Let A be the required column matrix
Then A =\(\begin{matrix} Peons \\ \begin{matrix} Clerks \\ \begin{matrix} Typist \\ Section \ officer \end{matrix} \end{matrix} \end{matrix}\left[ \begin{matrix} 15 \\ \begin{matrix} 6 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix} \end{matrix} \right] \)
Since there are 30 collegs,
30 A = 30\(\left[ \begin{matrix} 15 \\ \begin{matrix} 6 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix} \end{matrix} \right] =\left[ \begin{matrix} 450 \\ \begin{matrix} 180 \\ \begin{matrix} 30 \\ 30 \end{matrix} \end{matrix} \end{matrix} \right] \)

Given vertices are (k. 2). (2, 4) and (3, 2)
And also Given Area of a triangle = 4 sq,units
We know that, area of \(\triangle\)ABC =absolute value of \(\frac { 1 }{ 2 } \left| \begin{matrix} { x }_{ 1 } & { y }_{ 1 } & 1 \\ { x }_{ 2 } & { y }_{ 2 } & 1 \\ { x }_{ 3 } & { y }_{ 3 } & 1 \end{matrix} \right| \)
\(\Rightarrow\) 4 = absolute of \(\frac { 1 }{ 2 } \left| \begin{matrix} k & 2 & 1 \\ 2 & 4 & 1 \\ 3 & 2 & 1 \end{matrix} \right| \)
\(\Rightarrow\) 4 = absolute value of \(\frac{1}{2}\)[k(4 - 2)-2(2 - 3) + 1(4 - 12)]
[Expanded along R₁]
\(\Rightarrow\) 4 = absolute value of \(\frac{1}{2}\)[2k + 2 - 8]
\(\Rightarrow\) 4 =absolute value of \(\frac{1}{2}\)[2k - 6]
\(\Rightarrow\) 4 = \(\pm \frac { 1 }{ 2 } \)(2k - 6)
Case (i) when 4 =\(\frac{1}{2}\)(2k - 6)
\(\Rightarrow\) 8 = 2k - 6
\(\Rightarrow\)14 = 2k
\(\Rightarrow\) k =7
Case(ii)  When 4 = -\(\frac{1}{2}\)(2k - 6)
\(\Rightarrow\) 8 = -2k + 6
\(\Rightarrow\) 8 - 6 = -2k
\(\Rightarrow\) 2 = -2k
\(\Rightarrow\) k = -1
\(\therefore\) The values of k are -1 or 7.

To prove the given points are collinear, it suffices to prove |A| = \(\begin{vmatrix} a&b+c &1 \\ b &c+a &1 \\ c & a+b &1 \end{vmatrix}=0\)
Applying C₁\(\rightarrow\) C₁ + C₂, we deduce that 
|A| = \(\begin{vmatrix} a+b+c&b+c &1 \\ a+b+c &c+a &1 \\ a+b+c & a+b &1 \end{vmatrix}=(a+b+c)\begin{vmatrix} 1&b+c &1 \\ 1 &c+a &1 \\ 1 & a+b &1 \end{vmatrix}=(a+b+c)\times0=0\)
which shows that the given points are collinear.

