(I + A)²-3A = I
Alternative Mwthod:
Given A² = A,
(I+A)²-3A = I² + 2IA + A²-3A
= I + 2A + A - 3A = I

Let A be any square matrix. Then, 
\(A=\frac { 1 }{ 2 } \left( A+{ A }^{ T } \right) +\frac { 1 }{ 2 } \left( A-{ A }^{ T } \right) \)
= P + Q (say),
where, \(P=\frac { 1 }{ 2 } \left( A+{ A }^{ T } \right) \)
and \(Q=\frac { 1 }{ 2 } \left( A-{ A }^{ T } \right) \)
Now, \({ P }^{ T }=\left[ \frac { 1 }{ 2 } \left( A+{ A }^{ T } \right) \right] ^{ T }\)
\(=\frac { 1 }{ 2 } \left( A+{ A }^{ T } \right) ^{ T }\quad \left[ \because \quad \left( KA \right) ^{ T }=K.{ A }^{ T } \right] \)
\(\Rightarrow \ { P }^{ T }=\frac { 1 }{ 2 } \left[ { A }^{ T }+\left( { A }^{ T } \right) ^{ T } \right] \) \(\left[ \because \ \left( A+B \right) ^{ T }={ A }^{ T }+{ B }^{ T } \right] \)
\(\Rightarrow \ { P }^{ T }=\frac { 1 }{ 2 } \left( { A }^{ T }+A \right) \)  \(\left[ \because \ \left( { A }^{ I } \right) ^{ T }=A \right] \)
\(\Rightarrow { P }^{ T }=\frac { 1 }{ 2 } \left( A{ +A }^{ T } \right) =P\)
\(\therefore \)  P is symmetric matrix.
Also, \({ Q }^{ T }=\frac { 1 }{ 2 } \left( A-{ A }^{ T } \right) ^{ T }\)
\(=\frac { 1 }{ 2 } \left( A-{ A }^{ T } \right) ^{ T }\)
\(=\frac { 1 }{ 2 } \left[ { A }^{ T }-\left( { A }^{ T } \right) ^{ T } \right] \)
\(=\frac { 1 }{ 2 } \left[ { A }^{ T }-A \right] \)
\({ \Rightarrow Q }^{ T }=\frac { 1 }{ 2 } \left[ A-{ A }^{ T } \right] =-Q\)
\(\therefore \)  Q is skew symmetric matrix.
Thus, A = P + Q, where P is a symmetric matrix and Q is a skew symmetric matrix.
Hence, A is expressible as the sum of a symmetric and a skew symmetric matrix.
Uniqueness : If possible, let A = R + S, where R is symmetric and S is skew symmetric, then
A^T = (R + S)^T = R^T + S^T
\(\Rightarrow \)  A^T = R - S  (\(\because \)  R^T = R and S^T = - S)
Now, A = R + S and A^T = R - S
\(\Rightarrow R=\frac { 1 }{ 2 } \left[ A+{ A }^{ T } \right] =P\)
\(S=\frac { 1 }{ 2 } \left[ A-{ A }^{ T } \right] =Q\)
Hence, A is uniquely expressible as the sum of a symmetric and a skew symmetric matrix.

We have, \(A=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{matrix} \right] \)
\({ A }^{ \prime }=\left[ \begin{matrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{matrix} \right] \)
\({ A+A }^{ \prime }=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)
\(\Rightarrow \frac { 1 }{ 2 } { A+A }^{ \prime }=\frac { 1 }{ 2 } \left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] =0\)
and \({ A-A }^{ \prime }=\left[ \begin{matrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{matrix} \right] \)
\(\Rightarrow \frac { 1 }{ 2 } { A-A }^{ \prime }=\frac { 1 }{ 2 } \left[ \begin{matrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{matrix} \right] \)
\(\Rightarrow A=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{matrix} \right] \)

