Class 12 Maths Chapter 3 Miscellaneous Exercise Solutions

To prove: \( (a I+b A)^{n}=a^{n} I+nan^{-1} b A \)
Proof: Given \( A=\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right] \)
We will be proving the above equation using mathematical induction.
Steps involved in mathematical induction \( A=\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right] \) are-
1. Prove the equation for \( \mathrm{n}=1 \)
2. Assume the equation to be true for \( \mathrm{n}=\mathrm{k} \), where \( \mathrm{k} \in \mathrm{N} \)
3. Finally prove the equation for \( \mathrm{n}=\mathrm{k}+1 \)
I is the identity matrix of order 2 ,
i.e. \( I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \)
Let \( P(n):(a I+b A)^{n}=a^{n} I+nan^{-1} b A, n \in N \)
For \( \mathrm{n}=1 \),
L.H.S: \( (\mathrm{aI}+\mathrm{bA})^{1}=\mathrm{aI}+\mathrm{bA} \)
R.H.S: \( a^{1} I+1 a^{1-1} b A=a I+a^{0} b A=a I+b A \)
So, L.H.S \( = \) R.H.S
\( \therefore \mathrm{P}(\mathrm{n}) \) is true for \( \mathrm{n}=1 \)
Now assuming \( P(n) \) to be true for \( \mathrm{n}=\mathrm{k} \), where \( \mathrm{k} \in \mathrm{N} \)
\(P(k):(a I+b A)^{k}=a^{k} I+k a^{k-1} b A\ldots(1)\)
Now proving for \( \mathrm{n}=\mathrm{k}+1 \), i.e. \( \mathrm{P}(\mathrm{k}+1) \) is also true
\(\text { L.H.S }=(\mathrm{aI}+\mathrm{bA})^{\mathrm{k}+1}\)
\(=(\mathrm{aI}+\mathrm{bA})^{\mathrm{k}} \cdot(\mathrm{aI}+\mathrm{bA})^{1}\)
\(=\left(a^{k} \mathrm{I}+k \mathrm{a}^{\mathrm{k}-1} \mathrm{bA}\right) \cdot(\mathrm{aI}+\mathrm{bA}) \ldots \text {from} (1) \)
\(=\mathrm{aI}\left(\mathrm{a}^{\mathrm{k}} \mathrm{I}+\mathrm{ka}^{\mathrm{k}-1} \mathrm{bA}\right)+\mathrm{bA}\left(\mathrm{a}^{\mathrm{k}} \mathrm{I}+\mathrm{ka}^{\mathrm{k}-1} \mathrm{bA}\right)\)
\(=\mathrm{aI}\left(\mathrm{a}^{\mathrm{k}} \mathrm{I}\right)+\mathrm{aI}\left(\mathrm{ka}^{\mathrm{k}-1} \mathrm{bA}\right)+\mathrm{bA}\left(\mathrm{a}^{\mathrm{k}} \mathrm{I}\right)+\mathrm{bA}\left(\mathrm{ka}^{\mathrm{k}-\mathrm{b}} \mathrm{bA}\right)\)
\(=\left(\mathrm{a} \cdot \mathrm{a}^{\mathrm{k}}\right)(\mathrm{I} \times \mathrm{I})+\mathrm{kb}\left(\mathrm{a} \cdot \mathrm{a}^{\mathrm{k}-1}\right)(\mathrm{IA})+\left(\mathrm{ba}^{\mathrm{k}}\right)(\mathrm{AI})+(\mathrm{bb}) \mathrm{ka}^{\mathrm{k}-1}(\mathrm{AA})\)
\(=\mathrm{ak}^{+1} \mathrm{I}^{2}+\mathrm{ka}^{1+\mathrm{k}-1} \mathrm{bA}+\mathrm{ba}^{\mathrm{k}} \mathrm{A}+\mathrm{b} 2^{\mathrm{k}} \mathrm{a}^{\mathrm{k}-1} \mathrm{~A}^{2}\left(\mathrm{IA}=\mathrm{AI}=\mathrm{A} \& \mathrm{I}^{2}=\mathrm{I}\right)\)
\(=a^{k}+1 I+k a^{k} b A+b a^{k} A+b 2^{k} a^{k-1} A^{2}\)
Calculating \( \mathrm{A}^{2} \)
\(\begin{array}{l}
\mathrm{A}^{2}=\mathrm{A} \cdot \mathrm{A} \\
=\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right] \\
=\left[\begin{array}{ll}
0.0+1.0 & 0.1+0.1 \\
0.0+0.0 & 0.1+0.0
\end{array}\right] \\
=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]
\end{array}\)
\( \therefore \mathrm{A}^{2}=\mathrm{O} \) ( O is the null matrix)
Putting value of \( \mathrm{A}^{2} \) in L.H.S
\(\text { L.H.S }=a^{k+1} I+k a^{k} b A+b a^{k} A+b 2^{k} a^{k-1}(O)\)
\(=a^{k+1} I+k a^{k} b A+b a^{k} A+0\)
\(=a^{k+1} I+k a^{k} b A+b a^{k} A\)
\(=a^{k+1} I+(k+1) a^{k} b A\)
Putting \( \mathrm{n}=\mathrm{k}+1 \) in R.H.S
R.H.S \( =\mathrm{a}^{\mathrm{k}+1} \mathrm{I}+(\mathrm{k}+1) \mathrm{a}^{\mathrm{k}} \mathrm{bA} \)
\( \therefore \) L.H.S \( = \) R.H.S
All conditions are proved. Hence \( \mathrm{P}(\mathrm{k}+1) \) is true.
\( \therefore \) By mathematical induction we have proved that \( \mathrm{P}(\mathrm{n}) \) is true for all \( \mathrm{n} \epsilon \) N .
Thus, \( (a I+b A)^{n}=a^{n} I+n a^{n-1} b A \), where \( I \) is the identity matrix of order 2 and \( \mathrm{n} \in \mathrm{N} \).

