Types of matrices, Algebra of matrices Questions in English

(B) Given $A = \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix}$.
First,calculate $A^2 = A \times A = \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix} = \begin{bmatrix} 1(1)+2(3) & 1(2)+2(5) \\ 3(1)+5(3) & 3(2)+5(5) \end{bmatrix} = \begin{bmatrix} 7 & 12 \\ 18 & 31 \end{bmatrix}$.
Substitute $A^2$ and $A$ into the equation $\alpha A^2 - \beta A = 2I$:
$\alpha \begin{bmatrix} 7 & 12 \\ 18 & 31 \end{bmatrix} - \beta \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix} = 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
This gives the matrix equation:
$\begin{bmatrix} 7\alpha - \beta & 12\alpha - 2\beta \\ 18\alpha - 3\beta & 31\alpha - 5\beta \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$.
Comparing the elements,we get:
$7\alpha - \beta = 2$ $(i)$
$12\alpha - 2\beta = 0 \Rightarrow 6\alpha - \beta = 0$ $(ii)$
Subtracting $(ii)$ from $(i)$ gives $\alpha = 2$.
Substituting $\alpha = 2$ into $(ii)$,we get $6(2) - \beta = 0 \Rightarrow \beta = 12$.
Finally,$\alpha^2 + \beta = (2)^2 + 12 = 4 + 12 = 16$.

(C) Given: $A=\begin{bmatrix} a & 3 & 5 \\ 5 & -1 & 3 \\ 2 & 3 & -4 \end{bmatrix}$ and $B=\begin{bmatrix} b & 1 & 4 \\ 4 & c & 1 \\ -3 & 1 & d \end{bmatrix}$.
Since the trace of $A$ is $-4$,we have $a - 1 - 4 = -4$,which implies $a = 1$.
Now,compute the product $AB$:
$AB = \begin{bmatrix} 1 & 3 & 5 \\ 5 & -1 & 3 \\ 2 & 3 & -4 \end{bmatrix} \begin{bmatrix} b & 1 & 4 \\ 4 & c & 1 \\ -3 & 1 & d \end{bmatrix} = \begin{bmatrix} b+12-15 & 1+3c+5 & 4+3+5d \\ 5b-4-9 & 5-c+3 & 20-1+3d \\ 2b+12+12 & 2+3c-4 & 8+3-4d \end{bmatrix} = \begin{bmatrix} b-3 & 3c+6 & 5d+7 \\ 5b-13 & 8-c & 3d+19 \\ 2b+24 & 3c-2 & 11-4d \end{bmatrix}$.
Comparing this with the given matrix $\begin{bmatrix} -1 & 0 & 17 \\ -3 & 10 & 25 \\ 28 & -8 & 3 \end{bmatrix}$:
From $b-3 = -1$,we get $b = 2$.
From $3c+6 = 0$,we get $c = -2$.
From $5d+7 = 17$,we get $5d = 10$,so $d = 2$.
Thus,$a+b+c+d = 1 + 2 - 2 + 2 = 3$.

(D) Given $A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$.
First,calculate $A^2 = A \cdot A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{bmatrix}$.
Next,calculate $A^3 = A^2 \cdot A = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$.
Since $A^3 = I$,we can find $A^5$ as follows:
$A^5 = A^3 \cdot A^2 = I \cdot A^2 = A^2$.
Calculating $A^2$ again: $A^2 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{bmatrix}$.
However,checking the options provided,we re-evaluate $A^3$.
$A^3 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$.
Thus $A^5 = A^2$. Since $A^2$ is not listed,let us check $A^{-1}$.
$A^{-1} = \frac{1}{|A|} \text{adj}(A)$. $|A| = 1(0-0) - (-1)(0-0) + 1(0 - (-1)) = 1$.
$\text{adj}(A) = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{bmatrix}^T = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & -1 \\ 1 & 2 & 1 \end{bmatrix}$.
Since $A^3 = I$,$A^2 = A^{-1}$. Therefore $A^5 = A^2 = A^{-1}$.

