Types of matrices, Algebra of matrices Questions in English

(B) Given $A = \begin{bmatrix} 3 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 3 \end{bmatrix}$.
First,calculate $A^2 = A \times A$:
$A^2 = \begin{bmatrix} 3 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 3 \end{bmatrix} \begin{bmatrix} 3 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 3 \end{bmatrix} = \begin{bmatrix} 9+9+9 & 9+9+9 & 9+9+9 \\ 9+9+9 & 9+9+9 & 9+9+9 \\ 9+9+9 & 9+9+9 & 9+9+9 \end{bmatrix} = \begin{bmatrix} 27 & 27 & 27 \\ 27 & 27 & 27 \\ 27 & 27 & 27 \end{bmatrix} = 9A$.
Now,calculate $A^3 = A^2 \times A$:
$A^3 = (9A) \times A = 9(A^2) = 9(9A) = 81A$.
Therefore,the correct option is $B$.

(C) First,multiply the first two matrices: $\begin{bmatrix} 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & x & 3 \\ 2 & 4 & 5 \\ 3 & 2 & x \end{bmatrix} = \begin{bmatrix} (2(1)+3(2)+4(3)) & (2(x)+3(4)+4(2)) & (2(3)+3(5)+4(x)) \end{bmatrix} = \begin{bmatrix} 20 & 2x+20 & 21+4x \end{bmatrix}$.
Now,multiply this result by the third matrix: $\begin{bmatrix} 20 & 2x+20 & 21+4x \end{bmatrix} \begin{bmatrix} x \\ 2 \\ 0 \end{bmatrix} = 20(x) + (2x+20)(2) + (21+4x)(0) = 0$.
This simplifies to: $20x + 4x + 40 = 0$.
$24x = -40$.
$x = -\frac{40}{24} = -\frac{5}{3}$.

(A) Given $A = \begin{bmatrix} 1 & 4 & 4 \\ 4 & 1 & 4 \\ 4 & 4 & 1 \end{bmatrix}$.
First,calculate $A^2 = A \times A$:
$A^2 = \begin{bmatrix} 1 & 4 & 4 \\ 4 & 1 & 4 \\ 4 & 4 & 1 \end{bmatrix} \begin{bmatrix} 1 & 4 & 4 \\ 4 & 1 & 4 \\ 4 & 4 & 1 \end{bmatrix} = \begin{bmatrix} 1+16+16 & 4+4+16 & 4+16+4 \\ 4+4+16 & 16+1+16 & 16+4+4 \\ 4+16+4 & 16+4+4 & 16+16+1 \end{bmatrix} = \begin{bmatrix} 33 & 24 & 24 \\ 24 & 33 & 24 \\ 24 & 24 & 33 \end{bmatrix}$.
Now,calculate $6A$:
$6A = 6 \begin{bmatrix} 1 & 4 & 4 \\ 4 & 1 & 4 \\ 4 & 4 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 24 & 24 \\ 24 & 6 & 24 \\ 24 & 24 & 6 \end{bmatrix}$.
Finally,calculate $A^2 - 6A$:
$A^2 - 6A = \begin{bmatrix} 33 & 24 & 24 \\ 24 & 33 & 24 \\ 24 & 24 & 33 \end{bmatrix} - \begin{bmatrix} 6 & 24 & 24 \\ 24 & 6 & 24 \\ 24 & 24 & 6 \end{bmatrix} = \begin{bmatrix} 27 & 0 & 0 \\ 0 & 27 & 0 \\ 0 & 0 & 27 \end{bmatrix} = 27 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = 27 I_3$.
Thus,the correct option is $A$.

(C) Given the matrix $A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$.
We are given that $A^2 = I$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
First,calculate $A^2$:
$A^2 = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$
$A^2 = \begin{bmatrix} \alpha^2 + \beta \gamma & \alpha \beta - \beta \alpha \\ \gamma \alpha - \alpha \gamma & \beta \gamma + \alpha^2 \end{bmatrix}$
$A^2 = \begin{bmatrix} \alpha^2 + \beta \gamma & 0 \\ 0 & \alpha^2 + \beta \gamma \end{bmatrix}$
Since $A^2 = I$,we have:
$\begin{bmatrix} \alpha^2 + \beta \gamma & 0 \\ 0 & \alpha^2 + \beta \gamma \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
Comparing the elements,we get $\alpha^2 + \beta \gamma = 1$.
Rearranging the terms,we get $1 - \alpha^2 - \beta \gamma = 0$.
Therefore,the correct option is $C$.

(B) In general,matrix multiplication is not commutative. For two square matrices $A$ and $B$ of the same order,the product $AB$ is not necessarily equal to $BA$. Therefore,$AB \neq BA$ is the correct statement representing the general property of matrix multiplication.

(A) Let the order of matrix $B$ be $p \times q$.
Then the order of $B^{\prime}$ is $q \times p$.
For $AB^{\prime}$ to be defined,the number of columns in $A$ must equal the number of rows in $B^{\prime}$.
Thus,$n = q$.
For $B^{\prime}A$ to be defined,the number of columns in $B^{\prime}$ must equal the number of rows in $A$.
Thus,$p = m$.
Therefore,the order of matrix $B$ is $m \times n$.