For matrix to be symmetric,
\(\Rightarrow \quad { A }^{ \prime }=A\)
\(\therefore \ \left[ \begin{matrix} 0 & 3 & 3a \\ 2b & 1 & 3 \\ -2 & 3 & -1 \end{matrix} \right] =\left[ \begin{matrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{matrix} \right] \)
2b = 3, 3a = -2
\(\therefore\)By equality of matrices, \(a=\frac { -2 }{ 3 } \) and \(b=\frac { 3 }{ 2 }\)

We have, \({ A }_{ 3\times x }\) and \({ B }_{ 4\times y }\) = \({ AB }_{ 3\times 3 }\)
(i) \(\left( { A }_{ 3\times x }, \right) \left( { B }_{ 4\times y } \right) =\left( { AB }_{ 3\times 3 } \right) \)
\(\therefore\)   x = 4 and y = 3
(ii) \(\left( { A }_{ x\times 2 } \right) \left( { B }_{ y\times 4 } \right) =\left( { AB }_{ 3\times 4 } \right) \)
\(\therefore\)  y = 2, x = 3

We have, \(X+Y=\left[ \begin{matrix} 2 & 3 \\ 5 & 1 \end{matrix} \right] ,X-Y=\left[ \begin{matrix} 6 & 5 \\ 7 & 3 \end{matrix} \right] \)
\(\left( X+Y \right) +\left( X-Y \right) =\left[ \begin{matrix} 2 & 3 \\ 5 & 1 \end{matrix} \right] +\left[ \begin{matrix} 6 & 5 \\ 7 & 3 \end{matrix} \right] \)
\(2X=\left[ \begin{matrix} 8 & 8 \\ 12 & 4 \end{matrix} \right] \)
\(\Rightarrow X=\left[ \begin{matrix} 4 & 4 \\ 6 & 2 \end{matrix} \right] \)
\(\therefore \ X=\left[ \begin{matrix} 4 & 4 \\ 6 & 2 \end{matrix} \right] \)
and \(Y=\left[ \begin{matrix} 2 & 3 \\ 5 & 1 \end{matrix} \right] -\left[ \begin{matrix} 4 & 4 \\ 6 & 2 \end{matrix} \right] \)
\(=\left[ \begin{matrix} -2 & -1 \\ -1 & -1 \end{matrix} \right] \)

Let A = [a_ij] be a square matrix of order m. If possible, let B and C be two
inverses of A. We shall show that B = C.
Since B is the inverse of A
AB = BA = I ... (1)
Since C is also the inverse of A
AC = CA = I ... (2)
Thus B = BI = B (AC) = (BA) C = IC = C

Given
\(\left[ \begin{matrix} a+4 & 3b \\ 8 & -6 \end{matrix} \right] =\left[ \begin{matrix} 2a+2 & b+2 \\ 8 & a-8b \end{matrix} \right] \)
On equating the corresponding elements, we get  a + 4 = 2a + 2 ⇒   a = 2
3b = b + 2 ⇒   b = 1
∴  a - 2b = 2 - 2(1)
= 2 - 2
= 0

Let B = A + A′, then
B′ = (A + A′)′
= A′ + (A′)′ (as (A + B)′ = A′ + B′)
= A′ + A (as (A′)′ = A)
= A + A′ (as A + B = B + A)
= B
Therefore B = A + A′ is a symmetric matrix
Now let C = A – A′
C′ = (A – A′)′ = A′ – (A′)′ (Why?)
= A′ – A (Why?)
= – (A – A′) = – C
Therefore C = A – A′ is a skew symmetric matrix.

A square matrix \(A=[{ a }_{ ij }] \)is skew symmetric
if \({ a }_{ ij }=-{ a }_{ ji }\forall i,j \)
Let \(i=j\Rightarrow { a }_{ ii }=-{ a }_{ ii }\Rightarrow { 2a }_{ ii }=0\Rightarrow { a }_{ ii }=0\Rightarrow 0  \)
All the diagonal elements of a skew symmetric matrix are always zero.