(D) Given $A = \begin{bmatrix} 0 & 5 \\ 0 & 0 \end{bmatrix}$.
First,calculate $A^2$:
$A^2 = \begin{bmatrix} 0 & 5 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 5 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Since $A^2 = O$ (the zero matrix),all higher powers of $A$ will also be zero matrices,i.e.,$A^n = O$ for all $n \geq 2$.
Given $f(x) = x + x^2 + \dots + x^{2018}$,we have:
$f(A) = A + A^2 + A^3 + \dots + A^{2018}$.
Substituting the powers of $A$:
$f(A) = A + O + O + \dots + O = A$.
Now,calculate $f(A) + I$:
$f(A) + I = A + I = \begin{bmatrix} 0 & 5 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 5 \\ 0 & 1 \end{bmatrix}$.

(A) Given $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$.
We calculate the powers of $A$:
$A^2 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$
$A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$
By observation,we hypothesize that $A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$.
Now,evaluate the expression $nA - (n-1)I$:
$nA - (n-1)I = n \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} - (n-1) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$= \begin{bmatrix} n & n \\ 0 & n \end{bmatrix} - \begin{bmatrix} n-1 & 0 \\ 0 & n-1 \end{bmatrix}$
$= \begin{bmatrix} n - (n-1) & n - 0 \\ 0 - 0 & n - (n-1) \end{bmatrix} = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} = A^n$.
Thus,$A^n = nA - (n-1)I$ is correct.

(D) Given,$A=\left[\begin{array}{rr}i & -i \\ -i & i\end{array}\right]$ and $B=\left[\begin{array}{rr}1 & -1 \\ -1 & 1\end{array}\right]$.
First,calculate $A^2$:
$A^2 = \left[\begin{array}{rr}i & -i \\ -i & i\end{array}\right] \left[\begin{array}{rr}i & -i \\ -i & i\end{array}\right] = \left[\begin{array}{rr}i^2+i^2 & -i^2-i^2 \\ -i^2-i^2 & i^2+i^2\end{array}\right] = \left[\begin{array}{rr}-2 & 2 \\ 2 & -2\end{array}\right] = -2 \left[\begin{array}{rr}1 & -1 \\ -1 & 1\end{array}\right] = -2B$.
Now,calculate $A^8$ using $A^2 = -2B$:
$A^8 = (A^2)^4 = (-2B)^4 = (-2)^4 B^4 = 16 B^4$.
Next,calculate $B^2$:
$B^2 = \left[\begin{array}{rr}1 & -1 \\ -1 & 1\end{array}\right] \left[\begin{array}{rr}1 & -1 \\ -1 & 1\end{array}\right] = \left[\begin{array}{rr}2 & -2 \\ -2 & 2\end{array}\right] = 2B$.
Then $B^4 = (B^2)^2 = (2B)^2 = 4B^2 = 4(2B) = 8B$.
Finally,$A^8 = 16(8B) = 128B$.

(A) Given that,$A = \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix} \dots (i)$
First,we calculate $A^2$:
$A^2 = A \cdot A = \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} (-1)(-1) + (0)(0) & (-1)(0) + (0)(2) \\ (0)(-1) + (2)(0) & (0)(0) + (2)(2) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix}$
Next,we calculate $A^3$:
$A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} (1)(-1) + (0)(0) & (1)(0) + (0)(2) \\ (0)(-1) + (4)(0) & (0)(0) + (4)(2) \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & 8 \end{bmatrix}$
Now,calculate $A^3 - A^2$:
$A^3 - A^2 = \begin{bmatrix} -1 & 0 \\ 0 & 8 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} -1-1 & 0-0 \\ 0-0 & 8-4 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & 4 \end{bmatrix}$
Since $A = \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}$,we can see that $2A = 2 \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & 4 \end{bmatrix}$.
Therefore,$A^3 - A^2 = 2A$.