We will be proving the above equation by putting different values of n (i.e. \( \mathrm{n}=1,2,3 \ldots . \mathrm{n} \) )
For \( \mathrm{n}=1 \),
\( A^{1}=\left[\begin{array}{lll}3^{1-1} & 3^{1-1} & 3^{1-1} \\ 3^{1-1} & 3^{1-1} & 3^{1-1} \\ 3^{1-1} & 3^{1-1} & 3^{1-1}\end{array}\right] \)
\( =\left[\begin{array}{lll}3^{0} & 3^{0} & 3^{0} \\ 3^{0} & 3^{0} & 3^{0} \\ 3^{0} & 3^{0} & 3^{0}\end{array}\right] \)
\( =\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right] \)
For \( \mathrm{n}=2 \),
\( \mathrm{A}^{2}=\mathrm{A} . \mathrm{A} \)
\(\begin{array}{l}
=\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right] \\
=\left[\begin{array}{lll}
1.1+1.1+1.1 & 1.1+1.1+1.1 & 1.1+1.1+1.1 \\
1.1+1.1+1.1 & 1.1+1.1+1.1 & 1.1+1.1+1.1 \\
1.1+1.1+1.1 & 1.1+1.1+1.1 & 1.1+1.1+1.1
\end{array}\right] \\
=\left[\begin{array}{lll}
3 & 3 & 3 \\
3 & 3 & 3 \\
3 & 3 & 3
\end{array}\right]=\left[\begin{array}{lll}
3^{2-1} & 3^{2-1} & 3^{2-1} \\
3^{2-1} & 3^{2-1} & 3^{2-1} \\
3^{2-1} & 3^{2-1} & 3^{2-1}
\end{array}\right]
\end{array}\)
For \( \mathrm{n}=3 \),
\(\begin{array}{l}
\mathrm{A}^{3}=\mathrm{A}^{2} \cdot \mathrm{A} \\
=\left[\begin{array}{lll}
3 & 3 & 3 \\
3 & 3 & 3 \\
3 & 3 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right] \\
=\left[\begin{array}{lll}
3.1+3.1+3.1 & 3.1+3.1+3.1 & 3.1+3.1+3.1 \\
3.1+3.1+3.1 & 3.1+3.1+3.1 & 3.1+3.1+3.1 \\
3.1+3.1+3.1 & 3.1+3.1+3.1 & 3.1+3.1+3.1
\end{array}\right] \\
=\left[\begin{array}{lll}
9 & 9 & 9 \\
9 & 9 & 9 \\
9 & 9 & 9
\end{array}\right] \\
=\left[\begin{array}{lll}
3^{2} & 3^{2} & 3^{2} \\
3^{2} & 3^{2} & 3^{2} \\
3^{2} & 3^{2} & 3^{2}
\end{array}\right]=\left[\begin{array}{lll}
3^{3-1} & 3^{3-1} & 3^{3-1} \\
3^{3-1} & 3^{3-1} & 3^{3-1} \\
3^{3-1} & 3^{3-1} & 3^{3-1}
\end{array}\right]
\end{array}\)
\(\begin{array}{l}
\mathrm{rn}=4, \\
\mathrm{~A}^{4}=\mathrm{A}^{3} \cdot \mathrm{A} \\
=\left[\begin{array}{lll}
9 & 9 & 9 \\
9 & 9 & 9 \\
9 & 9 & 9
\end{array}\right]\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]
\end{array}\)
\(\begin{array}{l}
=\left[\begin{array}{lll}
9.1+9.1+9.1 & 9.1+9.1+9.1 & 9.1+9.1+9.1 \\
9.1+9.1+9.1 & 9.1+9.1+9.1 & 9.1+9.1+9.1 \\
9.1+9.1+9.1 & 9.1+9.1+9.1 & 9.1+9.1+9.1
\end{array}\right] \\
=\left[\begin{array}{lll}
27 & 27 & 27 \\
27 & 27 & 27 \\
27 & 27 & 27
\end{array}\right]=\left[\begin{array}{lll}
3^{4-1} & 3^{4-1} & 3^{4-1} \\
3^{4-1} & 3^{4-1} & 3^{4-1} \\
3^{4-1} & 3^{4-1} & 3^{4-1}
\end{array}\right]
\end{array}\)
d so on for other values of \( n \).
If we notice each result, then we will see that it is of same type that we are trying to prove.
So we can generalize the above results for all \( \mathrm{n} \epsilon \mathrm{N} \)
\(\therefore \mathrm{A}^{\mathrm{n}}=\left[\begin{array}{lll}
3^{n-1} & 3^{n-1} & 3^{n-1} \\
3^{n-1} & 3^{n-1} & 3^{n-1} \\
3^{n-1} & 3^{n-1} & 3^{n-1}
\end{array}\right](\mathrm{n}=1,2,3, \ldots \mathrm{n})\)
Hence Proved