Given A^T = \(\left[ \begin{matrix} 4 & 5 \\ -1 & 0 \\ 2 & 3 \end{matrix} \right] \) and B = \(\left[ \begin{matrix} 2 & -1 & 1 \\ 7 & 5 & -2 \end{matrix} \right] \)
(ii) Verify (A + B)^T= A^T + B^T = B^T = B^T + A^T
(A)^T =\({ \left[ \begin{matrix} 4 & 5 \\ -1 & 0 \\ 2 & 3 \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 4 & -1 & 2 \\ 5 & 0 & 3 \end{matrix} \right] \)
A = \(\left[ \begin{matrix} 4 & -2 & 2 \\ 5 & 0 & 3 \end{matrix} \right] \)
Now, A + B = \(\left[ \begin{matrix} 4 & -1 & 2 \\ 5 & 0 & 3 \end{matrix} \right] +\left[ \begin{matrix} 2 & -1 & 1 \\ 7 & 5 & -2 \end{matrix} \right] =\left[ \begin{matrix} 6 & -2 & 3 \\ 12 & 5 & 1 \end{matrix} \right] \)
\(\therefore\) (A + B)^T= \(\left[ \begin{matrix} 6 & 12 \\ -2 & 5 \\ 3 & 1 \end{matrix} \right] \) ---(1)
B^T = \(\left[ \begin{matrix} 6 & 12 \\ -2 & 5 \\ 3 & 1 \end{matrix} \right] \)
\(\therefore\) A^T + B^T = \(\left[ \begin{matrix} 4 & 5 \\ -1 & 0 \\ 2 & 3 \end{matrix} \right] +\left[ \begin{matrix} 2 & 7 \\ -1 & 5 \\ 1 & -2 \end{matrix} \right] =\left[ \begin{matrix} 6 & 12 \\ -2 & 5 \\ 3 & 1 \end{matrix} \right] \)----(2)
B^T + A^T = \(\left[ \begin{matrix} 2 & 7 \\ -1 & 5 \\ 1 & -2 \end{matrix} \right] +\left[ \begin{matrix} 4 & 5 \\ -1 & 0 \\ 2 & 3 \end{matrix} \right] =\left[ \begin{matrix} 6 & 12 \\ -2 & 5 \\ 3 & 1 \end{matrix} \right] \) ----------------------(3)
From (1), (2) and (3), (A + B)^T = A^T + B^T = B^T+ A^T

Let B be a 3 \(\times\) 4 matrix with entries as
B = \(\left( \begin{matrix} { a }_{ 11 } & a_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{matrix}\begin{matrix} { a }_{ 14 } \\ { a }_{ 24 } \\ { a }_{ 34 } \end{matrix} \right) \)
\({ a }_{ ij }=\frac { \left| 3i-4j \right| }{ 4 } \)
\({ a }_{ 11 }=\frac { \left| 3-4 \right| }{ 4 } =\frac { \left| -1 \right| }{ 4 } =\frac { 1 }{ 4 } \)
\({ a }_{ 12 }=\frac { \left| 3-8 \right| }{ 4 } =\frac { \left| -5 \right| }{ 4 } =\frac { 5 }{ 4 } \)
\({ a }_{ 13 }=\frac { \left| 3-12 \right| }{ 4 } =\frac { \left| -9 \right| }{ 4 } =\frac { 9 }{ 4 } \)
\({ a }_{ 14 }=\frac { \left| 3-16 \right| }{ 4 } =\frac { \left| -13 \right| }{ 4 } =\frac { 13 }{ 4 } \)
\({ a }_{ 21 }=\frac { \left| 3(2)-(4)1 \right| }{ 4 } =\frac { \left| 6-4 \right| }{ 4 } =\frac { 2 }{ 4 } \)
\({ a }_{ 22 }=\frac { \left| 3(2)-4(2) \right| }{ 4 } =\frac { \left| 6-8 \right| }{ 4 } =\frac { 2 }{ 4 } \)
\({ a }_{ 23 }=\frac { \left| 3(2)-4(3) \right| }{ 4 } =\frac { \left| 6-12 \right| }{ 4 } =\frac { 6 }{ 4 } \)
\({ a }_{ 24 }=\frac { \left| 3(2)-4(4) \right| }{ 4 } =\frac { \left| 6-16 \right| }{ 4 } =\frac { 10 }{ 4 } \)
\({ a }_{ 31 }=\frac { \left| 3(3)-4(1) \right| }{ 4 } =\frac { \left| 9-4 \right| }{ 4 } =\frac { 5 }{ 4 } \)
\({ a }_{ 32 }=\frac { \left| 3(3)-4(2) \right| }{ 4 } =\frac { \left| 9-8 \right| }{ 4 } =\frac { 1 }{ 4 } \)
\({ a }_{ 33 }=\frac { \left| 3(3)-4(3) \right| }{ 4 } =\frac { \left| 9-12 \right| }{ 4 } =\frac { 3 }{ 4 } \)
\({ a }_{ 34 }=\frac { \left| 3(3)-4(4) \right| }{ 4 } =\frac { \left| 9-16 \right| }{ 4 } =\frac { 7 }{ 4 } \)
\(\therefore B=\left[ \begin{matrix} 1/4 & 5/4 & 9/4 \\ 2/4 & 2/4 & 6/4 \\ 5/4 & 1/4 & 3/4 \end{matrix}\begin{matrix} 13/4 \\ 10/4 \\ 7/4 \end{matrix} \right] =\frac { 1 }{ 4 } \left[ \begin{matrix} 1 & 5 & 9 \\ 2 & 2 & 6 \\ 5 & 1&3 & \end{matrix}\begin{matrix} 13 \\ 10 \\ 7 \end{matrix} \right] \)