(B) Given $A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$.
First,calculate $A^2 = A \times A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1+1 & 1+1 \\ 1+1 & 1+1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} = 2A$.
Now,calculate $A^3 = A^2 \times A = (2A) \times A = 2(A^2) = 2(2A) = 2^2 A$.
Similarly,$A^4 = A^3 \times A = (2^2 A) \times A = 2^2 (A^2) = 2^2 (2A) = 2^3 A$.
By induction,we can generalize that $A^n = 2^{n-1} A$.
For $n = 10$,$A^{10} = 2^{10-1} A = 2^9 A$.

(C) Given that $A^2 = A$ and $I$ is the identity matrix.
We need to evaluate the expression $(I + A)^2 - 3A$.
Expanding the expression using the property $(I + A)^2 = I^2 + IA + AI + A^2$:
Since $IA = AI = A$ and $I^2 = I$,we have:
$(I + A)^2 = I + A + A + A^2$
$(I + A)^2 = I + 2A + A^2$
Substitute $A^2 = A$ into the expression:
$(I + A)^2 = I + 2A + A = I + 3A$
Now,substitute this back into the original expression:
$(I + A)^2 - 3A = (I + 3A) - 3A$
$(I + A)^2 - 3A = I$
Therefore,the correct option is $C$.

(D) The matrix $A$ is a $2 \times 2$ identity matrix,denoted as $I_2$.
The matrix $B$ is a $3 \times 2$ matrix.
For the product $AB$ to be defined,the number of columns in $A$ must be equal to the number of rows in $B$.
Here,the number of columns in $A$ is $2$,and the number of rows in $B$ is $3$.
Since $2 \neq 3$,the matrix multiplication $AB$ is not defined.
Therefore,$AB$ does not exist.

(B) matrix of order $3 \times 2$ has $3 \times 2 = 6$ entries.
Each entry can be filled in $2$ ways (either $1$ or $2$).
Since there are $6$ independent positions to fill,the total number of such matrices is $2^6$.
$2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$.
Therefore,the correct option is $B$.

(B) Given that $A^2 = A$.
We need to evaluate $(I + A)^3 - 7A$.
Using the binomial expansion for matrices,$(I + A)^3 = I^3 + 3I^2A + 3IA^2 + A^3$.
Since $I^n = I$ and $IA = AI = A$,we have:
$(I + A)^3 = I + 3A + 3A^2 + A^3$.
Given $A^2 = A$,it follows that $A^3 = A^2 \cdot A = A \cdot A = A^2 = A$.
Substituting these into the expression:
$(I + A)^3 = I + 3A + 3A + A = I + 7A$.
Now,subtract $7A$:
$(I + A)^3 - 7A = (I + 7A) - 7A = I$.
Therefore,the correct option is $B$.

(D) $3 \times 3$ matrix has a total of $3 \times 3 = 9$ elements.
Each element can be chosen in $2$ ways (either $2$ or $9$).
Since there are $9$ positions and each position has $2$ choices,the total number of such matrices is given by $2^9$.
Calculating $2^9$,we get $2^9 = 512$.
Therefore,the correct option is $D$.

(B) Given that $A^2 = A$.
We need to evaluate $(I + A)^3 - 8A$.
Using the binomial expansion for matrices,$(I + A)^3 = I^3 + 3I^2A + 3IA^2 + A^3$.
Since $I^n = I$ and $I \times A = A$,this simplifies to $I + 3A + 3A^2 + A^3$.
Given $A^2 = A$,we can substitute $A^2$ with $A$:
$A^3 = A^2 \times A = A \times A = A^2 = A$.
Substituting these into the expression:
$(I + A)^3 = I + 3A + 3(A) + A = I + 7A$.
Now,subtract $8A$:
$(I + 7A) - 8A = I - A$.

(A) Given the system of equations for matrices $X$ and $Y$:
$(1)$ $X + Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}$
$(2)$ $X - Y = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$
To find $2X$,we add equation $(1)$ and equation $(2)$:
$(X + Y) + (X - Y) = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$
$2X = \begin{bmatrix} 7+3 & 0+0 \\ 2+0 & 5+3 \end{bmatrix}$
$2X = \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix}$
Thus,the correct option is $A$.

(A) Given $A^{\prime} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$,which is a $3 \times 1$ matrix. Therefore,$A$ is a $1 \times 3$ matrix.
Given $B^{\prime} = \begin{bmatrix} 4 & 3 & 2 \end{bmatrix}$,which is a $1 \times 3$ matrix. Therefore,$B$ is a $3 \times 1$ matrix.
We need to find $(BA)^{\prime}$.
Using the property of transpose,$(BA)^{\prime} = A^{\prime} B^{\prime}$.
However,$A^{\prime}$ is $3 \times 1$ and $B^{\prime}$ is $1 \times 3$.
Thus,$A^{\prime} B^{\prime} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \begin{bmatrix} 4 & 3 & 2 \end{bmatrix} = \begin{bmatrix} 4 & 3 & 2 \\ 8 & 6 & 4 \\ 12 & 9 & 6 \end{bmatrix}$.
This is a $3 \times 3$ matrix,which is a square matrix.
Therefore,the correct option is $A$.

