Math worksheet with matrix problems involving construction, element calculation, and decomposition.

Here are the step-by-step solutions for the problems in your homework.

Q1: Matrix Orders

The "order" of a matrix is written as $m \times n$ (rows $\times$ columns). The total number of elements is $m \times n$. We need to find pairs of whole numbers that multiply to give the total number of elements.

(i) 12 Elements:
Factors of 12 are: $1, 2, 3, 4, 6, 12$.
Possible orders ($m \times n$):
* $1 \times 12$
* $2 \times 6$
* $3 \times 4$
* $4 \times 3$
* $6 \times 2$
* $12 \times 1$

7 Elements:
7 is a prime number (only divisible by 1 and itself).
Possible orders:
* $1 \times 7$
* $7 \times 1$

(ii) 8 Elements:
Factors of 8 are: $1, 2, 4, 8$.
Possible orders:
* $1 \times 8$
* $2 \times 4$
* $4 \times 2$
* $8 \times 1$

5 Elements:
5 is a prime number.
Possible orders:
* $1 \times 5$
* $5 \times 1$

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Q2: Construct a $2 \times 3$ Matrix

We need to fill a matrix with 2 rows and 3 columns using the given formulas for element $a_{ij}$ (where $i$ is the row number and $j$ is the column number).

(i) $a_{ij} = \frac{i + 3j}{3}$
* $a_{11} (i=1, j=1) = \frac{1+3}{3} = \frac{4}{3}$
* $a_{12} (i=1, j=2) = \frac{1+6}{3} = \frac{7}{3}$
* $a_{13} (i=1, j=3) = \frac{1+9}{3} = \frac{10}{3}$
* $a_{21} (i=2, j=1) = \frac{2+3}{3} = \frac{5}{3}$
* $a_{22} (i=2, j=2) = \frac{2+6}{3} = \frac{8}{3}$
* $a_{23} (i=2, j=3) = \frac{2+9}{3} = \frac{11}{3}$

Matrix: $\begin{bmatrix} 4/3 & 7/3 & 10/3 \\ 5/3 & 8/3 & 11/3 \end{bmatrix}$

(ii) $a_{ij} = \frac{2i + 3j}{2}$
* $a_{11} = \frac{2(1)+3(1)}{2} = \frac{5}{2}$
* $a_{12} = \frac{2(1)+3(2)}{2} = \frac{8}{2} = 4$
* $a_{13} = \frac{2(1)+3(3)}{2} = \frac{11}{2}$
* $a_{21} = \frac{2(2)+3(1)}{2} = \frac{7}{2}$
* $a_{22} = \frac{2(2)+3(2)}{2} = \frac{10}{2} = 5$
* $a_{23} = \frac{2(2)+3(3)}{2} = \frac{13}{2}$

Matrix: $\begin{bmatrix} 5/2 & 4 & 11/2 \\ 7/2 & 5 & 13/2 \end{bmatrix}$

(iii) $a_{ij} = \frac{3i + j}{2}$
* $a_{11} = \frac{3+1}{2} = 2$
* $a_{12} = \frac{3+2}{2} = \frac{5}{2}$
* $a_{13} = \frac{3+3}{2} = 3$
* $a_{21} = \frac{6+1}{2} = \frac{7}{2}$
* $a_{22} = \frac{6+2}{2} = 4$
* $a_{23} = \frac{6+3}{2} = \frac{9}{2}$

Matrix: $\begin{bmatrix} 2 & 5/2 & 3 \\ 7/2 & 4 & 9/2 \end{bmatrix}$

(iv) $a_{ij} = \frac{3i - j}{2}$
* $a_{11} = \frac{3-1}{2} = 1$
* $a_{12} = \frac{3-2}{2} = \frac{1}{2}$
* $a_{13} = \frac{3-3}{2} = 0$
* $a_{21} = \frac{6-1}{2} = \frac{5}{2}$
* $a_{22} = \frac{6-2}{2} = 2$
* $a_{23} = \frac{6-3}{2} = \frac{3}{2}$

Matrix: $\begin{bmatrix} 1 & 1/2 & 0 \\ 5/2 & 2 & 3/2 \end{bmatrix}$

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Q3: Construct a $4 \times 3$ Matrix

Rows $i = 1, 2, 3, 4$. Columns $j = 1, 2, 3$.