\(\begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix}+\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \)
\(\Rightarrow \begin{bmatrix} 2+y & 6 \\ 1 & 2x+2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \)
\(\Rightarrow 2+y=5,2x+2=8\Rightarrow x=3,y=3 \)
\(\therefore \ x+y=3+3=6\)

Given \(A=\left[ \begin{matrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ x & -3 & 0 \end{matrix} \right]  \)
For a skew symmetric matrix, a_ij = -a_ji ∀∀ i, j
\(\therefore \ x=-(-2)=2\)

\(y+2x=7,-x=-2\Rightarrow x=2,y=3\)

We know that\(A={ I }_{ 2 }A\)
i.e., \(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}A\)
\(\Rightarrow \begin{bmatrix} 1 & -1 \\ 0 & 5 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}A\)  [Applying \({ R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\)]
\(\Rightarrow \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ \frac { -2 }{ 5 } & \frac { 1 }{ 5 } \end{bmatrix}A\)   [Applying \({ R }_{ 2 }\rightarrow { 1 }/{ 5 }{ R }_{ 2 }\)]
\(\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} \frac { 3 }{ 5 } & \frac { 1 }{ 5 } \\ \frac { -2 }{ 5 } & \frac { 1 }{ 5 } \end{bmatrix}A.\)  [Applying \({ R }_{ 1 }\rightarrow { R }_{ 1 }+{ R }_{ 2 }\)]
Hence, \({ A }^{ -1 }=\frac { 1 }{ 5 } \begin{bmatrix} 3 & 1 \\ -2 & 1 \end{bmatrix}.\)

We have: \(A=\left[ \begin{matrix} 3 & 1 & -1 \\ 0 & 1 & 2 \end{matrix} \right] .\)
\(\therefore \ A\prime =\left[ \begin{matrix} 3 & 0 \\ 1 & 1 \\ -1 & 2 \end{matrix} \right]\)
\(\therefore \ AA\prime =\left[ \begin{matrix} 3 & 1 & -1 \\ 0 & 1 & 2 \end{matrix} \right] \left[ \begin{matrix} 3 & 0 \\ 1 & 1 \\ -1 & 2 \end{matrix} \right] \)
\(=\begin{bmatrix} 9+1+1 & 0+1-2 \\ 0+1-2 & 0+1+4 \end{bmatrix}=\begin{bmatrix} 11 & -1 \\ -1 & 5 \end{bmatrix}.\)
Which is a symmetric matrix.

(A) is the correct answer.
Reason: Since A and B are symmetric matrices,
\(\therefore \ A\prime =A\quad and\quad B\prime =B ...(1)\)
\(Now\ (AB-BA)\prime =(AB)\prime -(BA)\prime \)
\(=B\prime A\prime -A\prime B\prime\)
\(=BA-AB\)
\(=-(AB-BA)\)
\(\Rightarrow (AB-BA)\) is skew-symmetric.

We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 24 elements, we have to find all the ordered pairs of natural numbers whose product is 24.
The ordered pairs are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), and  (6, 4)
Hence, the possible orders of a matrix having 24 elements are:
1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, and 6 × 4
(1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13.
Hence, the possible orders of a matrix having 13 elements are 1 × 13 and 13 × 1.

\(cos\theta \begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}+sin\theta \begin{bmatrix} sin\theta & -cos\theta \\ cos\theta & sin\theta \end{bmatrix}\)
\(=\begin{bmatrix} { cos }^{ 2 }\theta & cos\theta sin\theta \\ -cos\theta sin\theta & { cos }^{ 2 }\theta \end{bmatrix}+\begin{bmatrix} { sin }^{ 2 }\theta & -sin\theta cos\theta \\ sin\theta cos\theta & { sin }^{ 2 }\theta \end{bmatrix}\)
\(=\begin{bmatrix} { cos }^{ 2 }\theta +{ sin }^{ 2 }\theta & cos\theta sin\theta -sin\theta cos\theta \\ -cos\theta sin\theta +sin\theta cos\theta & { cos }^{ 2 }\theta +{ sin }^{ 2 }\theta \end{bmatrix}\)
\(=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\).