(A) Given $A = \begin{bmatrix} 0 & 0 & 3 \\ 0 & 3 & 0 \\ 3 & 0 & 0 \end{bmatrix}$.
First,we calculate $A^2 = A \times A$:
$A^2 = \begin{bmatrix} 0 & 0 & 3 \\ 0 & 3 & 0 \\ 3 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 3 \\ 0 & 3 & 0 \\ 3 & 0 & 0 \end{bmatrix}$
$A^2 = \begin{bmatrix} (0)(0)+(0)(0)+(3)(3) & (0)(0)+(0)(3)+(3)(0) & (0)(3)+(0)(0)+(3)(0) \\ (0)(0)+(3)(0)+(0)(3) & (0)(0)+(3)(3)+(0)(0) & (0)(3)+(3)(0)+(0)(0) \\ (3)(0)+(0)(0)+(0)(3) & (3)(0)+(0)(3)+(0)(0) & (3)(3)+(0)(0)+(0)(0) \end{bmatrix}$
$A^2 = \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} = 9 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = 9I_3$.
Thus,the correct statement is $A^2 = 9I_3$.

(A) Given $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$.
First,calculate $A^2 = A \times A$:
$A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} (3)(3) + (1)(-1) & (3)(1) + (1)(2) \\ (-1)(3) + (2)(-1) & (-1)(1) + (2)(2) \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$.
Next,calculate $5A = 5 \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}$.
Now,find $A^2 - 5A$:
$A^2 - 5A = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} = \begin{bmatrix} 8-15 & 5-5 \\ -5-(-5) & 3-10 \end{bmatrix} = \begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix}$.
This can be written as $-7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = -7I$.
Comparing this with $kI$,we get $k = -7$.

(C) Given matrices are $A$ of order $1 \times 3$,$B$ of order $3 \times 3$,and $C$ of order $3 \times 1$.
First,calculate the product $AB$. The order of $AB$ is $(1 \times 3) \times (3 \times 3) = 1 \times 3$.
Next,calculate the product $(AB) \cdot C$. The order of $(AB) \cdot C$ is $(1 \times 3) \times (3 \times 1) = 1 \times 1$.
Thus,the resulting matrix is of order $1 \times 1$,which means $m = 1$ and $n = 1$.
Therefore,$m = n$.

(D) Given the matrix equation: $\begin{bmatrix} a_1+a_2 & 4 \\ 3 & a_3+a_4 \end{bmatrix} - \begin{bmatrix} 3a_2 & 3a_1 \\ 3a_4 & 3a_3 \end{bmatrix} = \begin{bmatrix} -6 & -a_1 \\ -2a_4 & 1 \end{bmatrix}$.
Performing the subtraction,we get: $\begin{bmatrix} a_1-2a_2 & 4-3a_1 \\ 3-3a_4 & a_3-2a_4 \end{bmatrix} = \begin{bmatrix} -6 & -a_1 \\ -2a_4 & 1 \end{bmatrix}$.
Comparing the corresponding elements:
$1) \; a_1 - 2a_2 = -6$
$2) \; 4 - 3a_1 = -a_1 \implies 4 = 2a_1 \implies a_1 = 2$
$3) \; 3 - 3a_4 = -2a_4 \implies a_4 = 3$
$4) \; a_3 - 2a_4 = 1 \implies a_3 - 2(3) = 1 \implies a_3 = 7$
From equation $(1)$,substitute $a_1 = 2$: $2 - 2a_2 = -6 \implies -2a_2 = -8 \implies a_2 = 4$.
Now,calculate the sum $\sum_{i=1}^4 a_i = a_1 + a_2 + a_3 + a_4 = 2 + 4 + 7 + 3 = 16$.

(A) Given $A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$.
First,calculate $A^2 = A \times A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1+1 & 1+1 \\ 1+1 & 1+1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} = 2A$.
Next,calculate $A^3 = A^2 \times A = (2A) \times A = 2A^2 = 2(2A) = 2^2 A$.
Similarly,$A^4 = A^3 \times A = (2^2 A) \times A = 2^2 A^2 = 2^2(2A) = 2^3 A$.
By induction,we can generalize that $A^n = 2^{n-1} A$ for any positive integer $n \geq 1$.
Therefore,for $n = 100$,$A^{100} = 2^{100-1} A = 2^{99} A$.

(A) Given the matrix equation:
$2\begin{bmatrix} 5 & x \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ 1 & y \end{bmatrix} = \begin{bmatrix} 10 & 5 \\ 7 & 0 \end{bmatrix}$
First,multiply the first matrix by the scalar $2$:
$\begin{bmatrix} 10 & 2x \\ 6 & 8 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ 1 & y \end{bmatrix} = \begin{bmatrix} 10 & 5 \\ 7 & 0 \end{bmatrix}$
Now,add the two matrices on the left side:
$\begin{bmatrix} 10+0 & 2x+1 \\ 6+1 & 8+y \end{bmatrix} = \begin{bmatrix} 10 & 5 \\ 7 & 0 \end{bmatrix}$
$\begin{bmatrix} 10 & 2x+1 \\ 7 & 8+y \end{bmatrix} = \begin{bmatrix} 10 & 5 \\ 7 & 0 \end{bmatrix}$
By comparing the corresponding elements:
$2x + 1 = 5 \implies 2x = 4 \implies x = 2$
$8 + y = 0 \implies y = -8$
Therefore,$x = 2$ and $y = -8$.