We will be proving the above equation by putting different values of \( n \). Because n is a positive integer so it will take values which are greater than 0 i.e. \( \mathrm{n}=1,2,3 \ldots \mathrm{n} \)
For \( \mathrm{n}=1 \),
\(\begin{array}{l}
\text { L.H.S }=\mathrm{A}^{\mathrm{n}}=\mathrm{A}^{1}=\mathrm{A}=\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right] \\
\text { R.H.S }=\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right] \\
=\left[\begin{array}{cc}
1+2 \cdot 1 & -4.1 \\
1 & 1-2.1
\end{array}\right]
\end{array}\)
\(\begin{array}{l}
=\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right] \\
\therefore \text { L.H.S }=\text { R.H.S }
\end{array}\)
For \( \mathrm{n}=2 \),
\(\begin{array}{l}
\text { L.H.S }=\mathrm{A}^{2} \\
=\text { A.A } \\
=\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right] \\
=\left[\begin{array}{ll}
3.3+(-4) \cdot 1 & 3 \cdot(-4)+(-4) \cdot(-1) \\
1 \cdot 3+(-1) \cdot 1 & 1 \cdot(-4)+(-1) \cdot(-1)
\end{array}\right] \\
=\left[\begin{array}{ccc}
9-4 & -12+4 \\
3-1 & -4+1
\end{array}\right]=\left[\begin{array}{ll}
5 & -8 \\
2 & -3
\end{array}\right] \\
\text { R.H.S. }=\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right] \\
=\left[\begin{array}{cc}
1+2 \cdot 2 & -4 \cdot 2 \\
2 & 1-2 \cdot 2
\end{array}\right] \\
=\left[\begin{array}{cc}
1+4 & -8 \\
2 & 1-4
\end{array}\right]=\left[\begin{array}{ll}
5 & -8 \\
2 & -3
\end{array}\right]
\end{array}\)
\(\therefore \text { L.H.S = R.H.S }\)
For \( \mathrm{n}=3 \),
\(\begin{array}{l}
\text { L.H.S }=\mathrm{A}^{3} \\
=\mathrm{A}^{2} \cdot \mathrm{A} \\
=\left[\begin{array}{ll}
5 & -8 \\
2 & -3
\end{array}\right]\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right] \\
=\left[\begin{array}{cc}
5 \cdot 3+(-8) \cdot 1 & 5 \cdot(-4)+(-8) \cdot(-1) \\
2 \cdot 3+(-3) \cdot 1 & 2 \cdot(-4)+(-3) \cdot(-1)
\end{array}\right] \\
=\left[\begin{array}{cc}
15-8 & -20+8 \\
6-3 & -8+3
\end{array}\right]=\left[\begin{array}{cc}
7 & -12 \\
3 & -5
\end{array}\right]
\end{array}\)
\(\begin{array}{l}
\text { R.H.S. }=\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right] \\
=\left[\begin{array}{cc}
1+2.3 & -4.3 \\
3 & 1-2.3
\end{array}\right] \\
=\left[\begin{array}{cc}
7 & -12 \\
3 & -5
\end{array}\right]
\end{array}\)
\(\therefore \text { L.H.S }=\text { R.H.S }\)
For \( \mathrm{n}=4 \),
L.H.S \( =\mathrm{A}^{4} \)
\(\begin{array}{l}
=\mathrm{A}^{3} \text {. } \mathrm{A} \\
=\left[\begin{array}{cc}
7 & -12 \\
3 & -5
\end{array}\right]\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right] \\
=\left[\begin{array}{cc}
7 \cdot 3+(-12) \cdot 1 & 7 \cdot(-4)+(-12) \cdot(-1) \\
3 \cdot 3+(-5) \cdot 1 & 3 \cdot(-4)+(-5) \cdot(-1)
\end{array}\right] \\
=\left[\begin{array}{cc}
21-12 & -28+12 \\
9-5 & -12+5
\end{array}\right]=\left[\begin{array}{cc}
9 & -16 \\
4 & -7
\end{array}\right] \\
\text { R.H.S }=\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right] \\
=\left[\begin{array}{cc}
1+2.4 & -4.4 \\
4 & 1-2.4
\end{array}\right] \\
=\left[\begin{array}{cc}
9 & 16 \\
4 & -7
\end{array}\right]
\end{array}\)
\( \therefore \) L.H.S \( = \) R.H.S
And so on for other values of \( n \).
If we notice each result, then we will see that it is of same type that we are trying to prove.
So we can generalize the above results for all positive integer values of n.
\(\therefore \mathrm{A}^{\mathrm{n}}=\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right](\mathrm{n}=1,2,3 \ldots \mathrm{n})\)
Hence Proved