(B) Given equations are:
$(1)$ $A + 2B = \begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{bmatrix}$
$(2)$ $2A - B = \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix}$
Multiply equation $(2)$ by $2$:
$4A - 2B = \begin{bmatrix} 4 & -2 & 10 \\ 4 & -2 & 12 \\ 0 & 2 & 4 \end{bmatrix}$ $(3)$
Adding $(1)$ and $(3)$:
$5A = \begin{bmatrix} 1+4 & 2-2 & 0+10 \\ 6+4 & -3-2 & 3+12 \\ -5+0 & 3+2 & 1+4 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 10 \\ 10 & -5 & 15 \\ -5 & 5 & 5 \end{bmatrix}$
$A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & -1 & 3 \\ -1 & 1 & 1 \end{bmatrix}$
$\operatorname{tr}(A) = 1 + (-1) + 1 = 1$
From $(1)$,$2B = \begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{bmatrix} - A = \begin{bmatrix} 0 & 2 & -2 \\ 4 & -2 & 0 \\ -4 & 2 & 0 \end{bmatrix}$
$B = \begin{bmatrix} 0 & 1 & -1 \\ 2 & -1 & 0 \\ -2 & 1 & 0 \end{bmatrix}$
$\operatorname{tr}(B) = 0 + (-1) + 0 = -1$
$\operatorname{tr}(A) - \operatorname{tr}(B) = 1 - (-1) = 2$

(D) Given matrices: $P_{2 \times 3}$,$X_{4 \times 3}$,$Y_{4 \times 3}$.
$X^T$ has order $3 \times 4$.
$P^T$ has order $3 \times 2$.
$X^T Y$ has order $(3 \times 4) \times (4 \times 3) = 3 \times 3$.
$(X^T Y)^{-1}$ has order $3 \times 3$.
$P(X^T Y)^{-1}$ has order $(2 \times 3) \times (3 \times 3) = 2 \times 3$.
$P(X^T Y)^{-1} P^T$ has order $(2 \times 3) \times (3 \times 2) = 2 \times 2$.
Finally,the transpose of a $2 \times 2$ matrix remains $2 \times 2$.

(B) Given that $A$ and $B$ are symmetric matrices,we have $A^{\prime} = A$ and $B^{\prime} = B$.
Consider the transpose of the matrix $(AB - BA)$:
$(AB - BA)^{\prime} = (AB)^{\prime} - (BA)^{\prime}$
Using the property $(XY)^{\prime} = Y^{\prime}X^{\prime}$,we get:
$(AB - BA)^{\prime} = B^{\prime}A^{\prime} - A^{\prime}B^{\prime}$
Substituting $A^{\prime} = A$ and $B^{\prime} = B$:
$(AB - BA)^{\prime} = BA - AB$
$(AB - BA)^{\prime} = -(AB - BA)$
Since the transpose of the matrix $(AB - BA)$ is equal to its negative,the matrix $(AB - BA)$ is a skew-symmetric matrix.

(C) Given $P = \begin{bmatrix} \cos \frac{\pi}{4} & -\sin \frac{\pi}{4} \\ \sin \frac{\pi}{4} & \cos \frac{\pi}{4} \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix}$.
Note that $P$ is a rotation matrix $R_{\theta}$ where $\theta = \frac{\pi}{4}$.
The property of a rotation matrix is $R_{\theta}^n = R_{n\theta}$.
Therefore,$P^3 = R_{3 \times \frac{\pi}{4}} = R_{\frac{3\pi}{4}} = \begin{bmatrix} \cos \frac{3\pi}{4} & -\sin \frac{3\pi}{4} \\ \sin \frac{3\pi}{4} & \cos \frac{3\pi}{4} \end{bmatrix} = \begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$.
Now,$P^3 X = \begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix}$.
$P^3 X = \begin{bmatrix} (-\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}) \\ (\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}) \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} - \frac{1}{2} \\ \frac{1}{2} - \frac{1}{2} \end{bmatrix} = \begin{bmatrix} -1 \\ 0 \end{bmatrix}$.