(i) $a_{ij} = 2i + \frac{e^j}{j}$
* Row 1 ($i=1$): $2 + e/1$, $2 + e^2/2$, $2 + e^3/3$
* Row 2 ($i=2$): $4 + e/1$, $4 + e^2/2$, $4 + e^3/3$
* Row 3 ($i=3$): $6 + e/1$, $6 + e^2/2$, $6 + e^3/3$
* Row 4 ($i=4$): $8 + e/1$, $8 + e^2/2$, $8 + e^3/3$

Matrix:
$\begin{bmatrix} 2+e & 2+\frac{e^2}{2} & 2+\frac{e^3}{3} \\ 4+e & 4+\frac{e^2}{2} & 4+\frac{e^3}{3} \\ 6+e & 6+\frac{e^2}{2} & 6+\frac{e^3}{3} \\ 8+e & 8+\frac{e^2}{2} & 8+\frac{e^3}{3} \end{bmatrix}$

(ii) $a_{ij} = \frac{i-j}{i+j}$
* $a_{11}=0/2=0$, $a_{12}=-1/3$, $a_{13}=-2/4=-1/2$
* $a_{21}=1/3$, $a_{22}=0/4=0$, $a_{23}=-1/5$
* $a_{31}=2/4=1/2$, $a_{32}=1/5$, $a_{33}=0/6=0$
* $a_{41}=3/5$, $a_{42}=2/6=1/3$, $a_{43}=1/7$

Matrix:
$\begin{bmatrix} 0 & -1/3 & -1/2 \\ 1/3 & 0 & -1/5 \\ 1/2 & 1/5 & 0 \\ 3/5 & 1/3 & 1/7 \end{bmatrix}$

(iii) $a_{ij} = i$
The value depends only on the row number.
Row 1 is all 1s, Row 2 is all 2s, etc.
Matrix:
$\begin{bmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 3 & 3 & 3 \\ 4 & 4 & 4 \end{bmatrix}$

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Q4: Find values of a, b, c, x, y, z

Two matrices are equal if their corresponding elements are equal.

Left Matrix: $\begin{pmatrix} 2+3 & 2+4 & 2y-7 \\ 4x+6 & a-1 & 0 \\ b-3 & 3b & z-2c \end{pmatrix} = \begin{pmatrix} 5 & 6 & 2y-7 \\ 4x+6 & a-1 & 0 \\ b-3 & 3b & z-2c \end{pmatrix}$

Right Matrix: $\begin{pmatrix} 0 & 6 & 3y-2 \\ 2x & -3 & 2c-2 \\ 2b+4 & -21 & 0 \end{pmatrix}$

Equating elements:
1. Row 1, Col 1: $5 = 0$ ... Wait, looking at the image, the first element on the left is `2+3` which is 5. The first element on the right is `0`. This implies $5=0$, which is impossible. Let me re-read the image carefully.
* Ah, the top left element on the left side is just `2+3`? No, usually these are variables. Let's look closer. It says `2+3`. That equals 5. The right side is `0`.
* Let's check other positions to see if I misread a variable.
* Row 1, Col 2: Left `2+4` = 6. Right `6`. Matches.
* Row 1, Col 3: Left `2y-7`. Right `3y-2`.
$2y - 7 = 3y - 2 \Rightarrow -7 + 2 = 3y - 2y \Rightarrow y = -5$.
* Row 2, Col 1: Left `4x+6`. Right `2x`.
$4x + 6 = 2x \Rightarrow 2x = -6 \Rightarrow x = -3$.
* Row 2, Col 2: Left `a-1`. Right `-3`.
$a - 1 = -3 \Rightarrow a = -2$.
* Row 2, Col 3: Left `0`. Right `2c-2`.
$0 = 2c - 2 \Rightarrow 2c = 2 \Rightarrow c = 1$.
* Row 3, Col 1: Left `b-3`. Right `2b+4`.
$b - 3 = 2b + 4 \Rightarrow -3 - 4 = 2b - b \Rightarrow b = -7$.
* Row 3, Col 2: Left `3b`. Right `-21`.
Check with $b=-7$: $3(-7) = -21$. Matches.
* Row 3, Col 3: Left `z-2c`. Right `0`.
$z - 2c = 0$. Since $c=1$, $z - 2(1) = 0 \Rightarrow z = 2$.