The number of elements is the product of number of rows and number of columns.
Therefore, we will find all ordered pairs of natural numbers whose product is 12.
Thus, all the possible orders of the matrix are 1 \(\times\) 12, 12 \(\times\) 1, 2 \(\times\)6, 6 \(\times\)2, 3 \(\times\) 4 and 4 \(\times\) 3.
Since 7 is prime, the only possible orders of the matrix are 1 \(\times\) 7 and 7 \(\times\) 1.

Given A² = B
\(\Rightarrow \left[ \begin{matrix} \alpha & 0 \\ 1 & 1 \end{matrix} \right] \left[ \begin{matrix} \alpha & 0 \\ 1 & 1 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 5 & 1 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} { \alpha }^{ 2 }+0 & 0+0 \\ \alpha +1 & 0+1 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 5 & 1 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} { \alpha }^{ 2 } & 0 \\ \alpha +1 & 1 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 5 & 1 \end{matrix} \right] \)
\(\Rightarrow\) \({ \alpha }^{ 2 }\) = 1 or \(\alpha +1\) = 5
\(\Rightarrow\) = \(\pm \) 1 or \(\alpha\) = 4 which is not possible.
Hence, there is no value of  for which A² = B is true.

\(A=\left[\begin{array}{cc} 4 & 2 \\ -1 & x \end{array}\right] ; I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\)
\(A-2 I=\left[\begin{array}{cc} 4 & 2 \\ -1 & x \end{array}\right]-2\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 4-2 & 2+0 \\ -1+0 & x-2 \end{array}\right]\)
\(=\left[\begin{array}{cc} 2 & 2 \\ -1 & x-2 \end{array}\right]\)
\(A-3 I=\left[\begin{array}{rr} 4 & 2 \\ -1 & x \end{array}\right]-3\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 4-3 & 2+0 \\ -1+0 & x-3 \end{array}\right]\)
\(=\left[\begin{array}{cc} 1 & 2 \\ -1 & x-3 \end{array}\right]\)
Given, \((A-2 I)(A-3 I)=\left[\begin{array}{cc} 2 & 2 \\ -1 & x-2 \end{array}\right]\left[\begin{array}{cc} 1 & 2 \\ -1 & x-3 \end{array}\right]\)
\(=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]\)
\(\left[\begin{array}{cc} 2-2 & 4+2 x-6 \\ -1-x+2 & -2+(x-2)(x-3) \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]\)
\(-1-x+2=0\)
\(-x+1=0\)
\(x=1\)