(A) Given $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$.
By the property of rotation matrices,$A^n = \begin{bmatrix} \cos n \theta & -\sin n \theta \\ \sin n \theta & \cos n \theta \end{bmatrix}$.
For $n = 100$,we have $A^{100} = \begin{bmatrix} \cos 100 \theta & -\sin 100 \theta \\ \sin 100 \theta & \cos 100 \theta \end{bmatrix}$.
Given $\theta = \frac{2 \pi}{7}$,so $100 \theta = \frac{200 \pi}{7}$.
We can write $\frac{200 \pi}{7} = \frac{196 \pi + 4 \pi}{7} = 28 \pi + \frac{4 \pi}{7}$.
Since $\cos(28 \pi + \alpha) = \cos \alpha$ and $\sin(28 \pi + \alpha) = \sin \alpha$,we get $A^{100} = \begin{bmatrix} \cos \frac{4 \pi}{7} & -\sin \frac{4 \pi}{7} \\ \sin \frac{4 \pi}{7} & \cos \frac{4 \pi}{7} \end{bmatrix} = A^2$.
Comparing this with the options,none of the provided options match $A^2$ exactly. However,evaluating the expression,$A^{100} = A^2$ is the correct result.

(D) Given $A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$.
First,we check if $A$ is symmetric or skew-symmetric. The transpose $A^T = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} = A$. Since $A^T = A$,$A$ is a symmetric matrix.
Next,we calculate $A^2 = A \times A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$.
Since $A^2 = I$,the correct option is $D$.

(B) Given $A = \begin{bmatrix} 1 & 1 \\ 0 & i \end{bmatrix}$.
Calculate the powers of $A$:
$A^2 = \begin{bmatrix} 1 & 1 \\ 0 & i \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & i \end{bmatrix} = \begin{bmatrix} 1 & 1+i \\ 0 & i^2 \end{bmatrix} = \begin{bmatrix} 1 & 1+i \\ 0 & -1 \end{bmatrix}$.
$A^3 = A^2 \times A = \begin{bmatrix} 1 & 1+i \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & i \end{bmatrix} = \begin{bmatrix} 1 & 1+i+i \\ 0 & -i \end{bmatrix} = \begin{bmatrix} 1 & 1+2i \\ 0 & -i \end{bmatrix}$.
Wait,let us re-evaluate $A^4$:
$A^4 = A^2 \times A^2 = \begin{bmatrix} 1 & 1+i \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1+i \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 1+i-1-i \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.
Since $A^4 = I$,we have $A^{2018} = A^{2016} \times A^2 = (A^4)^{504} \times A^2 = I^{504} \times A^2 = A^2$.
Thus,$A^{2018} = \begin{bmatrix} 1 & 1+i \\ 0 & -1 \end{bmatrix}$.
Comparing with $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$,we get $a=1$ and $d=-1$.
Therefore,$a+d = 1 + (-1) = 0$.

(C) Let the given equations be:
$(0\,1\,2) M = (1\,0\,0)$ --- $(i)$
$(3\,4\,5) M = (0\,1\,0)$ --- $(ii)$
We want to find $(6\,7\,8) M$.
Observe the linear combination of the row vectors $(0\,1\,2)$ and $(3\,4\,5)$.
Let $x(0\,1\,2) + y(3\,4\,5) = (6\,7\,8)$.
Comparing components:
$3y = 6 \Rightarrow y = 2$
$x + 4y = 7 \Rightarrow x + 8 = 7 \Rightarrow x = -1$
$2x + 5y = 2(-1) + 5(2) = -2 + 10 = 8$. This matches the third component.
Thus, $(6\,7\,8) = -1(0\,1\,2) + 2(3\,4\,5)$.
Multiplying by $M$ on the right:
$(6\,7\,8) M = -1((0\,1\,2) M) + 2((3\,4\,5) M)$
$(6\,7\,8) M = -1(1\,0\,0) + 2(0\,1\,0)$
$(6\,7\,8) M = (-1\,0\,0) + (0\,2\,0) = (-1\,2\,0)$.

(D) In matrix algebra,the product of two non-zero matrices can result in a null matrix.
For example,consider $A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$.
Both $A$ and $B$ are non-zero matrices,but their product $AB = O_{3}$.
Therefore,it is possible that $A \neq O_{3}$ and $B \neq O_{3}$.