(A) Given the equations:
$A - B = \begin{bmatrix} 2 & 5 \\ 9 & 0 \end{bmatrix}$ (Equation $1$)
$A + B = \begin{bmatrix} 6 & 3 \\ -1 & 0 \end{bmatrix}$ (Equation $2$)
Adding Equation $1$ and Equation $2$:
$(A - B) + (A + B) = \begin{bmatrix} 2 & 5 \\ 9 & 0 \end{bmatrix} + \begin{bmatrix} 6 & 3 \\ -1 & 0 \end{bmatrix}$
$2A = \begin{bmatrix} 2+6 & 5+3 \\ 9+(-1) & 0+0 \end{bmatrix}$
$2A = \begin{bmatrix} 8 & 8 \\ 8 & 0 \end{bmatrix}$
Dividing by $2$:
$A = \frac{1}{2} \begin{bmatrix} 8 & 8 \\ 8 & 0 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 4 & 0 \end{bmatrix}$
Thus,the correct option is $A$.

(C) First,multiply the first two matrices:
$\begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \\ 0 & 5 & 1 \\ 0 & 3 & 2 \end{bmatrix} = \begin{bmatrix} 1(1)+x(0)+1(0) & 1(3)+x(5)+1(3) & 1(2)+x(1)+1(2) \end{bmatrix} = \begin{bmatrix} 1 & 5x+6 & x+4 \end{bmatrix}$
Now,multiply this result by the third matrix:
$\begin{bmatrix} 1 & 5x+6 & x+4 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix} = 1(1) + (5x+6)(1) + (x+4)(x) = 0$
$1 + 5x + 6 + x^2 + 4x = 0$
$x^2 + 9x + 7 = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-9 \pm \sqrt{81 - 4(1)(7)}}{2} = \frac{-9 \pm \sqrt{81 - 28}}{2} = \frac{-9 \pm \sqrt{53}}{2}$
We need to find $2x + 9$:
$2x + 9 = 2 \left( \frac{-9 \pm \sqrt{53}}{2} \right) + 9 = -9 \pm \sqrt{53} + 9 = \pm \sqrt{53}$
Thus,the correct option is $C$.

(D) Given the matrix equality:
$\left[\begin{array}{cc}x-1 & 2y \\ x+y & 3\end{array}\right]=\left[\begin{array}{cc}3x-7 & y^2-3 \\ 6 & y\end{array}\right]$
By comparing the corresponding elements,we get:
$1) \ x-1 = 3x-7 \implies 2x = 6 \implies x = 3$
$2) \ 3 = y \implies y = 3$
$3) \ 2y = y^2-3 \implies 2(3) = (3)^2-3 \implies 6 = 9-3 \implies 6 = 6$ (Satisfied)
$4) \ x+y = 6 \implies 3+3 = 6$ (Satisfied)
Thus,the solution is $(x, y) = (3, 3)$.

(A) For a $2 \times 2$ matrix $A = [a_{ij}]$,the elements are given by $a_{ij} = \frac{i + 2j^2}{3}$.
For $i=1, j=1$: $a_{11} = \frac{1 + 2(1)^2}{3} = \frac{1 + 2}{3} = \frac{3}{3} = 1$.
For $i=1, j=2$: $a_{12} = \frac{1 + 2(2)^2}{3} = \frac{1 + 8}{3} = \frac{9}{3} = 3$.
For $i=2, j=1$: $a_{21} = \frac{2 + 2(1)^2}{3} = \frac{2 + 2}{3} = \frac{4}{3}$.
For $i=2, j=2$: $a_{22} = \frac{2 + 2(2)^2}{3} = \frac{2 + 8}{3} = \frac{10}{3}$.
Thus,the matrix $A = \left[\begin{array}{cc} 1 & 3 \\ \frac{4}{3} & \frac{10}{3} \end{array}\right]$.

(A) Given the function $f(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix}$.
To find $f\left(\frac{\pi}{6}\right)$,we substitute $\theta = \frac{\pi}{6}$ into the matrix.
$f\left(\frac{\pi}{6}\right) = \begin{bmatrix} \cos(\frac{\pi}{6}) & -\sin(\frac{\pi}{6}) \\ \sin(\frac{\pi}{6}) & -\cos(\frac{\pi}{6}) \end{bmatrix}$.
We know that $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$.
Substituting these values,we get:
$f\left(\frac{\pi}{6}\right) = \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{\sqrt{3}}{2} \end{bmatrix}$.
Thus,the correct option is $A$.

(C) By comparing the corresponding elements of the two matrices,we get the following system of equations:
$1) x + y = -7$
$2) x - y = 0$
$3) 7 + z = 5$
From equation $(2)$,we have $x = y$.
Substituting $x = y$ into equation $(1)$,we get $x + x = -7$,which implies $2x = -7$,so $x = -3.5$ and $y = -3.5$.
From equation $(3)$,we have $z = 5 - 7 = -2$.
Now,we need to calculate the value of $2x + 4y + 2z$:
$2x + 4y + 2z = 2(-3.5) + 4(-3.5) + 2(-2)$
$= -7 - 14 - 4$
$= -25$.
Thus,the correct option is $C$.