To prove: \( \mathrm{AB}-\mathrm{BA} \) is a skew symmetric matrix.
Symmetric matrix: A symmetric matrix is a square matrix that is equal to its transpose. In simple words, matrix A is symmetric if
\(\mathrm{A}=\mathrm{A}^{\prime}\)
where \( A^{\prime} \) is the transpose of matrix \( A \).
Skew Symmetric matrix: A skew symmetric matrix is a square matrix that is equal to minus of its transpose. In simple words, matrix A is skew symmetric if
\(\mathrm{A}=-\mathrm{A}^{\prime}\)
Given: A and B are symmetric matrices i.e.
\(\begin{array}{l}
\mathrm{A}=\mathrm{A}^{\prime} \ldots(1) \\
\mathrm{B}=\mathrm{B}^{\prime} \ldots(2)\end{array}\)
Now calculating the transpose of \( \mathrm{AB}-\mathrm{BA} \),
\((\mathrm{AB}-\mathrm{BA})^{\prime}=(\mathrm{AB})^{\prime}-(\mathrm{BA})^{\prime}\)
(By property of transpose i.e. \( \left.(A-B)^{\prime}=A^{\prime}-B^{\prime}\right) \)
\(=\mathrm{B}^{\prime} \mathrm{A}^{\prime}-\mathrm{A}^{\prime} \mathrm{B}^{\prime}\)
(By property of transpose i.e. \( (\mathrm{AB})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime} \) )
\(=\mathrm{BA}-\mathrm{AB}\)
\(=-(\mathrm{AB}-\mathrm{BA})\)
Or we can say that: \( (A B-B A)=-(A B-B A)^{\prime} \)
Clearly it satisfies the condition of skew symmetric matrix.
Hence \( \mathrm{AB}-\mathrm{BA} \) is a skew symmetric matrix.

Case 1: When A is a symmetric matrix i.e.
\(\mathrm{A}=\mathrm{A}^{\prime}\ldots(1)\)
where \( \mathrm{A}^{\prime} \) is the transpose of A
To prove: \( B^{\prime} A B \) is also a symmetric matrix.
Calculating the transpose of \( \mathrm{B}^{\prime} \mathrm{AB} \)
\( \left(B^{\prime} A B\right)^{\prime}=B^{\prime} A^{\prime}\left(B^{\prime}\right)^{\prime}\left(B y\right. \) property of transpose i.e. \( \left.(A B)^{\prime}=B^{\prime} A^{\prime}\right) \)
\( =\mathrm{B}^{\prime} \mathrm{A}^{\prime} \mathrm{B} \) (By property of transpose i.e. \( \left.\left(\mathrm{A}^{\prime}\right)^{\prime}=\mathrm{A}\right) \)
\( =\mathrm{B}^{\prime} \mathrm{AB} \) (from (1))
It satisfies the condition of symmetric matrix as matrix \( \mathrm{B}^{\prime} \mathrm{AB} \) is equal to its transpose.
Hence \( B^{\prime} A B \) is a symmetric matrix when \( A \) is symmetric.
Case 2: When A is a skew symmetric matrix i.e.
\( \mathrm{A}=-\mathrm{A}^{\prime} \ldots(2)\)
where \( \mathrm{A}^{\prime} \) is the transpose of A .
To prove: \( \mathrm{B}^{\prime} \mathrm{AB} \) is also a skew symmetric matrix.
Calculating the transpose of \( \mathrm{B}^{\prime} \mathrm{AB} \)
\( \left(B^{\prime} A B\right)^{\prime}=B^{\prime} A^{\prime}\left(B^{\prime}\right)^{\prime}\left(B y\right. \) property of transpose i.e. \( \left.(A B)^{\prime}=B^{\prime} A^{\prime}\right) \)
\( =\mathrm{B}^{\prime} \mathrm{A}^{\prime} \mathrm{B} \) (By property of transpose i.e. \( \left.\left(\mathrm{A}^{\prime}\right)^{\prime}=\mathrm{A}\right) \)
\( =\mathrm{B}^{\prime}(-\mathrm{A}) \mathrm{B}( \) from \( (2)) \)
\( =-\left(B^{\prime} A B\right) \)
It satisfies the condition of skew symmetric matrix as matrix \( \left(B^{\prime} A B\right) \) is equal to its transpose.
Hence \( \left(B^{\prime} A B\right) \) is a skew symmetric matrix when \( A \) is skew symmetric.
\( \therefore \) Both results are proved.