*Note on Top-Left Element:* The image shows `2+3` on the left and `0` on the right. This is likely a typo in the question paper (perhaps it was meant to be a variable like $x+3$ or similar), but based on the other solvable equations, we can find all variables except checking that specific cell. Assuming the rest of the system holds:

Values:
$a = -2$
$b = -7$
$c = 1$
$x = -3$
$y = -5$
$z = 2$

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Q5: Find Matrices x and y

Given:
1) $2x - y = \begin{pmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{pmatrix}$
2) $x + 2y = \begin{pmatrix} 3 & 2 & 5 \\ -2 & 1 & -7 \end{pmatrix}$

Let $A = \begin{pmatrix} 6 & -6 & 0 \\ -4 & 2 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & 2 & 5 \\ -2 & 1 & -7 \end{pmatrix}$.

Multiply equation (2) by 2:
$2(x + 2y) = 2B \Rightarrow 2x + 4y = 2B$

Subtract equation (1) from this new equation:
$(2x + 4y) - (2x - y) = 2B - A$
$5y = 2B - A$

Calculate $2B - A$:
$2B = \begin{pmatrix} 6 & 4 & 10 \\ -4 & 2 & -14 \end{pmatrix}$
$2B - A = \begin{pmatrix} 6-6 & 4-(-6) & 10-0 \\ -4-(-4) & 2-2 & -14-1 \end{pmatrix} = \begin{pmatrix} 0 & 10 & 10 \\ 0 & 0 & -15 \end{pmatrix}$

So, $5y = \begin{pmatrix} 0 & 10 & 10 \\ 0 & 0 & -15 \end{pmatrix}$
Divide by 5:
$y = \begin{pmatrix} 0 & 2 & 2 \\ 0 & 0 & -3 \end{pmatrix}$

Now find $x$ using $x = B - 2y$:
$2y = \begin{pmatrix} 0 & 4 & 4 \\ 0 & 0 & -6 \end{pmatrix}$
$x = \begin{pmatrix} 3 & 2 & 5 \\ -2 & 1 & -7 \end{pmatrix} - \begin{pmatrix} 0 & 4 & 4 \\ 0 & 0 & -6 \end{pmatrix} = \begin{pmatrix} 3-0 & 2-4 & 5-4 \\ -2-0 & 1-0 & -7-(-6) \end{pmatrix}$
$x = \begin{pmatrix} 3 & -2 & 1 \\ -2 & 1 & -1 \end{pmatrix}$

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Q6: Find the value of x

Equation: $\begin{bmatrix} 1 & 1 & x \end{bmatrix} \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 0$

First, multiply the two matrices on the right:
$\begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1(1)+0(1)+2(1) \\ 0(1)+2(1)+1(1) \\ 2(1)+1(1)+0(1) \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 3 \end{pmatrix}$

Now multiply the row vector by this result:
$\begin{bmatrix} 1 & 1 & x \end{bmatrix} \begin{pmatrix} 3 \\ 3 \\ 3 \end{pmatrix} = 0$
$1(3) + 1(3) + x(3) = 0$
$3 + 3 + 3x = 0$
$6 + 3x = 0$
$3x = -6$
$x = -2$

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Q7: Prove $A^n$

$A = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}$. We want to prove $A^n = \begin{pmatrix} 1+2n & -4n \\ n & 1-2n \end{pmatrix}$.

Base Case ($n=1$):
Formula gives: $\begin{pmatrix} 1+2(1) & -4(1) \\ 1 & 1-2(1) \end{pmatrix} = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}$. This matches $A$.

Inductive Step:
Assume true for $n=k$: $A^k = \begin{pmatrix} 1+2k & -4k \\ k & 1-2k \end{pmatrix}$.
Show true for $n=k+1$: $A^{k+1} = A^k \cdot A$.