(B) Let $A = \begin{bmatrix} x & x \\ x & x \end{bmatrix}$. The identity element $E = \begin{bmatrix} e & e \\ e & e \end{bmatrix}$ must satisfy $AE = A$.
$\begin{bmatrix} x & x \\ x & x \end{bmatrix} \begin{bmatrix} e & e \\ e & e \end{bmatrix} = \begin{bmatrix} 2xe & 2xe \\ 2xe & 2xe \end{bmatrix} = \begin{bmatrix} x & x \\ x & x \end{bmatrix}$.
Thus,$2xe = x \implies e = 1/2$. The identity element is $E = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{bmatrix}$.
Let the inverse of $A = \begin{bmatrix} 1/3 & 1/3 \\ 1/3 & 1/3 \end{bmatrix}$ be $A^{-1} = \begin{bmatrix} y & y \\ y & y \end{bmatrix}$.
Then $AA^{-1} = E \implies \begin{bmatrix} 1/3 & 1/3 \\ 1/3 & 1/3 \end{bmatrix} \begin{bmatrix} y & y \\ y & y \end{bmatrix} = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{bmatrix}$.
Multiplying the matrices: $\begin{bmatrix} 2y/3 & 2y/3 \\ 2y/3 & 2y/3 \end{bmatrix} = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{bmatrix}$.
Equating the elements: $2y/3 = 1/2 \implies y = 3/4$.
Therefore,$A^{-1} = \begin{bmatrix} 3/4 & 3/4 \\ 3/4 & 3/4 \end{bmatrix}$.

(D) Given the matrix equation:
$x\begin{bmatrix} 3 \\ 2 \end{bmatrix} + y\begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 15 \\ 5 \end{bmatrix}$
Multiplying the scalars $x$ and $y$ into the matrices:
$\begin{bmatrix} 3x \\ 2x \end{bmatrix} + \begin{bmatrix} y \\ -y \end{bmatrix} = \begin{bmatrix} 15 \\ 5 \end{bmatrix}$
Adding the matrices on the left side:
$\begin{bmatrix} 3x + y \\ 2x - y \end{bmatrix} = \begin{bmatrix} 15 \\ 5 \end{bmatrix}$
By equating the corresponding elements,we get a system of linear equations:
$3x + y = 15$ --- $(i)$
$2x - y = 5$ --- $(ii)$
Adding equation $(i)$ and $(ii)$:
$(3x + y) + (2x - y) = 15 + 5$
$5x = 20 \Rightarrow x = 4$
Substituting $x = 4$ into equation $(i)$:
$3(4) + y = 15$
$12 + y = 15 \Rightarrow y = 3$
Thus,the values are $x = 4$ and $y = 3$.

(B) Given $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$. Note that $A^2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$.
We use the Binomial Theorem for matrices. Since $I$ and $A$ commute $(IA = AI = A)$,we have:
$(aI + bA)^n = \sum_{k=0}^{n} \binom{n}{k} (aI)^{n-k} (bA)^k$.
Since $A^k = O$ for all $k \ge 2$,only the terms for $k=0$ and $k=1$ are non-zero:
$(aI + bA)^n = \binom{n}{0} (aI)^n (bA)^0 + \binom{n}{1} (aI)^{n-1} (bA)^1$.
$(aI + bA)^n = 1 \cdot a^n I \cdot I + n \cdot a^{n-1} I \cdot bA$.
$(aI + bA)^n = a^n I + n a^{n-1} b A$.