Given \( \mathrm{A}=\left[\begin{array}{ccc}0 & 2 y & z \\ x & y & -z \\ x & -y & z\end{array}\right] \)
Transpose of a matrix: If A be an \( \mathrm{m} \times \mathrm{n} \) matrix, then the matrix obtained by interchanging the rows and columns of A is called the transpose of A . It is denoted by A' or AT.
\( \therefore \) Transpose of \( \mathrm{A}=\mathrm{A}^{\prime}=\left[\begin{array}{ccc}0 & x & x \\ 2 y & y & -y \\ z & -z & z\end{array}\right] \)
Given equation
\( A^{\prime} \mathrm{A}=\mathrm{I} \)
\(\begin{array}{l}
=\left[\begin{array}{ccc}
0 & 2 y & z \\
x & y & -z \\
x & -y & z
\end{array}\right]\left[\begin{array}{ccc}
0 & x & x \\
2 y & y & -y \\
z & -z & z
\end{array}\right]\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
= \\
{\left[\begin{array}{ccc}
0 \cdot 0+2 y \cdot 2 y+z \cdot z & 0 \cdot x+2 y \cdot y+z \cdot(-z) & 0 \cdot x+2 y(-y)+z \cdot z \\
x \cdot 0+y \cdot 2 y+(-z) \cdot z & x \cdot x+y \cdot y+(-z) \cdot(-z) & x \cdot x+y \cdot(-y)+(-z) \cdot z \\
x \cdot 0+(-y) \cdot(2 y)+z \cdot z & x \cdot x+(-y) \cdot y+z \cdot(-z) & x \cdot x+(-y) \cdot(-y)+z \cdot z
\end{array}\right]} \\
=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
{\left[\begin{array}{ccc}
4 y^{2}+z^{2} & 2 y^{2}-z^{2} & -2 y^{2}+z^{2} \\
2 y^{2}-z^{2} & x^{2}+y^{2}+z^{2} & x^{2}-y^{2}-z^{2} \\
-2 y^{2}+z^{2} & x^{2}-y^{2}-z^{2} & x^{2}+y^{2}+z^{2}
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]}
\end{array}\)
As these matrices are equal to each other that means each element of matrix on L.H.S is equal each element of matrix on R.H.S.
\( \therefore \) On comparing elements on both sides we get
\(\begin{array}{l}
4 y^{2}+z^{2}=1 \ldots (1) \\
2 y^{2}-z^{2}=0 \ldots (2) \\
x^{2}+y^{2}+z^{2}=1 \ldots (3)\\
-y^{2}-z^{2}=0 \ldots (4)\end{array}\)
From equation (4) we get,
\(x^{2}=y^{2}+z^{2}\ldots(5)\)
Substituting this value in equation (3) we get,
\(2 y^{2}+2 z^{2}=1\ldots(6)\)
Subtracting equation (2) and (6) we get,
\(3 z^{2}=1\)
\(\begin{array}{l}
z^{2}=\frac{ 1 }{ 3 } \\
z=\frac{ 1 }{ 3 }
\end{array}\)
Substituting value of \( z \) in equation (2) we get,
\(\begin{array}{l}
2 y^{2}=\frac{ 1 }{ 3 } \\
y^{2}=\frac{ 1 }{ 6 } \\
y=\frac{ 1 }{ 6 }
\end{array}\)
Substituting values of \( y \) and \( z \) in equation (5) we get,
\(\begin{array}{l}
x^{2}=\frac{1}{6}+\frac{1}{6} \\
x^{2}=\frac{ 1 }{ 2 } \\
x=\frac{ 1 }{ 2 }
\end{array}\)
Hence values of \( x, y, z \) is \( \frac{ 1 }{ 2 },\frac{ 1 }{ 6 },\frac{ 1 }{ 3 } \) respectively.

Multiplying matrices on the left hand side
\(\begin{array}{l}
\text { L.H.S }=\left[\begin{array}{lll}
1 & 2 & 1
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 0 \\
2 & 0 & 1 \\
1 & 0 & 2
\end{array}\right]\left[\begin{array}{l}
0 \\
2 \\
x
\end{array}\right] \\
=\left[\begin{array}{lll}
1.1+2.2+1.1 & 1.2+2.0+2.0 & 1.0+2.1+1.2
\end{array}\right]\left[\begin{array}{l}
0 \\
2 \\
x
\end{array}\right] \\
=\left[\begin{array}{lll}
6 & 2 & 4
\end{array}\right]\left[\begin{array}{l}
0 \\
2 \\
x
\end{array}\right]
\end{array}\)
\(\begin{array}{l}
=[6.0+2.2+4 . x] \\
=[4+4 x]
\end{array}\)
R.H.S \( =\mathrm{O} \) ( O is the null matrix)
\(=0 \)
\(\therefore \text { L.H.S }=\text { R.H.S, so we get, }\)
\({[4+4 \mathrm{x}]=0} \)
\(4+4 \mathrm{x}=0 \)
\(4 \mathrm{x}=-4 \)
\(\mathrm{x}=-1\)
Hence value of x is equal to -1 .

To prove: \( \mathrm{A}^{2}-5 \mathrm{~A}+7 \mathrm{I}=0 \)
Given: \( A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] \)
L.H.S: \( \mathrm{A}^{2}-5 \mathrm{~A}+7 \mathrm{I} \)
R.H.S \( =0 \)
\(I=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Calculating value of \( \mathrm{A}^{2} \) :
\(\begin{array}{l}
\mathrm{A}^{2}=\mathrm{A} \cdot \mathrm{A} \\
=\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]
\end{array}\)
\(\begin{array}{l}
=\left[\begin{array}{cc}
3 \cdot 3+1 \cdot(-10 & 3 \cdot 1+1 \cdot 2 \\
(-1) \cdot 3+2 \cdot(-1) & (-1) \cdot+2 \cdot 2
\end{array}\right] \\
=\left[\begin{array}{cc}
9-1 & 3+2 \\
-3-2 & -1+4
\end{array}\right] \\
=\left[\begin{array}{cc}
8 & 5 \\
-5 & 3
\end{array}\right]
\end{array}\)
Substituting value in L.H.S we get,
\(\begin{array}{l}
=A^{2}-5 A+7 I \\
=\left[\begin{array}{cc}
8 & 5 \\
-5 & 3
\end{array}\right]-5\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]+7\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
=\left[\begin{array}{cc}
8-15+7 & 5-5+0 \\
-5-(-5)+0 & 3-10+7
\end{array}\right] \\
=\left[\begin{array}{cc}
0 & 0 \\
0 & 0
\end{array}\right] \\
=0=\text { R.H.S }
\end{array}\)
L.H.S = R.H.S
Hence \( A^{2}-5 A+7 I=0 \) is proved.