We shall prove the result by using principle of mathematical induction.
Let \(P\left( n \right) { :A }^{ n }=\left[ \begin{matrix} { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \\ { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \\ { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \end{matrix} \right] \)
Now, \(P\left( 1 \right) { :A }^{ 1 }=\left[ \begin{matrix} { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \\ { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \\ { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \end{matrix} \right] =\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right] \)
The result is true for n = 1.
Let the result be true for n = k.
So, \({ A }^{ k }=\left[ \begin{matrix} { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \end{matrix} \right] \)
Now, we prove that P(k + 1) is true.
Now, A^k+1 = A. A^k
\(=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right] \left[ \begin{matrix} { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \\ { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \\ { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \\ { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \\ { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \end{matrix} \right] \)
= A^k+1
Hence, it is true n = k + 1.
Hence, by principle of mathematical induction P(n) is true for all \(n\epsilon N\)

We know that A = I₃A
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] A\)
Applying \({ R }_{ 3 }\rightarrow 2{ R }_{ 3 }\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 2 & 4 & 4 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{matrix} \right] A\)
Applying \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 0 & 3 & 3 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 2 \end{matrix} \right] A\)
Applying \({ R }_{ 2 }\rightarrow 4{ R }_{ 2 }\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 8 & 4 & 4 \\ 0 & 3 & 3 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & -1 & 2 \end{matrix} \right] A\)
Applying \({ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 0 & 0 & 1 \\ 0 & 3 & 3 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ -1 & 4 & 0 \\ 0 & -1 & 2 \end{matrix} \right] A\)
Applying \({ R }_{ 2 }\rightarrow { R }_{ 2 }+{ R }_{ 3 }\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 0 & 3 & 4 \\ 0 & 3 & 3 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ -1 & 3 & 2 \\ 0 & -1 & 2 \end{matrix} \right] A\)
Applying \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2}\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 0 & 3 & 4 \\ 0 & 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ -1 & 3 & 2 \\ 1 & -4 & 0 \end{matrix} \right] A\)
Applying \({ R }_{ 1 }\rightarrow { R }_{ 1 }+{ 3R }_{ 3 }\)
\(\left[ \begin{matrix} 8 & 4 & 0 \\ 0 & 3 & 4 \\ 0 & 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 4 & -12 & 0 \\ -1 & 3 & 2 \\ 1 & -4 & 0 \end{matrix} \right] A\)
Applying \({ R }_{ 2 }\rightarrow { R }_{ 2 }+{ 4R }_{ 3 }\)
\(\left[ \begin{matrix} 8 & 4 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 4 & -12 & 0 \\ 3 & -13 & 2 \\ 1 & -4 & 0 \end{matrix} \right] A\)
Applying \({ R }_{ 1 }\rightarrow { R }_{ 1 }-\frac { 4 }{ 3 } { R }_{ 2 }\)
\(\left[ \begin{matrix} 8 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 0 & \frac { 16 }{ 3 } & -\frac { 8 }{ 3 } \\ 3 & -13 & 2 \\ 1 & -4 & 0 \end{matrix} \right] \)
Applying \({ R }_{ 1 }\rightarrow \frac { 1 }{ 8 } { R }_{ 1 },{ R }_{ 2 }\rightarrow \frac { 1 }{ 3 } { R }_{ 2 }\) and R₃ = - R₃
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} 0 & \frac { 2 }{ 3 } & -\frac { 1 }{ 3 } \\ 3 & -\frac { 13 }{ 3 } & \frac { 2 }{ 3 } \\ -1 & 4 & 0 \end{matrix} \right] A\)
\({ A }^{ -1 }=\left[ \begin{matrix} 0 & \frac { 2 }{ 3 } & -\frac { 1 }{ 3 } \\ 3 & -\frac { 13 }{ 3 } & \frac { 2 }{ 3 } \\ -1 & 4 & 0 \end{matrix} \right] \)
Matrix representation of given linear equation is
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 19 \\ 5 \\ 7 \end{matrix} \right] \)
\(\Rightarrow \)  AX = B
\(\Rightarrow \)  X = A^-1B
\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0 & \frac { 2 }{ 3 } & -\frac { 1 }{ 3 } \\ 1 & -\frac { 13 }{ 3 } & \frac { 2 }{ 3 } \\ -1 & 4 & 0 \end{matrix} \right] \left[ \begin{matrix} 19 \\ 5 \\ 7 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0+\frac { 10 }{ 3 } -\frac { 7 }{ 3 } \\ 19-\frac { 65 }{ 3 } +\frac { 14 }{ 3 } \\ -19+20+0 \end{matrix} \right] \)
\(\therefore \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 2 \\ 1 \end{matrix} \right] \)
\(\therefore \)  x = 1, y = 2, z = 1