A -B =\(\begin{bmatrix} 4 & 6 & 2 \\ 0 & 1 & 5 \\ 0 & 3 & 2 \end{bmatrix}\)-\(\begin{bmatrix} 0 & 1 & -1 \\ 3 & -1 & 4 \\ -1 & 2 & 1 \end{bmatrix}\)=\(\begin{bmatrix} 4 & 5 & 3 \\ -3 & 2 & 1 \\1 & 1 & 1 \end{bmatrix}\)
(A - B)^T =\(\begin{bmatrix} 4 & -3 & 1 \\ 5 & 2 & 1 \\3 & 1 & 1 \end{bmatrix}\) ..(1)
A^T - B^T =\(\begin{bmatrix} 4 & 0 & 0 \\ 6 & 1 & 3 \\ 2 & 5 & 2 \end{bmatrix}\)-\(\begin{bmatrix} 0 & 3 & -1 \\ 1 & -1 & 2 \\ -1 & 4 & 1 \end{bmatrix}\)=\(\begin{bmatrix} 4 & -3 & 1 \\ 5 & 2 & 1 \\3 & 1 & 1 \end{bmatrix}\).....(2)
From (1) and (2), (A - B)^T= A^T - B^T.

AB =\(\begin{bmatrix} 4 & 6 & 2 \\ 0 & 1 & 5 \\ 0 & 3 & 2 \end{bmatrix}\)\(\begin{bmatrix} 0 & 1 & -1 \\ 3 & -1 & 4 \\ -1 & 2 & 1 \end{bmatrix}\)=\(\begin{bmatrix} 16 & 2 & -1 \\ -2 & 9 & 9 \\ 7 & 1 & 14 \end{bmatrix}\)
(AB)^T =\(\begin{bmatrix} 16 & -2 & 7 \\ 2 & 9 & 1 \\ 22 & 9 & 14 \end{bmatrix}\)...(1)
B^T = \(\begin{bmatrix} 0 & 3 & -1 \\ 1 & -1 & 2 \\ -1 & 4 & 1 \end{bmatrix}\), A^T=\(\begin{bmatrix} 4 & 0 & 0 \\ 6 & 1 & 3 \\ 2 & 5 & 2 \end{bmatrix}\)
B^TA^T=\(\begin{bmatrix} 0 & 3 & -1 \\ 1 & -1 & 2 \\ -1 & 4 & 1 \end{bmatrix}\)\(\begin{bmatrix} 4 & 0 & 0 \\ 6 & 1 & 3 \\ 2 & 5 & 2 \end{bmatrix}\)=\(\begin{bmatrix} 16 & -2 & 7 \\ 2 & 9 & 1 \\ 22 & 9 & 14 \end{bmatrix}\)...(2)
From (1) and (2), (AB)^T = B^TA^T.

A² = AA =\(\begin{bmatrix} 0 &c &b \\ c & 0 &a \\ b & a & 0 \end{bmatrix}\)\(\begin{bmatrix} 0 &c &b \\ c & 0 &a \\ b & a & 0 \end{bmatrix}\)=\(\begin{bmatrix} C_{11} &C_{12} &C_{13} \\ C_{21} & C_{22} &C_{23} \\ C_{31} & C_{32} & C_{33} \end{bmatrix}\)

Given \(\lambda \)=-2
Let A= \(\begin{vmatrix} 0& 2\lambda &1 \\ \lambda^2 &0 &3\lambda^3+1 \\ -1 &6\lambda-1 &0 \end{vmatrix}\)=\(\left| \begin{matrix} 0 & -4 & 1 \\ 4 & 0 & 13 \\ -1 & -13 & 0 \end{matrix} \right| \) [put \(\lambda \)=-2]
\(=0+4(0+13)+1(-52-0)=4(13)-52\)
\(=52-52=0\)