(B) Given $A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$.
We calculate the powers of $A$:
$A^2 = A \cdot A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & \frac{2(2+1)}{2} \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix}$.
$A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 & 6 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 & \frac{3(3+1)}{2} \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix}$.
By observing the pattern,for any positive integer $n$,we have:
$A^n = \begin{bmatrix} 1 & n & \frac{n(n+1)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1 \end{bmatrix}$.

(C) Given $Q = \begin{bmatrix} \cos \frac{\pi}{4} & -\sin \frac{\pi}{4} \\ \sin \frac{\pi}{4} & \cos \frac{\pi}{4} \end{bmatrix}$ and $x = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix}$.
We know that the matrix $Q(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$ represents a rotation matrix.
By the property of rotation matrices,$Q^{n}(\theta) = Q(n\theta)$.
Therefore,$Q^{3} = Q\left(3 \times \frac{\pi}{4}\right) = Q\left(\frac{3\pi}{4}\right) = \begin{bmatrix} \cos \frac{3\pi}{4} & -\sin \frac{3\pi}{4} \\ \sin \frac{3\pi}{4} & \cos \frac{3\pi}{4} \end{bmatrix}$.
Evaluating the trigonometric values: $\cos \frac{3\pi}{4} = -\frac{1}{\sqrt{2}}$ and $\sin \frac{3\pi}{4} = \frac{1}{\sqrt{2}}$.
So,$Q^{3} = \begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$.
Now,calculate $Q^{3}x = \begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix}$.
$Q^{3}x = \begin{bmatrix} (-\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}) \\ (\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}) \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} - \frac{1}{2} \\ \frac{1}{2} - \frac{1}{2} \end{bmatrix} = \begin{bmatrix} -1 \\ 0 \end{bmatrix}$.

(D) Given $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 2 & 0 & 2 \end{bmatrix}$.
We can write $A = 2I + B$,where $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 0 \end{bmatrix}$.
Note that $B^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O$.
Since $2I$ and $B$ commute,we use the Binomial Theorem:
$A^n = (2I + B)^n = \sum_{k=0}^n \binom{n}{k} (2I)^{n-k} B^k = \binom{n}{0} (2I)^n + \binom{n}{1} (2I)^{n-1} B + 0 + ...$
$A^n = 2^n I + n(2^{n-1}) B = 2^n \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + n 2^{n-1} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 0 \end{bmatrix}$
$A^n = \begin{bmatrix} 2^n & 0 & 0 \\ 0 & 2^n & 0 \\ 0 & 0 & 2^n \end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ n 2^n & 0 & 0 \end{bmatrix} = \begin{bmatrix} 2^n & 0 & 0 \\ 0 & 2^n & 0 \\ n 2^n & 0 & 2^n \end{bmatrix}$.
Comparing this with the given form,we get $a = 2^n$ and $b = n 2^n$.

(C) Given $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $P = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2 \end{bmatrix}$.
We calculate $P^2$ and $P^3$:
$P^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{bmatrix}$.
$P^3 = P^2 \cdot P = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -8 \end{bmatrix}$.
Now,calculate $P^3 + 2P^2$:
$P^3 + 2P^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -8 \end{bmatrix} + 2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} 1+2 & 0 & 0 \\ 0 & -1+2 & 0 \\ 0 & 0 & -8+8 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$.
Comparing this with the options,we check $2I + P$:
$2I + P = 2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2 \end{bmatrix} = \begin{bmatrix} 2+1 & 0 & 0 \\ 0 & 2-1 & 0 \\ 0 & 0 & 2-2 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$.
Thus,$P^3 + 2P^2 = 2I + P$.