$A^{k+1} = \begin{pmatrix} 1+2k & -4k \\ k & 1-2k \end{pmatrix} \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}$

Element (1,1): $(1+2k)(3) + (-4k)(1) = 3 + 6k - 4k = 3 + 2k = 1 + 2(k+1)$. (Matches)
Element (1,2): $(1+2k)(-4) + (-4k)(-1) = -4 - 8k + 4k = -4 - 4k = -4(k+1)$. (Matches)
Element (2,1): $(k)(3) + (1-2k)(1) = 3k + 1 - 2k = 1 + k = (k+1)$. (Matches)
Element (2,2): $(k)(-4) + (1-2k)(-1) = -4k - 1 + 2k = -1 - 2k = 1 - 2(k+1)$. (Matches)

Thus, proven by induction.

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Q8: Find $A^2 - 4A + 3I_3$

$A = \begin{pmatrix} 1 & 2 & 0 \\ 3 & -4 & 5 \\ 0 & -1 & 3 \end{pmatrix}$

First, calculate $A^2 = A \cdot A$:
Row 1:
$1(1)+2(3)+0(0) = 7$
$1(2)+2(-4)+0(-1) = -6$
$1(0)+2(5)+0(3) = 10$
Row 2:
$3(1)-4(3)+5(0) = -9$
$3(2)-4(-4)+5(-1) = 6+16-5 = 17$
$3(0)-4(5)+5(3) = -20+15 = -5$
Row 3:
$0(1)-1(3)+3(0) = -3$
$0(2)-1(-4)+3(-1) = 4-3 = 1$
$0(0)-1(5)+3(3) = -5+9 = 4$

$A^2 = \begin{pmatrix} 7 & -6 & 10 \\ -9 & 17 & -5 \\ -3 & 1 & 4 \end{pmatrix}$

Next, calculate $4A$:
$4A = \begin{pmatrix} 4 & 8 & 0 \\ 12 & -16 & 20 \\ 0 & -4 & 12 \end{pmatrix}$

Next, calculate $3I_3$ (Identity matrix scaled by 3):
$3I_3 = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix}$

Finally, compute $A^2 - 4A + 3I_3$:
Element (1,1): $7 - 4 + 3 = 6$
Element (1,2): $-6 - 8 + 0 = -14$
Element (1,3): $10 - 0 + 0 = 10$
Element (2,1): $-9 - 12 + 0 = -21$
Element (2,2): $17 - (-16) + 3 = 17 + 16 + 3 = 36$
Element (2,3): $-5 - 20 + 0 = -25$
Element (3,1): $-3 - 0 + 0 = -3$
Element (3,2): $1 - (-4) + 0 = 5$
Element (3,3): $4 - 12 + 3 = -5$

Result: $\begin{pmatrix} 6 & -14 & 10 \\ -21 & 36 & -25 \\ -3 & 5 & -5 \end{pmatrix}$

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Q9: Express A as sum of Symmetric and Skew-Symmetric

Any square matrix $A$ can be written as $A = P + Q$, where $P = \frac{1}{2}(A + A^T)$ is symmetric and $Q = \frac{1}{2}(A - A^T)$ is skew-symmetric.

$A = \begin{pmatrix} 4 & 2 & -1 \\ 3 & 5 & 7 \\ 1 & -2 & 1 \end{pmatrix}$
$A^T = \begin{pmatrix} 4 & 3 & 1 \\ 2 & 5 & -2 \\ -1 & 7 & 1 \end{pmatrix}$

Symmetric Part ($P$):
$A + A^T = \begin{pmatrix} 8 & 5 & 0 \\ 5 & 10 & 5 \\ 0 & 5 & 2 \end{pmatrix}$
$P = \frac{1}{2} \begin{pmatrix} 8 & 5 & 0 \\ 5 & 10 & 5 \\ 0 & 5 & 2 \end{pmatrix} = \begin{pmatrix} 4 & 2.5 & 0 \\ 2.5 & 5 & 2.5 \\ 0 & 2.5 & 1 \end{pmatrix}$

Skew-Symmetric Part ($Q$):
$A - A^T = \begin{pmatrix} 0 & -1 & -2 \\ 1 & 0 & 9 \\ 2 & -9 & 0 \end{pmatrix}$
$Q = \frac{1}{2} \begin{pmatrix} 0 & -1 & -2 \\ 1 & 0 & 9 \\ 2 & -9 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -0.5 & -1 \\ 0.5 & 0 & 4.5 \\ 1 & -4.5 & 0 \end{pmatrix}$

Check: $P+Q = A$. Correct.

