A: $\operatorname{adj}A=C^T=\begin{pmatrix}-7&-9&5\\6&12&-10\\5&5&-5\end{pmatrix}$

Inverse of A: $A^{-1}=\frac{1}{|A|}\operatorname{adj}A$ $A^{-1}=-\frac{1}{10}\begin{pmatrix}-7&-9&5\\6&12&-10\\5&5&-5\end{pmatrix}$ $A^{-1}=\begin{pmatrix}\frac{7}{10}&\frac{9}{10}&-\frac12\\[4pt]-\frac35&-\frac65&1\\[4pt]-\frac12&-\frac12&\frac12\end{pmatrix}$

Verification:

$$A\cdot A^{-1}=\begin{pmatrix}1&2&-3\\2&-1&4\\3&1&3\end{pmatrix} \begin{pmatrix}\frac{7}{10}&\frac{9}{10}&-\frac12\\[4pt]-\frac35&-\frac65&1\\[4pt]-\frac12&-\frac12&\frac12\end{pmatrix}$$ $$=\begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}=I_3$$

Thus, $\boxed{A^{-1}=\begin{pmatrix}\frac{7}{10}&\frac{9}{10}&-\frac12\\[4pt]-\frac35&-\frac65&1\\[4pt]-\frac12&-\frac12&\frac12\end{pmatrix}}$ and it is verified that $A\cdot A^{-1}=I_3$.

Let the ages of the three students be
$\frac{a}{r},\ a,\ ar$ where $a$ is the middle term and $r$ is the common ratio.

Given product: $\frac{a}{r}\times a\times ar=a^3=64$ $a=4$

Given sum: $\frac{a}{r}+a+ar=21$ $4\left(\frac1r+1+r\right)=21$ $\frac1r+1+r=\frac{21}{4}$

Multiplying by $r$: $1+r+r^2=\frac{21r}{4}$ $4r^2+4r+4=21r$ $4r^2-17r+4=0$

Solving: $r=\frac{17\pm\sqrt{289-64}}{8}=\frac{17\pm15}{8}$ $r=4 \quad \text{or} \quad r=\frac14$

For $r=4$, the ages are: $\frac{4}{4},\ 4,\ 4\times4 = 1,\ 4,\ 16$

For $r=\frac14$, the ages are the same in reverse order: $16,\ 4,\ 1$

Hence, the ages of the three students are 1 year, 4 years and 16 years.