\(Given A=\left[\begin{array}{ccc}1 & 2 & 2 \\ 2 & 1 & -2 \\ x & 2 & y\end{array}\right]\)
\(A^T=\left[\begin{array}{ccc} 1 & 2 & x \\ 2 & 1 & 2 \\ 2 & -2 & y \end{array}\right]\)
\(A A^T=9 I\)
\(\left[\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & -2 \\ x & 2 & y \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & x \\ 2 & 1 & 2 \\ 2 & -2 & y \end{array}\right]=9\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\)
\(\left[\begin{array}{ccc} 1+4+4 & 2+2-4 & x+4+2 y \\ 2+2-4 & 4+1+4 & 2 x+2-2 y \\ x+4+2 y & 2 x+2-2 y & x^2+4+y^2 \end{array}\right]=\left[\begin{array}{lll} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{array}\right]\)
\(\left[\begin{array}{ccc} 9 & 0 & x+2 y+4 \\ 0 & 9 & 2 x-2 y+2 \\ x+4+2 y & 2 x+2-2 y & x^2+4+y^2 \end{array}\right]=\left[\begin{array}{lll} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{array}\right]\)
\(x+2 y+4=0 \Rightarrow x+2 y=-4 ....(1)\)
\(2 x-2 y+2=0 \Rightarrow 2 x-2 y=-2 ....(2)\)
\(\begin{aligned} &\text { Add (1) & (2) } \Rightarrow 3 x=-6 \\ &x=-2 \end{aligned}\)
\(\text { Put } x=-2 \text { in }(1) \Rightarrow-2+2 y=-4\)
\(-2+4=-2 y\)
\(2=-2 y\)
\(y=-1\)
\(\therefore x=-2\)

Cost matrix A = [30 15 45], Fruit matrix

Cost of packages are obtained by computing AB. That is, by multiplying cost of each item in A (cost matrix A) with number of items in B (Fruit matrix B).
\(A B=\left[\begin{array}{lll} 30 & 15 & 45 \end{array}\right]\left[\begin{array}{lll} 6 & 5 & 6 \\ 3 & 4 & 6 \\ 3 & 4 & 6 \end{array}\right]=\left[\begin{array}{lll} 360 & 390 & 540 \end{array}\right]\)
Pack-I cost Rs. 360, Pack-II cost Rs. 390, Pack-III costs Rs. 540.

\(\left[\begin{array}{lll} x & 2 & -1 \end{array}\right]\) \(\begin{bmatrix} 1&1 &2 \\ -1 & -4 &1 \\ -1 &-1 &-2 \end{bmatrix}\)\(\begin{bmatrix} x \\ 2 \\ 1 \end{bmatrix}\)= 0
That is, [x - 2 +1  x - 8 + 1 2x + 2 + 2] \(\begin{bmatrix} x \\ 2 \\ 1 \end{bmatrix}\)= 0
 [x - 1  x - 7 2x + 4] \(\begin{bmatrix} x \\ 2 \\ 1 \end{bmatrix}\)= 0
x (x - 1) + 2(x - 7) + 1(2 x +4) = 0
x² + 3x − 10 = 0 \(\Rightarrow\) x = -5, 2.

\(|A|=\left|\begin{array}{ccc} 0 & a-b & k \\ b-a & 0 & 5 \\ -k & -5 & 0 \end{array}\right|\)
\(=0-(a-b)[0+5 k]+k(-5(b-a)-0)\)
\(=(-a+b)(5 k)+k(-5 b+5 a)\)
\(=-5 k a+5 k b-5 k b+5 a k\)
= 0
\(|A|=0\)
\(\therefore\) A is singular

\(A \& B \text { are symmetric } \Rightarrow A+B \text { is symmetric }\)

\(A^{2}=A \times A=\left[\begin{array}{cc} \lambda & 1 \\ -1 & -\lambda \end{array}\right]\left[\begin{array}{cc} \lambda & 1 \\ -1 & -\lambda \end{array}\right]\)
\(=\left[\begin{array}{cc} \lambda^{2}-1 & \lambda-\lambda \\ -\lambda+\lambda & -1+\lambda^{2} \end{array}\right]=0\)
\(\therefore \lambda^{2} =1 \)
\(\lambda =\pm 1\)