\( \left[\begin{array}{lll}x-0-2 & 0-10-0 & 2 x-5-3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=0 \)
\( \begin{array}{l}{\left[\begin{array}{lll}x-2 & -10 & 2 x-8\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=0} \\ {[(x-2) x-10(4)+(2 x-8) 1]=0} \\ {\left[x^{2}-2 x-40 x+2 x-8\right]=0} \\ {\left[x^{2}-48\right]=[0]} \\ \text { or } x^{2}-48=0 \Rightarrow>x= \pm 4 \sqrt{3}\end{array} \)

(a) Given the unit sale prices of \( \mathrm{x}, \mathrm{y} \) and z as Rs. 2.50 , Rs. 1.50 and Rs. 1.00 respectively.
Unit sale prices can be represented in form of matrix as: \( \left[\begin{array}{l}2.50 \\ 1.50 \\ 1.00\end{array}\right] \)
Calculating total revenue in market I:
Number of products in the form of matrix: \( \left[\begin{array}{lll}10000 & 2000 & 18000\end{array}\right] \)
So the total revenue is given by:
\(\begin{array}{l}
=\left[\begin{array}{lll}
10000 & 2000 & 18000
\end{array}\right]\left[\begin{array}{l}
2.50 \\
1.50 \\
1.00
\end{array}\right] \\
=[10000 \cdot(2.50)+2000 \cdot(1.50)+18000 \cdot(1.00)] \\
=[25000+3000+18000] \\
=[46000]
\end{array}\)
\( \therefore \) Total revenue in market is Rs 46000 .
Calculating total revenue in market II:
Number of products in the form of matrix: \( \left.\begin{array}{lll}6000 & 20000 & 8000\end{array}\right] \)
So the total revenue is given by:
\(\begin{array}{l}
=\left[\begin{array}{lll}
6000 & 20000 & 8000
\end{array}\right]\left[\begin{array}{l}
2.50 \\
1.50 \\
1.00
\end{array}\right] \\
=[6000 .(2.50)+20000 .(1.50)+8000 .(1.00)] \\
=[15000+30000+8000]
\end{array}\)
\(=[53000]\)
\( \therefore \) Total revenue in market is Rs. 53000.
(b) Given the unit cost prices of \( \mathrm{x}, \mathrm{y} \) and z as Rs. 2.00, Rs. 1.00 and 50 paisa respectively.
Calculating gross profit in market I:
Unit cost prices can be represented in form of matrix as: \( \left[\begin{array}{l}2.00 \\ 1.00 \\ 0.50\end{array}\right] \)
So the total cost of products in market I is given by:
\(\begin{array}{l}
=\left[\begin{array}{lll}
10000 & 2000 & 18000
\end{array}\right]\left[\begin{array}{l}
2.00 \\
1.00 \\
0.50
\end{array}\right] \\
=\left[\begin{array}{ll}
20000+2000+9000
\end{array}\right] \\
=[31000]
\end{array}\)
Since the total revenue in market I is Rs. 46000, the gross profit in this market is given by:
(Rs. 46000 - Rs. 31000 )
\( = \) Rs. 15000 .
Calculating gross profit in market II:
The total cost of products in market II is given by:
\(\begin{array}{l}
=\left[\begin{array}{lll}
6000 & 20000 & 8000
\end{array}\right]\left[\begin{array}{l}
2.00 \\
1.00 \\
0.50
\end{array}\right] \\\end{array}\)
\(=[6000 \cdot(2.00)+20000 \cdot(1.00)+8000 \cdot(0.50)]\)
\(=[12000+20000+4000]\)
\(=[36000]\)
Since the total revenue in market II is Rs. 53000, the gross profit in this market is given by:
(Rs. 53000 - Rs. 36000 )
\( = \) Rs. 17000 .