(A) Given,$P = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}$.
First,we calculate $P^2 = P \cdot P$:
$P^2 = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}$
$P^2 = \begin{bmatrix} (4+2-4) & (-4-6+8) & (-8-8+12) \\ (-2-3+4) & (2+9-8) & (4+12-12) \\ (2+2-3) & (-2-6+6) & (-4-8+9) \end{bmatrix}$
$P^2 = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} = P$.
Since $P^2 = P$,it follows that $P^3 = P^2 \cdot P = P \cdot P = P^2 = P$.
By induction,$P^n = P$ for all positive integers $n \ge 1$.
Therefore,$P^5 = P$.

(A) Given,$P = \begin{bmatrix} 1 & 2 & 1 \\ 1 & 3 & 1 \end{bmatrix}$.
We need to find $Q = P P^{T}$.
$P^{T} = \begin{bmatrix} 1 & 1 \\ 2 & 3 \\ 1 & 1 \end{bmatrix}$.
$Q = \begin{bmatrix} 1 & 2 & 1 \\ 1 & 3 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 2 & 3 \\ 1 & 1 \end{bmatrix}$.
$Q = \begin{bmatrix} (1 \times 1 + 2 \times 2 + 1 \times 1) & (1 \times 1 + 2 \times 3 + 1 \times 1) \\ (1 \times 1 + 3 \times 2 + 1 \times 1) & (1 \times 1 + 3 \times 3 + 1 \times 1) \end{bmatrix}$.
$Q = \begin{bmatrix} (1 + 4 + 1) & (1 + 6 + 1) \\ (1 + 6 + 1) & (1 + 9 + 1) \end{bmatrix} = \begin{bmatrix} 6 & 8 \\ 8 & 11 \end{bmatrix}$.
Now,the determinant of $Q$ is $|Q| = (6 \times 11) - (8 \times 8)$.
$|Q| = 66 - 64 = 2$.

(C) For the sum $A+B$ to be defined,the matrices $A$ and $B$ must have the same order,say $m \times n$.
For the product $AB$ to be defined,the number of columns in $A$ must be equal to the number of rows in $B$.
Since $A$ is of order $m \times n$,it has $n$ columns. Since $B$ is of order $m \times n$,it has $m$ rows.
Therefore,for $AB$ to be defined,we must have $n = m$.
Since $m = n$,both matrices have the same order $n \times n$,which means they are square matrices of the same order.

(A) To find the product $AB$,we multiply the rows of matrix $A$ by the columns of matrix $B$:
$AB = \begin{bmatrix} 2 & 1 & 3 \\ 4 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 0 & 2 \\ 5 & 0 \end{bmatrix}$
$AB = \begin{bmatrix} (2 \times 1 + 1 \times 0 + 3 \times 5) & (2 \times -1 + 1 \times 2 + 3 \times 0) \\ (4 \times 1 + 1 \times 0 + 0 \times 5) & (4 \times -1 + 1 \times 2 + 0 \times 0) \end{bmatrix}$
$AB = \begin{bmatrix} (2 + 0 + 15) & (-2 + 2 + 0) \\ (4 + 0 + 0) & (-4 + 2 + 0) \end{bmatrix}$
$AB = \begin{bmatrix} 17 & 0 \\ 4 & -2 \end{bmatrix}$

(A) Given the conditions for the elements of the $3 \times 3$ matrix $A$:
$a_{ij} = 0$ if $i = j$
$a_{ij} = 1$ if $i > j$
$a_{ij} = -1$ if $i < j$
Constructing the matrix $A$:
$A = \begin{bmatrix} 0 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & 1 & 0 \end{bmatrix}$
Now,find the transpose $A^T$:
$A^T = \begin{bmatrix} 0 & 1 & 1 \\ -1 & 0 & 1 \\ -1 & -1 & 0 \end{bmatrix} = -A$
Since $A^T = -A$,the matrix $A$ is a skew-symmetric matrix.
Now,calculate the determinant $|A|$:
$|A| = 0(0 - (-1)) - (-1)(0 - (-1)) + (-1)(1 - 0)$
$|A| = 0 + 1(1) - 1(1) = 1 - 1 = 0$
Since the determinant $|A| = 0$,the matrix is singular and therefore not invertible.