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Q10: Express matrices as sum of Symmetric and Skew-Symmetric

(i) $A = \begin{pmatrix} 6 & 1 \\ 3 & 4 \end{pmatrix}$
$A^T = \begin{pmatrix} 6 & 3 \\ 1 & 4 \end{pmatrix}$
Symmetric $P = \frac{1}{2}(A+A^T) = \frac{1}{2}\begin{pmatrix} 12 & 4 \\ 4 & 8 \end{pmatrix} = \begin{pmatrix} 6 & 2 \\ 2 & 4 \end{pmatrix}$
Skew $Q = \frac{1}{2}(A-A^T) = \frac{1}{2}\begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$

(ii) $A = \begin{pmatrix} 3 & 2 & 3 \\ 4 & 5 & 3 \\ 2 & 4 & 5 \end{pmatrix}$
$A^T = \begin{pmatrix} 3 & 4 & 2 \\ 2 & 5 & 4 \\ 3 & 3 & 5 \end{pmatrix}$
$A+A^T = \begin{pmatrix} 6 & 6 & 5 \\ 6 & 10 & 7 \\ 5 & 7 & 10 \end{pmatrix} \Rightarrow P = \begin{pmatrix} 3 & 3 & 2.5 \\ 3 & 5 & 3.5 \\ 2.5 & 3.5 & 5 \end{pmatrix}$
$A-A^T = \begin{pmatrix} 0 & -2 & 1 \\ 2 & 0 & -1 \\ -1 & 1 & 0 \end{pmatrix} \Rightarrow Q = \begin{pmatrix} 0 & -1 & 0.5 \\ 1 & 0 & -0.5 \\ -0.5 & 0.5 & 0 \end{pmatrix}$

(iii) $A = \begin{pmatrix} 2 & -1 & -1 \\ -1 & 4 & 2 \\ -6 & 0 & 8 \end{pmatrix}$
$A^T = \begin{pmatrix} 2 & -1 & -6 \\ -1 & 4 & 0 \\ -1 & 2 & 8 \end{pmatrix}$
$A+A^T = \begin{pmatrix} 4 & -2 & -7 \\ -2 & 8 & 2 \\ -7 & 2 & 16 \end{pmatrix} \Rightarrow P = \begin{pmatrix} 2 & -1 & -3.5 \\ -1 & 4 & 1 \\ -3.5 & 1 & 8 \end{pmatrix}$
$A-A^T = \begin{pmatrix} 0 & 0 & 5 \\ 0 & 0 & 2 \\ -5 & -2 & 0 \end{pmatrix} \Rightarrow Q = \begin{pmatrix} 0 & 0 & 2.5 \\ 0 & 0 & 1 \\ -2.5 & -1 & 0 \end{pmatrix}$

(iv) $A = \begin{pmatrix} 6 & 1 & -5 \\ -2 & -5 & 4 \\ -3 & 3 & -1 \end{pmatrix}$
$A^T = \begin{pmatrix} 6 & -2 & -3 \\ 1 & -5 & 3 \\ -5 & 4 & -1 \end{pmatrix}$
$A+A^T = \begin{pmatrix} 12 & -1 & -8 \\ -1 & -10 & 7 \\ -8 & 7 & -2 \end{pmatrix} \Rightarrow P = \begin{pmatrix} 6 & -0.5 & -4 \\ -0.5 & -5 & 3.5 \\ -4 & 3.5 & -1 \end{pmatrix}$
$A-A^T = \begin{pmatrix} 0 & 3 & -2 \\ -3 & 0 & 1 \\ 2 & -1 & 0 \end{pmatrix} \Rightarrow Q = \begin{pmatrix} 0 & 1.5 & -1 \\ -1.5 & 0 & 0.5 \\ 1 & -0.5 & 0 \end{pmatrix}$

Final Answer:
Q1:
(i) 12 elements: $1\times12, 2\times6, 3\times4, 4\times3, 6\times2, 12\times1$. 7 elements: $1\times7, 7\times1$.
(ii) 8 elements: $1\times8, 2\times4, 4\times2, 8\times1$. 5 elements: $1\times5, 5\times1$.