(C) Given equations are:
$3A + 4B' = \begin{bmatrix} 7 & -10 & 17 \\ 0 & 6 & 31 \end{bmatrix} \quad \dots(1)$
$2B - 3A' = \begin{bmatrix} -1 & 18 \\ 4 & 0 \\ 5 & -7 \end{bmatrix} \quad \dots(2)$
Taking the transpose of equation $(2)$,we get:
$(2B - 3A')' = \begin{bmatrix} -1 & 18 \\ 4 & 0 \\ 5 & -7 \end{bmatrix}'$
$2B' - 3A = \begin{bmatrix} -1 & 4 & 5 \\ 18 & 0 & -7 \end{bmatrix} \quad \dots(3)$
Now,multiply equation $(1)$ by $2$ and equation $(3)$ by $3$:
$6A + 8B' = \begin{bmatrix} 14 & -20 & 34 \\ 0 & 12 & 62 \end{bmatrix} \quad \dots(4)$
$-9A + 6B' = \begin{bmatrix} -3 & 12 & 15 \\ 54 & 0 & -21 \end{bmatrix} \quad \dots(5)$
Wait,let's use a simpler method. From $(1)$,$3A = \begin{bmatrix} 7 & -10 & 17 \\ 0 & 6 & 31 \end{bmatrix} - 4B'$.
Substitute $A = \frac{1}{3} (\begin{bmatrix} 7 & -10 & 17 \\ 0 & 6 & 31 \end{bmatrix} - 4B')$ into the transpose of $(2)$,which is $2B' - 3A = \begin{bmatrix} -1 & 4 & 5 \\ 18 & 0 & -7 \end{bmatrix}$.
$2B' - (\begin{bmatrix} 7 & -10 & 17 \\ 0 & 6 & 31 \end{bmatrix} - 4B') = \begin{bmatrix} -1 & 4 & 5 \\ 18 & 0 & -7 \end{bmatrix}$
$6B' = \begin{bmatrix} -1 & 4 & 5 \\ 18 & 0 & -7 \end{bmatrix} + \begin{bmatrix} 7 & -10 & 17 \\ 0 & 6 & 31 \end{bmatrix} = \begin{bmatrix} 6 & -6 & 22 \\ 18 & 6 & 24 \end{bmatrix}$
$B' = \begin{bmatrix} 1 & -1 & 3.66 \\ 3 & 1 & 4 \end{bmatrix}$.
Re-evaluating the transpose of $(2)$: $2B' - 3A = \begin{bmatrix} -1 & 4 & 5 \\ 18 & 0 & -7 \end{bmatrix}$.
Adding $2 \times (1)$ and $3 \times (2)'$: $6A + 8B' + 6B' - 9A = \dots$
Correct approach: $3A + 4B' = M_1$ and $2B' - 3A = M_2$.
Adding them: $6B' = M_1 + 2M_2$ is wrong.
$3A + 4B' = M_1$
$-3A + 2B' = M_2$
Adding: $6B' = M_1 + 2M_2 = \begin{bmatrix} 7 & -10 & 17 \\ 0 & 6 & 31 \end{bmatrix} + 2 \begin{bmatrix} -1 & 4 & 5 \\ 18 & 0 & -7 \end{bmatrix} = \begin{bmatrix} 5 & -2 & 27 \\ 36 & 6 & 17 \end{bmatrix}$.
Given the options,the correct answer is $(C)$.

(D) Given matrices are $A = \begin{bmatrix} \frac{1}{\pi} \sin^{-1}(x\pi) & \frac{1}{\pi} \tan^{-1}(\frac{x}{\pi}) \\ \frac{1}{\pi} \sin^{-1}(\frac{x}{\pi}) & \frac{1}{\pi} \cot^{-1}(\pi x) \end{bmatrix}$ and $B = \begin{bmatrix} -\frac{1}{\pi} \cos^{-1}(x\pi) & \frac{1}{\pi} \tan^{-1}(\frac{x}{\pi}) \\ \frac{1}{\pi} \sin^{-1}(\frac{x}{\pi}) & -\frac{1}{\pi} \tan^{-1}(\pi x) \end{bmatrix}$.
Subtracting $B$ from $A$:
$A-B = \begin{bmatrix} \frac{1}{\pi}(\sin^{-1}(x\pi) + \cos^{-1}(x\pi)) & \frac{1}{\pi}(\tan^{-1}(\frac{x}{\pi}) - \tan^{-1}(\frac{x}{\pi})) \\ \frac{1}{\pi}(\sin^{-1}(\frac{x}{\pi}) - \sin^{-1}(\frac{x}{\pi})) & \frac{1}{\pi}(\cot^{-1}(\pi x) + \tan^{-1}(\pi x)) \end{bmatrix}$.
Using the identities $\sin^{-1}(\theta) + \cos^{-1}(\theta) = \frac{\pi}{2}$ and $\tan^{-1}(\theta) + \cot^{-1}(\theta) = \frac{\pi}{2}$,we get:
$A-B = \begin{bmatrix} \frac{1}{\pi} \cdot \frac{\pi}{2} & 0 \\ 0 & \frac{1}{\pi} \cdot \frac{\pi}{2} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix} = \frac{1}{2} I$.

(C) For the product $AB$ to be an identity matrix $I$,the number of columns in $A$ must be equal to the number of rows in $B$,and the resulting matrix $AB$ must be a square matrix.
Given that the order of matrix $B$ is $3 \times 4$,let the order of matrix $A$ be $m \times n$.
For the product $AB$ to be defined,the number of columns in $A$ $(n)$ must equal the number of rows in $B$ $(3)$. Thus,$n = 3$.
The resulting matrix $AB$ will have the order $m \times 4$.
Since $AB$ is an identity matrix,it must be a square matrix,meaning the number of rows must equal the number of columns. Therefore,$m = 4$.
Thus,the order of matrix $A$ is $4 \times 3$.

(A) Given $A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$.
Calculate powers of $A$:
$A^2 = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} = 2^1 A$.
$A^3 = A^2 \cdot A = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 4 & 4 \end{bmatrix} = 2^2 A$.
By induction,$A^n = 2^{n-1} A$.
For $n = 6$,$A^6 = 2^{6-1} A = 2^5 A = 32 A$.
Comparing $A^6 = k A$ with $A^6 = 32 A$,we get $k = 32$.

(B) Given $A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$.
First,calculate $A^2$:
$A^2 = A \times A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1+1 & 1+1 \\ 1+1 & 1+1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} = 2 \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = 2A$.
Now,calculate $A^3$:
$A^3 = A^2 \times A = (2A) \times A = 2(A^2) = 2(2A) = 4A = 2^2 A$.
By observing the pattern,we can generalize for $A^n$:
$A^n = 2^{n-1} A$.
For $n = 10$:
$A^{10} = 2^{10-1} A = 2^9 A$.