\(\begin{matrix} A \\ B \\ C \end{matrix}\left[ \begin{matrix} \overset { HF }{ 40 } & \overset { M }{ 50 } & \overset { P }{ 20 } \\ 25 & 40 & 30 \\ 35 & 50 & 40 \end{matrix} \right] \left[ \begin{matrix} 25 \\ 100 \\ 50 \end{matrix} \right] =\left[ \begin{matrix} 7,000 \\ 6,125 \\ 7,875 \end{matrix} \right] =\left[ \begin{matrix} 1000+5000+1000 \\ 625+4000+1500 \\ 875+5000+2000 \end{matrix} \right] \)
Funds collected by school A : Rs. 7,000
Funds collected by school B : Rs. 6,125
Funds collected by school C : Rs. 7,875
Total collection : Rs. 7,000 + Rs. 6,125 + Rs. 7,875 = Rs. 21,000
Value : Fund will help us to provide relief to those affected by flooding.

A matrix is a collection of numbers arranged into a fixed number of rows and columns. Usually the numbers are real numbers. In general, matrices can contain complex numbers but we won't see those here. Here is an example of a matrix with three rows and three columns:

The top row is row 1. The leftmost column is column 1. This matrix is a 3x3 matrix because it has three rows and three columns. In describing matrices, the format is rows X columns
Each number that makes up a matrix is called an element of the matrix. The elements in a matrix have specific locations.

\({\left[\begin{array}{ccc} x & -5 & -1 \end{array}\right]\left[\begin{array}{c} x+2 \\ 9 \\ 2 x+3 \end{array}\right]=0} \)
\(\Rightarrow \left[x^{2}+2 x-45-2 x-3\right]=[0] \Rightarrow x^{2}-48=0 \Rightarrow x=\pm 4 \sqrt{3} \)

\(\mathrm{LHS}=(a I+b A)^{3}=\left\{\left[\begin{array}{ll} a & 0 \\ 0 & a \end{array}\right]+\left[\begin{array}{ll} 0 & b \\ 0 & 0 \end{array}\right]\right\}^{3} \)
\(=\left[\left.\begin{array}{ll} a & b \\ 0 & a \end{array}\right|^{3}=\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]\right. \)
\(=\left[\begin{array}{cc} a^{2} & 2 a b \\ 0 & a^{2} \end{array}\right]\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]=\left[\begin{array}{cc} a^{3}+0 & a^{2} b+2 a^{2} b \\ 0+0 & 0+a^{3} \end{array}\right]=\left[\begin{array}{cc} a^{3} & 3 a^{2} b \\ 0 & a^{3} \end{array}\right] \)
\(\mathrm{RHS}=a^{3} I+3 a^{2} b A=\left[\begin{array}{cc} a^{3} & 0 \\ 0 & a^{3} \end{array}\right]+\left[\begin{array}{cc} 0 & 3 a^{2} b \\ 0 & 0 \end{array}\right]=\left[\begin{array}{cc} a^{3} & 3 a^{2} b \\ 0 & a^{3} \end{array}\right] \)
\(\mathrm{LHS}=\mathrm{RHS} \)
\(\text {Hence, }(a I+b l)^{3}=a^{3} I+3 a^{2} b A \)