(A) For a matrix $A = [a_{ij}]$ to be skew-symmetric,it must satisfy the condition $a_{ij} = -a_{ji}$ for all $i, j$ and $a_{ii} = 0$ for all $i$.
$1$. The diagonal elements $a_{ii}$ must be $0$. There is only $1$ choice for each of the $n$ diagonal elements.
$2$. For the off-diagonal elements,we only need to choose the elements $a_{ij}$ where $i < j$. Once these are chosen,the elements $a_{ji}$ are automatically fixed as $a_{ji} = -a_{ij}$.
$3$. The number of pairs $(i, j)$ such that $i < j$ is given by the combination formula $\binom{n}{2} = \frac{n(n-1)}{2}$.
$4$. Each of these $\frac{n(n-1)}{2}$ positions can be filled with any of the $3$ values: $\{0, 1, -1\}$.
$5$. Therefore,the total number of such skew-symmetric matrices is $3^{\frac{n(n-1)}{2}}$.

(B) matrix $A$ is orthogonal if $A^T A = I$.
Taking the determinant on both sides,we get $|A^T A| = |I|$.
Since $|A^T| = |A|$,we have $|A|^2 = 1$,which implies $|A| = \pm 1$.
Since $|A| \neq 0$,every orthogonal matrix is non-singular.
Thus,$Q \subseteq P$.
Since there exist non-singular matrices that are not orthogonal (e.g.,any diagonal matrix with entries other than $1$ or $-1$),$Q$ is a proper subset of $P$.

(A) The matrix $A$ represents a rotation matrix in $3D$ space about the $z$-axis by an angle $\theta = \frac{2\pi}{n}$.
By the property of rotation matrices,$A^k = \begin{bmatrix} \cos (k\theta) & \sin (k\theta) & 0 \\ -\sin (k\theta) & \cos (k\theta) & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
For $A^n$,we have $k = n$,so $k\theta = n \times \frac{2\pi}{n} = 2\pi$.
Thus,$A^n = \begin{bmatrix} \cos (2\pi) & \sin (2\pi) & 0 \\ -\sin (2\pi) & \cos (2\pi) & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$.
For $A^{n-1}$,the angle is $(n-1) \times \frac{2\pi}{n} = 2\pi - \frac{2\pi}{n}$.
Since $n \geq 2$,$\frac{2\pi}{n}$ is not a multiple of $2\pi$,so $A^{n-1} \neq I$.
Therefore,the correct option is $A$.

(B) The identity matrix $I$ is given by $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Rearranging the columns of $I$ results in permutation matrices.
For a $3 \times 3$ matrix,the number of ways to rearrange the $3$ columns is given by $3! = 3 \times 2 \times 1 = 6$.
Thus,there are $6$ distinct choices for $P$.
Since $P$ is a permutation matrix,its determinant is equal to the sign of the permutation,which is either $1$ or $-1$.
Therefore,$\operatorname{det}(P) = \pm 1$.
Hence,there are six distinct choices for $P$ and $\operatorname{det}(P) = \pm 1$.

(B) The set $A = \{-1, 0, 1\}$ is a finite set with cardinality $n(A) = 3$.
The set $T$ consists of all orthogonal matrices of order $3 \times 3$. The group of orthogonal matrices $O(3)$ is an infinite set.
The set $U$ consists of all non-singular matrices of order $3 \times 3$,which is the general linear group $GL(3, \mathbb{R})$. This is also an infinite set.
$A$ bijective mapping between two sets exists if and only if they have the same cardinality.
Since $n(A) = 3$ and both $T$ and $U$ are infinite sets,$n(A) \neq n(T)$ and $n(A) \neq n(U)$.
Therefore,there does not exist a bijective mapping between $A$ and $T$,nor between $A$ and $U$.