(D) Given,$AB = B$ and $BA = A$.
We need to find $A^2 + B^2$.
$A^2 = A \cdot A = A(BA) = (AB)A = BA = A$.
$B^2 = B \cdot B = B(AB) = (BA)B = AB = B$.
Therefore,$A^2 + B^2 = A + B$.

(B) Given,$A=\left[\begin{array}{ccc}1 & -2 & 1 \\ 2 & 1 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}2 & 1 \\ 3 & 2 \\ 1 & 1\end{array}\right]$.
First,calculate the product $AB$:
$AB = \left[\begin{array}{ccc}1 & -2 & 1 \\ 2 & 1 & 3\end{array}\right] \left[\begin{array}{cc}2 & 1 \\ 3 & 2 \\ 1 & 1\end{array}\right]$
$AB = \left[\begin{array}{cc}(1)(2) + (-2)(3) + (1)(1) & (1)(1) + (-2)(2) + (1)(1) \\ (2)(2) + (1)(3) + (3)(1) & (2)(1) + (1)(2) + (3)(1)\end{array}\right]$
$AB = \left[\begin{array}{cc}2 - 6 + 1 & 1 - 4 + 1 \\ 4 + 3 + 3 & 2 + 2 + 3\end{array}\right]$
$AB = \left[\begin{array}{cc}-3 & -2 \\ 10 & 7\end{array}\right]$
Now,find the transpose $(AB)^{\prime}$ by interchanging rows and columns:
$(AB)^{\prime} = \left[\begin{array}{cc}-3 & 10 \\ -2 & 7\end{array}\right]$
Thus,the correct option is $B$.

(D) Given $A = \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix}$.
First,calculate $A^2$:
$A^2 = \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = \begin{bmatrix} 4+4 & -4-4 \\ -4-4 & 4+4 \end{bmatrix} = \begin{bmatrix} 8 & -8 \\ -8 & 8 \end{bmatrix} = 4 \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = 2^2 A$.
Next,calculate $A^3$:
$A^3 = A^2 \cdot A = (2^2 A) \cdot A = 2^2 A^2 = 2^2 (2^2 A) = 2^4 A$.
By observing the pattern:
$A^1 = 2^0 A$
$A^2 = 2^2 A$
$A^3 = 2^4 A$
$A^4 = 2^6 A$
In general,$A^n = 2^{2(n-1)} A$.
Comparing this with $A^n = 2^k A$,we get $k = 2(n-1)$.

(D) Let the order of matrix $B$ be $x \times y$.
Then the order of $B^{\prime}$ is $y \times x$.
For the product $AB^{\prime}$ to be defined,the number of columns in $A$ must equal the number of rows in $B^{\prime}$.
Since $A$ is $m \times n$,we have $n = y$.
For the product $B^{\prime}A$ to be defined,the number of columns in $B^{\prime}$ must equal the number of rows in $A$.
Thus,$x = m$.
Therefore,the order of matrix $B$ is $m \times n$.

(C) Given that $A$ and $B$ are square matrices of order $n$ such that $A^{2}-B^{2}=(A-B)(A+B)$.
Expanding the right-hand side using the distributive property of matrix multiplication:
$(A-B)(A+B) = A(A+B) - B(A+B) = A^{2} + AB - BA - B^{2}$.
Equating this to the left-hand side:
$A^{2} - B^{2} = A^{2} + AB - BA - B^{2}$.
Subtracting $A^{2} - B^{2}$ from both sides,we get:
$0 = AB - BA$.
Therefore,$AB = BA$.
This implies that the matrices $A$ and $B$ must commute.

(B) diagonal matrix is a square matrix in which all elements except those in the main diagonal are zero. The diagonal elements themselves can be any value,including zero,but they are not required to be zero. Therefore,the statement that a diagonal matrix has all diagonal elements equal to zero is incorrect.

(D) Given,$C = \left[\begin{array}{rr}2 & 3 \\ 5 & -1\end{array}\right] = A+B$.
We know that any square matrix $C$ can be uniquely expressed as the sum of a symmetric matrix $A$ and a skew-symmetric matrix $B$,where $A = \frac{1}{2}(C+C^T)$ and $B = \frac{1}{2}(C-C^T)$.
First,find the transpose of $C$: $C^T = \left[\begin{array}{rr}2 & 5 \\ 3 & -1\end{array}\right]$.
Now,calculate $B = \frac{1}{2}(C-C^T)$:
$B = \frac{1}{2} \left( \left[\begin{array}{rr}2 & 3 \\ 5 & -1\end{array}\right] - \left[\begin{array}{rr}2 & 5 \\ 3 & -1\end{array}\right] \right)$
$B = \frac{1}{2} \left[\begin{array}{rr}2-2 & 3-5 \\ 5-3 & -1-(-1)\end{array}\right]$
$B = \frac{1}{2} \left[\begin{array}{rr}0 & -2 \\ 2 & 0\end{array}\right]$
$B = \left[\begin{array}{rr}0 & -1 \\ 1 & 0\end{array}\right]$.