(i) \(A=\frac { 1 }{ 2 } (A+A')+\frac { 1 }{ 2 } (A-A');\) 
where \(A=\frac { 1 }{ 2 } (A+A')\) is symmetric
and \(\frac { 1 }{ 2 } (A-A')\) is skew symmetric Proceed.
\(\text { Let } A=\left[\begin{array}{rr} 3 & 5 \\ 1 & -1 \end{array}\right] \text { , then } A^{\prime}=\left[\begin{array}{ll} 3 & 1 \\ 5 & -1 \end{array}\right]\)
\(\text { Now, } A+A^{\prime}=\left[\begin{array}{rr} 3 & 5 \\ 1 & -1 \end{array}\right]+\left[\begin{array}{rr} 3 & 1 \\ 5 & -1 \end{array}\right]=\left[\begin{array}{rr} 6 & 6 \\ 6 & -2 \end{array}\right]\)
\(\text { Let } P=\frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{rr} 6 & 6 \\ 6 & -2 \end{array}\right]=\left[\begin{array}{rr} 3 & 3 \\ 3 & -1 \end{array}\right]\)
\(\text { Now, } P^{\prime}=\left[\begin{array}{rr} 3 & 3 \\ 3 & -1 \end{array}\right]=P\)
\(\text { Thus, }P=\frac{1}{2}\left(A+A^{\prime}\right) \text { is a symmetric matrix. }\)
\(\text { Now, } A-A^{\prime}=\left[\begin{array}{rr} 3 & 5 \\ 1 & -1 \end{array}\right]-\left[\begin{array}{cc} 3 & 1 \\ 5 & -1 \end{array}\right]=\left[\begin{array}{rr} 0 & 4 \\ -4 & 0 \end{array}\right]\)
\(\text { Let } Q=\frac{1}{2}\left(A-A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{rr} 0 & 4 \\ -4 & 0 \end{array}\right]=\left[\begin{array}{rr} 0 & 2 \\ -2 & 0 \end{array}\right]\)
\(\text { Now, } Q^{\prime}=\left[\begin{array}{rr} 0 & 2 \\ -2 & 0 \end{array}\right]=-Q\)
Thus \(Q=\frac{1}{2}\left(A-A^{\prime}\right) \text { is a skew-symmetric matrix. }\)
\(\text { Representing } A \text { as the sum of } P \text { and } Q \text { : }\)
\(P+Q=\left[\begin{array}{rr} 3 & 3 \\ 3 & -1 \end{array}\right]+\left[\begin{array}{rr} 0 & 2 \\ -2 & 0 \end{array}\right]=\left[\begin{array}{rr} 3 & 5 \\ 1 & -1 \end{array}\right]=A\)
(ii) \(\text { Let } A=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right], \text { then } A^{\prime}=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]\)
\(\text { Now, } A+A^{\prime}=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]+\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]=\left[\begin{array}{rrr} 12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6 \end{array}\right]\)
\(\text { Now, } P^{\prime}=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]=P\)
Thus,\(P=\frac{1}{2}\left(A+A^{\prime}\right) \text { is a symmetric matrix. }\)
\(\text { Now, } A-A^{\prime}=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]+\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\)
\(\text { Let } Q=\frac{1}{2}\left(A-A^{\prime}\right)=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\)
\(\text { Now, } Q^{\prime}=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]=-Q\)
Thus \(Q=\frac{1}{2}\left(A-A^{\prime}\right) \text { is a skew-symmetric matrix. }\)
\(\text { Representing } A \text { as the sum of } P \text { and } Q \text { : }\)
\(P+Q=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]+\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]=A\)
(iii) Let \(A=\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]\), then \(A^{\prime}=\left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]\)