(A) For a symmetric matrix,$A^T = A$,which implies $A_{ij} = A_{ji}$ for all $i, j$.
Equating the corresponding elements:
$A_{12} = A_{21} \Rightarrow x^2 + 3x = -2x - 6$
$x^2 + 5x + 6 = 0$
$(x + 2)(x + 3) = 0 \Rightarrow x = -2$ or $x = -3$.
$A_{23} = A_{32} \Rightarrow -4x - 2 = x^2 + 2$
$x^2 + 4x + 4 = 0$
$(x + 2)^2 = 0 \Rightarrow x = -2$.
Since both conditions must be satisfied simultaneously,the common value is $x = -2$.

(A) Given $A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$.
First,calculate $A^2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.
Since $A^2 = I$,it follows that $A^3 = A^2 \cdot A = I \cdot A = A$.
Now,expand $(A+I)^3 = A^3 + 3A^2I + 3AI^2 + I^3 = A + 3I + 3A + I = 4A + 4I$.
Next,expand $(A-I)^3 = A^3 - 3A^2I + 3AI^2 - I^3 = A - 3I + 3A - I = 4A - 4I$.
Finally,add the two expressions: $(4A+4I) + (4A-4I) = 8A$.

(D) Given $A = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$.
First,calculate $A^2 = A \times A = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} (2 \times 2 + 3 \times 4) & (2 \times 3 + 3 \times 5) \\ (4 \times 2 + 5 \times 4) & (4 \times 3 + 5 \times 5) \end{bmatrix} = \begin{bmatrix} 16 & 21 \\ 28 & 37 \end{bmatrix}$.
Now,calculate $A^2 - 2I = \begin{bmatrix} 16 & 21 \\ 28 & 37 \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 14 & 21 \\ 28 & 35 \end{bmatrix}$.
We can factor out $7$ from the resulting matrix: $\begin{bmatrix} 14 & 21 \\ 28 & 35 \end{bmatrix} = 7 \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} = 7A$.
Comparing this with $A^2 - 2I = KA$,we get $K = 7$.

(B) Let $C = AB - BA$.
Taking the transpose of $C$,we get $C^T = (AB - BA)^T$.
Using the property $(X - Y)^T = X^T - Y^T$ and $(XY)^T = Y^T X^T$,we have $C^T = (AB)^T - (BA)^T = B^T A^T - A^T B^T$.
Since $A$ and $B$ are skew-symmetric,$A^T = -A$ and $B^T = -B$.
Substituting these values,$C^T = (-B)(-A) - (-A)(-B) = BA - AB$.
Factoring out a negative sign,$C^T = -(AB - BA) = -C$.
Since $C^T = -C$,the matrix $AB - BA$ is a skew-symmetric matrix.

(C) Given $A = \begin{bmatrix} a & b \\ c & -a \end{bmatrix}$.
We calculate $A^2 = A \times A = \begin{bmatrix} a & b \\ c & -a \end{bmatrix} \begin{bmatrix} a & b \\ c & -a \end{bmatrix}$.
Performing matrix multiplication: $A^2 = \begin{bmatrix} a^2 + bc & ab - ab \\ ac - ac & bc + a^2 \end{bmatrix} = \begin{bmatrix} a^2 + bc & 0 \\ 0 & a^2 + bc \end{bmatrix}$.
Given that $A^2 = I$,where $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,we equate the matrices:
$\begin{bmatrix} a^2 + bc & 0 \\ 0 & a^2 + bc \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Comparing the elements,we get $a^2 + bc = 1$.
Rearranging the equation,we get $1 - a^2 - bc = 0$.

(A) Given $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}$,then $A^2 = \begin{bmatrix} 2^2 & 0 & 0 \\ 0 & 3^2 & 0 \\ 0 & 0 & 4^2 \end{bmatrix} = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 16 \end{bmatrix}$.
Given $B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -3 \end{bmatrix}$,then $B^2 = \begin{bmatrix} 1^2 & 0 & 0 \\ 0 & (-2)^2 & 0 \\ 0 & 0 & (-3)^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 9 \end{bmatrix}$.
Therefore,$A^2 + B^2 = \begin{bmatrix} 4+1 & 0 & 0 \\ 0 & 9+4 & 0 \\ 0 & 0 & 16+9 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 13 & 0 \\ 0 & 0 & 25 \end{bmatrix}$.