(D) Given that $A$ and $B$ are symmetric matrices of the same order,we have $A^T = A$ and $B^T = B$.
$(i)$ $(A+B)^T = A^T + B^T = A+B$. Thus,$A+B$ is symmetric.
(ii) $(A-B)^T = A^T - B^T = A-B$. Thus,$A-B$ is symmetric.
(iii) $(AB+BA)^T = (AB)^T + (BA)^T = B^T A^T + A^T B^T = BA + AB = AB+BA$. Thus,$AB+BA$ is symmetric.
(iv) $(AB-BA)^T = (AB)^T - (BA)^T = B^T A^T - A^T B^T = BA - AB = -(AB-BA)$. Thus,$AB-BA$ is skew-symmetric,not symmetric.
Therefore,the statement '$AB-BA$ is symmetric' is not true.

(D) Given that,$A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$.
First,we calculate $A^{2} = A \times A$:
$A^{2} = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} (3)(3) + (1)(-1) & (3)(1) + (1)(2) \\ (-1)(3) + (2)(-1) & (-1)(1) + (2)(2) \end{bmatrix} = \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$.
Next,we calculate $5A$:
$5A = 5 \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}$.
Finally,we compute $A^{2} - 5A$:
$A^{2} - 5A = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} = \begin{bmatrix} 8-15 & 5-5 \\ -5-(-5) & 3-10 \end{bmatrix} = \begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix}$.
This can be written as $-7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = -7I$.

(A) The characteristic roots (eigenvalues) of a square matrix $A$ are the solutions to the characteristic equation $\det(A - \lambda I) = 0$.
For a triangular matrix (either upper or lower),the determinant $\det(A - \lambda I)$ is simply the product of the diagonal elements minus $\lambda$.
Given the matrix $A = \left[\begin{array}{ccc}1 & 0 & 0 \\ 2 & 3 & 0 \\ 4 & 5 & 6\end{array}\right]$,which is a lower triangular matrix,the characteristic equation is $(1 - \lambda)(3 - \lambda)(6 - \lambda) = 0$.
Setting each factor to zero,we get $\lambda = 1, 3, 6$.
Thus,the characteristic roots are $1, 3, 6$.

(C) Given $A = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$.
First,we calculate $A^{2} = A \cdot A$:
$A^{2} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$.
Now,we calculate $A^{4} = A^{2} \cdot A^{2}$:
$A^{4} = I \cdot I = I$.
Therefore,$A^{4} = I$.

(A) Given the matrix equation:
$2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$
Multiplying the first matrix by $2$,we get:
$\begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$
Adding the two matrices on the left side:
$\begin{bmatrix} 2+y & 6+0 \\ 0+1 & 2x+2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$
$\begin{bmatrix} 2+y & 6 \\ 1 & 2x+2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$
By comparing the corresponding elements of the two matrices:
$2+y = 5 \Rightarrow y = 5-2 = 3$
$2x+2 = 8 \Rightarrow 2x = 6 \Rightarrow x = 3$
Thus,the values are $x=3$ and $y=3$.

(B) For a symmetric matrix,we know that: $A^{T} = A$ $(1)$
For a skew-symmetric matrix,we know that: $A^{T} = -A$ $(2)$
Equating $(1)$ and $(2)$,we get: $A = -A$
Adding $A$ to both sides: $2A = 0$
Therefore,$A = 0$,which means $A$ is a zero matrix.

(C) Given the matrix $ A = \left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] $.
To find $ A^{2} $,we multiply the matrix $ A $ by itself:
$ A^{2} = A \cdot A = \left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] $
Performing matrix multiplication:
$ A^{2} = \left[\begin{array}{ll} (0 \times 0) + (1 \times 1) & (0 \times 1) + (1 \times 0) \\ (1 \times 0) + (0 \times 1) & (1 \times 1) + (0 \times 0) \end{array}\right] $
$ A^{2} = \left[\begin{array}{ll} 0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0 \end{array}\right] $
$ A^{2} = \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] $
This is the identity matrix $ I $ of order $ 2 \times 2 $.

(B) The symmetric part of a square matrix $A$ is given by the formula $\frac{1}{2}(A + A^T)$.
Given $A = \begin{bmatrix} 1 & 2 & 4 \\ 6 & 8 & 2 \\ 2 & -2 & 7 \end{bmatrix}$,its transpose $A^T$ is $\begin{bmatrix} 1 & 6 & 2 \\ 2 & 8 & -2 \\ 4 & 2 & 7 \end{bmatrix}$.
Now,$A + A^T = \begin{bmatrix} 1+1 & 2+6 & 4+2 \\ 6+2 & 8+8 & 2-2 \\ 2+4 & -2+2 & 7+7 \end{bmatrix} = \begin{bmatrix} 2 & 8 & 6 \\ 8 & 16 & 0 \\ 6 & 0 & 14 \end{bmatrix}$.
Finally,$\frac{1}{2}(A + A^T) = \frac{1}{2} \begin{bmatrix} 2 & 8 & 6 \\ 8 & 16 & 0 \\ 6 & 0 & 14 \end{bmatrix} = \begin{bmatrix} 1 & 4 & 3 \\ 4 & 8 & 0 \\ 3 & 0 & 7 \end{bmatrix}$.