Now, \(A+A=\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]+\left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]=\left[\begin{array}{ccc}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{array}\right]\)
\(P=\frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{ccc}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{array}\right]=\left[\begin{array}{ccc}3 & \frac{1}{2} & -\frac{-5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2\end{array}\right]\)
Now, \(P^{\prime}=\left[\begin{array}{ccc}3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2\end{array}\right]=\left[\begin{array}{ccc}3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2\end{array}\right]=P\)
Thus, \(P=\frac{1}{2}\left(A+A^{\prime}\right)\) is a symmetric matrix. 
Now, \(A-A^{\prime}=\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]-\left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]=\left[\begin{array}{ccc}0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0\end{array}\right]\)
Let \(Q=\frac{1}{2}\left(A-A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{ccc}0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0\end{array}\right]=\left[\begin{array}{ccc}0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0\end{array}\right]\)
Now, \(Q=\left[\begin{array}{ccc}0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0\end{array}\right]=\left[\begin{array}{ccc}0 & -\frac{5}{2} & -\frac{3}{2} \\ \frac{5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0\end{array}\right]=-Q\)
Thus, \(Q=\frac{1}{2}\left(A-A^{\prime}\right)\) is a skew -symmetric matrix.
Representing A as the sum of P and Q
\(P+Q=\left[\begin{array}{ccc} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -22 & ] \end{array}\right]+\left[\begin{array}{ccc} 0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0 \end{array}\right]=\left[\begin{array}{ccc} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{array}\right]=A\)
(iv)
\(\text { Let } A=\left[\begin{array}{rr} 1 & 5 \\ -1 & 2 \end{array}\right] \text { , then } A^{\prime}=\left[\begin{array}{lr} 1 & -1 \\ 5 & 2 \end{array}\right]\)
\(\text { Now } A+A^{\prime}=\left[\begin{array}{rr} 1 & 5 \\ -1 & 2 \end{array}\right]+\left[\begin{array}{rr} 1 & -1 \\ 5 & 2 \end{array}\right]=\left[\begin{array}{ll} 2 & 4 \\ 4 & 4 \end{array}\right]\)
\(\text { Let } P=\frac{1}{2}\left(A+A^{\prime}\right)=\left[\begin{array}{ll} 1 & 2 \\ 2 & 2 \end{array}\right]\)
\(P=\frac{1}{2}\left(A+A^{\prime}\right) \text { is a symmetric matrix. }\)
\(\text { Now, } A-A^{\prime}=\left[\begin{array}{rr} 1 & 5 \\ -1 & 2 \end{array}\right]-\left[\begin{array}{rr} 1 & -1 \\ 5 & 2 \end{array}\right]=\left[\begin{array}{rr} 0 & 6 \\ -6 & 0 \end{array}\right]\)
\(\text { Let } Q=\frac{1}{2}\left(A-A^{\prime}\right)=\left[\begin{array}{rr} 0 & 3 \\ -3 & 0 \end{array}\right]\)
\(\text { Now, } Q^{\prime}=\left[\begin{array}{rr} 0 & -3 \\ 3 & 0 \end{array}\right]=-Q\)
\(Q=\frac{1}{2}\left(A-A^{\prime}\right) \text { is a skew-symmetric matrix. }\)
\(\text { Representing } A \text { as the sum of } P \text { and } Q \text { : }\)
\(P+Q=\left[\begin{array}{ll} 1 & 2 \\ 2 & 2 \end{array}\right]+\left[\begin{array}{rr} 0 & 3 \\ -3 & 0 \end{array}\right]=\left[\begin{array}{rr} 1 & 5 \\ -1 & 2 \end{array}\right]=A\)