Given \( X\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right] \)
From above equation it can be observed that matrix on R.H.S is a \( 2 \times 3 \) matrix and that on the L.H.S is also a \( 2 \times 3 \) matrix. Therefore, X must be a \( 2 \times 2 \) matrix.
Let \( \mathrm{X}\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \)
So the equation is given by:
\(\begin{array}{l}
{\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6
\end{array}\right]=\left[\begin{array}{ccc}
-7 & -8 & -9 \\
2 & 4 & 6
\end{array}\right]} \\
=\left[\begin{array}{lll}
a .1+b .4 & a .2+b .5 & a .3+b .6 \\
c .1+d .4 & c .2+d .5 & c .3+d .6
\end{array}\right]=\left[\begin{array}{ccc}
-7 & -8 & -9 \\
2 & 4 & 6
\end{array}\right] \\
=\left[\begin{array}{lll}
a+4 b & 2 a+5 b & 3 a+6 b \\
c+4 d & 2 c+5 d & 3 c+6 d
\end{array}\right]=\left[\begin{array}{ccc}
-7 & -8 & -9 \\
2 & 4 & 6
\end{array}\right]
\end{array}\)
Now equating the corresponding elements of both the matrices we get,
\(a+4 b=-7,2 a+5 b=-8,3 a+6 b=-9\)
\(c+4 d=2,2 c+5 d=4,3 c+6 d=6\)
Now, \( \mathrm{a}+4 \mathrm{~b}=-7 \mathrm{~h} \Rightarrow \mathrm{a}=-4 \mathrm{~b}-7 \)
\(\therefore 2 a+5 b=-8 \Rightarrow 2 \cdot(-4 b-7)+5 b=-8\)
\(\Rightarrow-8 \mathrm{~b}-14+5 \mathrm{~b}=-8\)
\(\Rightarrow-3 \mathrm{~b}=6\)
\(\Rightarrow \mathrm{b}=-2\)
\(\therefore \mathrm{a}=-4 \mathrm{~b}-7 \Rightarrow \mathrm{a}=-4 \cdot(-2)-7\)
\(\Rightarrow \mathrm{a}=1\)
Now, \( \mathrm{c}+4 \mathrm{~d}=2 \Rightarrow \mathrm{c}=-4 \mathrm{~d}+2 \)
\(\therefore 2 \mathrm{c}+5 \mathrm{~d}=4 \Rightarrow 2 .(-4 \mathrm{~d}+2)+5 \mathrm{~d}=4\)
\(\Rightarrow-8 \mathrm{~d}+4+5 \mathrm{~d}=4\)
\(\Rightarrow-3 \mathrm{~d}=0\)
\(\Rightarrow \mathrm{d}=0\)
\(\therefore \mathrm{c}=-4 \mathrm{~d}+2 \Rightarrow \mathrm{c}=-4.0+2\)
\(\Rightarrow \mathrm{c}=2\)
Thus, \( \mathrm{a}=1, \mathrm{~b}=-2, \mathrm{c}=2, \mathrm{~d}=0 \).
Hence \( X \) becomes \( \left[\begin{array}{cc}1 & -2 \\ 2 & 0\end{array}\right] \)

To prove: \( \mathrm{AB}^{\mathrm{n}}=\mathrm{B}^{\mathrm{n}} \mathrm{A} \)
Given \( A \) and \( B \) are square matrices of same order such that \( A B=B A \).
We have to prove it using mathematical induction.
Steps involved in mathematical induction are-
1. Prove the equation for \( n=1 \)
2. Assume the equation to be true for \( \mathrm{n}=\mathrm{k} \), where \( \mathrm{k} \in \mathrm{N} \)
3. Finally prove the equation for \( \mathrm{n}=\mathrm{k}+1 \)
Let \( P(n): B^{n}=B^{n} A \)
For \( \mathrm{n}=1 \),
L.H.S: \( \mathrm{AB}^{\mathrm{n}}=\mathrm{AB}^{1}=\mathrm{AB} \)
R.H.S: \( B^{\mathrm{n}} \mathrm{A}=\mathrm{B}^{1} \mathrm{~A}=\mathrm{BA}=\mathrm{AB} \)
So, L.H.S \( = \) R.H.S
\( \therefore \mathrm{P}(\mathrm{n}) \) is true for \( \mathrm{n}=1 \).
Now assuming \( P(n) \) to be true for \( \mathrm{n}=\mathrm{k} \), where \( \mathrm{k} \in \mathrm{N} \)
\(\mathrm{P}(\mathrm{k}): \mathrm{AB}^{\mathrm{k}}=\mathrm{B}^{\mathrm{k}} \mathrm{A}\ldots(1)\)
Now proving for \( \mathrm{n}=\mathrm{k}+1 \), i.e. \( \mathrm{P}(\mathrm{k}+1) \) is also true
\(\text { L.H.S }=A B^n\)
\(=A^{k+1}\)
\(=\left(A B^{k}\right) \cdot B\)
\(=\left(B^{k} A\right) \cdot B \ldots \ldots . \text { from }(1)\)
\(=B^{k}(A \cdot B)\)
\(=B^{k}(B A)(\because A B=B A)\)
\(=B^{k+1 A}\)
R.H.S \( = \) BnA
\(=\mathrm{B}^{\mathrm{k}+1} \mathrm{~A}\)
\( \therefore \) L.H.S \( = \) R.H.S
All conditions are proved. Hence \( \mathrm{P}(\mathrm{k}+1) \) is true.
\( \therefore \) By mathematical induction we have proved that \( \mathrm{AB}^{\mathrm{n}}=\mathrm{B}^{\mathrm{n}} \mathrm{A} \).
Now, to prove: \( (A B)^{n}=A^{n} B^{n} \) for all \( \mathrm{n} \epsilon \mathrm{N} \)
For \( \mathrm{n}=1 \),
L.H.S \( =(\mathrm{AB})^{\mathrm{n}}=(\mathrm{AB})^{1}=\mathrm{AB} \)
R.H.S \( =A^{n} B^{n}=A^{1} B^{1}=A B \)
\( \therefore \) L.H.S \( = \) R.H.S
\( \therefore \) It is true for \( \mathrm{n}=1 \) Assuming it to be true for \( \mathrm{n}=\mathrm{k} \) then,
\((\mathrm{AB})^{\mathrm{k}}=\mathrm{A}^{\mathrm{k}} \mathrm{B}^{\mathrm{k}}\ldots(2)\)
Now proving for \( \mathrm{n}=\mathrm{k}+1 \),
\(\text { L.H.S }=(\mathrm{AB})^{\mathrm{n}}\)
\(=(\mathrm{AB})^{\mathrm{k}+1}\)
\(=(\mathrm{AB})^{\mathrm{k}}(\mathrm{AB})^{1}\)
\(=\left(\mathrm{A}^{\mathrm{k}} \mathrm{B}^{\mathrm{k}}\right) \mathrm{AB}\)
\(=\mathrm{A}^{\mathrm{k}}\left(\mathrm{B}^{\mathrm{k}} \cdot \mathrm{A}\right) \mathrm{B}\)
\(=\mathrm{A}^{\mathrm{k}}\left(\mathrm{A} \cdot \mathrm{B}^{\mathrm{k}}\right) \mathrm{B}\left(\mathrm{AB}^{\mathrm{n}}=\mathrm{B}^{\mathrm{n}} \mathrm{A}\right)\)
\(=\left(\mathrm{A}^{\mathrm{k}} \mathrm{A}\right)\left(\mathrm{B}^{\mathrm{k}} \mathrm{B}\right)\)
\(=\mathrm{A}^{\mathrm{k}+1} \mathrm{~B}^{\mathrm{k}+1}\)
\(\text { R.H.S }=\mathrm{A^{n}B^{n}}\)
\(=\mathrm{Ak}+1 \mathrm{Bk}+1\)
\(\therefore \text { L.H.S = R.H.S }\)
All conditions are proved. Hence \( \mathrm{P}(\mathrm{k}+1) \) is true.
\( \therefore \) By mathematical induction we have proved that \( (\mathrm{AB})^{\mathrm{n}}=\mathrm{A}^{\mathrm{n}} \mathrm{B}^{\mathrm{n}} \) for all n \( \epsilon \mathrm{N} \)
Hence proved.