Q2:
(i) $\begin{bmatrix} 4/3 & 7/3 & 10/3 \\ 5/3 & 8/3 & 11/3 \end{bmatrix}$
(ii) $\begin{bmatrix} 5/2 & 4 & 11/2 \\ 7/2 & 5 & 13/2 \end{bmatrix}$
(iii) $\begin{bmatrix} 2 & 5/2 & 3 \\ 7/2 & 4 & 9/2 \end{bmatrix}$
(iv) $\begin{bmatrix} 1 & 1/2 & 0 \\ 5/2 & 2 & 3/2 \end{bmatrix}$

Q3:
(i) $\begin{bmatrix} 2+e & 2+e^2/2 & 2+e^3/3 \\ 4+e & 4+e^2/2 & 4+e^3/3 \\ 6+e & 6+e^2/2 & 6+e^3/3 \\ 8+e & 8+e^2/2 & 8+e^3/3 \end{bmatrix}$
(ii) $\begin{bmatrix} 0 & -1/3 & -1/2 \\ 1/3 & 0 & -1/5 \\ 1/2 & 1/5 & 0 \\ 3/5 & 1/3 & 1/7 \end{bmatrix}$
(iii) $\begin{bmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 3 & 3 & 3 \\ 4 & 4 & 4 \end{bmatrix}$

Q4: $a=-2, b=-7, c=1, x=-3, y=-5, z=2$.

Q5: $x = \begin{pmatrix} 3 & -2 & 1 \\ -2 & 1 & -1 \end{pmatrix}, y = \begin{pmatrix} 0 & 2 & 2 \\ 0 & 0 & -3 \end{pmatrix}$

Q6: $x = -2$

Q7: Proof provided via mathematical induction.

Q8: $\begin{pmatrix} 6 & -14 & 10 \\ -21 & 36 & -25 \\ -3 & 5 & -5 \end{pmatrix}$

Q9: Symmetric: $\begin{pmatrix} 4 & 2.5 & 0 \\ 2.5 & 5 & 2.5 \\ 0 & 2.5 & 1 \end{pmatrix}$, Skew: $\begin{pmatrix} 0 & -0.5 & -1 \\ 0.5 & 0 & 4.5 \\ 1 & -4.5 & 0 \end{pmatrix}$

Q10:
(i) Sym: $\begin{pmatrix} 6 & 2 \\ 2 & 4 \end{pmatrix}$, Skew: $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$
(ii) Sym: $\begin{pmatrix} 3 & 3 & 2.5 \\ 3 & 5 & 3.5 \\ 2.5 & 3.5 & 5 \end{pmatrix}$, Skew: $\begin{pmatrix} 0 & -1 & 0.5 \\ 1 & 0 & -0.5 \\ -0.5 & 0.5 & 0 \end{pmatrix}$
(iii) Sym: $\begin{pmatrix} 2 & -1 & -3.5 \\ -1 & 4 & 1 \\ -3.5 & 1 & 8 \end{pmatrix}$, Skew: $\begin{pmatrix} 0 & 0 & 2.5 \\ 0 & 0 & 1 \\ -2.5 & -1 & 0 \end{pmatrix}$
(iv) Sym: $\begin{pmatrix} 6 & -0.5 & -4 \\ -0.5 & -5 & 3.5 \\ -4 & 3.5 & -1 \end{pmatrix}$, Skew: $\begin{pmatrix} 0 & 1.5 & -1 \\ -1.5 & 0 & 0.5 \\ 1 & -0.5 & 0 \end{pmatrix}$