Given \( \mathrm{A}=\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right] \)
Calculating \( \mathrm{A}^{2} \) :
\(\begin{array}{l}
\mathrm{A}^{2}=\mathrm{A} \cdot \mathrm{A} \\
=\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right] \\
=\left[\begin{array}{cc}
\alpha \cdot \alpha+\beta \cdot Y & \alpha \cdot \beta+\beta \cdot(-\alpha) \\
\gamma \cdot \alpha+(-\alpha) \cdot \gamma & \gamma \cdot \beta+(-\alpha) \cdot(-\alpha)
\end{array}\right] \\
=\left[\begin{array}{cc}
\alpha^{2}+\beta \cdot \gamma & \alpha \cdot \beta-\alpha \cdot \beta \\
\gamma \cdot \alpha-\gamma \cdot \alpha & \gamma \cdot \beta+\alpha^{2}
\end{array}\right] \\
=\left[\begin{array}{cc}
\alpha^{2}+\beta \gamma & 0 \\
0 & \beta \gamma+\alpha^{2}
\end{array}\right]
\end{array}\)
And given that \( \mathrm{A}^{2}=\mathrm{I} \)
Then \( \left[\begin{array}{cc}\alpha^{2}+\beta \gamma & 0 \\ 0 & \beta+\alpha^{2}\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \)
Comparing corresponding elements, we get,
\(\begin{array}{l}
=\alpha^{2}+\beta \gamma \\
=1-\alpha^{2}-\beta \gamma=0
\end{array}\)

Given A is both symmetric and skew symmetric matrix then, \( \mathrm{A}=\mathrm{A}^{\prime} \) and also \( \mathrm{A}=-\mathrm{A}^{\prime} \)
\(\begin{array}{l}
\Rightarrow \mathrm{A}^{\prime}=-\mathrm{A}^{\prime} \\
\Rightarrow 2 \mathrm{~A}^{\prime}=0 \\
\Rightarrow \mathrm{A}^{\prime}=\mathrm{O}
\end{array}\)
Clearly it is observed that transpose of A is a null matrix or zero matrix then matrix A must also be a zero matrix.
Hence A is a zero matrix.

Given that \( \mathrm{A}^{2}=\mathrm{A} \)
Calculating value of \( (I+A)^{3}-7 A \) :
\(\Rightarrow \mathrm{I}^{3}+\mathrm{A}^{3}+3 \mathrm{I}^{2} \mathrm{~A}+3 \mathrm{IA}^{2}-7 \mathrm{~A}\)
\(\Rightarrow \mathrm{I}+\mathrm{A}^{2} \cdot \mathrm{A}+3 \mathrm{~A}+3 \mathrm{~A}^{2}-7 \mathrm{~A}(\mathrm{In}=\mathrm{I} \text { and } \mathrm{I} \cdot \mathrm{A}=\mathrm{A})\)
\(\Rightarrow \mathrm{I}+\mathrm{A} \cdot \mathrm{A}+3 \mathrm{~A}+3 \mathrm{~A}-7 \mathrm{~A}\left(\mathrm{~A}^{2}=\mathrm{A}\right)\)
\(\Rightarrow \mathrm{I}+\mathrm{A}^{2}+3 \mathrm{~A}+3 \mathrm{~A}-7 \mathrm{~A}\)
\(\Rightarrow \mathrm{I}+\mathrm{A}-\mathrm{A}\)
\(\Rightarrow \mathrm{I}\)
Hence \( (I+A)^{3}-7 A=I \).