Ncert Class 12 Maths Exercise 3.3 Solutions

We know that transpose of a matrix is obtained by interchanging the elements of the rows and columns. In other words, we can say, if \( \mathrm{A}= \) \( \left[a_{i j}\right]_{m \times n} \) then \( \mathrm{A}^{\prime}=\left[a_{i j}\right]_{n \times m} \)
So, let \( \left[\begin{array}{c}5 \\ \frac{1}{2} \\ -1\end{array}\right]=\mathrm{A} \)
Therefore, transpose of the given matrix A is denoted by A '
Hence, \( A^{\prime}=\left[\begin{array}{lll}5 & \frac{1}{2} & -1\end{array}\right] \)
The transpose of the given matrix is \( \left[\begin{array}{lll}5 & \frac{1}{2} & -1\end{array}\right] \)

We know that transpose of a matrix is obtained by interchanging the elements of the rows and columns. In other words, we can say, if \( \mathrm{A}= \) \( \left[a_{i j}\right]_{m \times n} \) then \( \mathrm{A}^{\prime}=\left[a_{i j}\right]_{n \times m} \)
So, let \( \left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]=B \)
Therefore, transpose of the given matrix B is denoted by B'.
Hence, \( B^{\prime}=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right] \)
The transpose of the given matrix is \( \left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right] \).

We know that transpose of a matrix is obtained by interchanging the elements of the rows and columns. In other words, we can say, if \( \mathrm{A}= \) \( \left[a_{i j}\right]_{m \times n} \) then \( \mathrm{A}^{\prime}=\left[a_{i j}\right]_{n \times m} \)
So, let \( \left[\begin{array}{ccc}-1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 6 & -1\end{array}\right]=\mathrm{C}, \mathrm{B}=\left[\begin{array}{ccc}-4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1\end{array}\right] \)
Therefore, the transpose of the given matrix C is denoted by C'.
Hence, \( C^{\prime}=\left[\begin{array}{ccc}-1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1\end{array}\right] \)
The transpose of the given matrix is \( \left[\begin{array}{ccc}-1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1\end{array}\right] \)

\( \mathrm{A}=\left[\begin{array}{ccc}-1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{array}\right] \) and \( \mathrm{B}=\left[\begin{array}{ccc}-4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1\end{array}\right] \)
\( (\mathrm{A}+\mathrm{B})^{\prime}=\mathrm{A}^{\prime}+\mathrm{B}^{\prime} \)
Explanation: We will first calculate L.H.S i.e. (A+B)' and then consecutively we will calculate R.H.S and verify that both are equal.
So, \( A+B=\left[\begin{array}{ccc}-1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{array}\right]+\left[\begin{array}{ccc}-4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1\end{array}\right] \)
\( =\mathrm{A}+\mathrm{B}=\left[\begin{array}{ccc}-1+(-4) & 2+1 & 3+(-5) \\ 5+1 & 7+2 & 9+0 \\ -2+1 & 1+3 & 1+1\end{array}\right] \)
\( =A+B=\left[\begin{array}{ccc}-5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2\end{array}\right] \)
Therefore, \( (A+B)=\left[\begin{array}{ccc}-5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2\end{array}\right] \rightarrow \ldots (1)\)
Now, \( A^{\prime}=\left[\begin{array}{ccc}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{array}\right] \) and \( B^{\prime}=\left[\begin{array}{ccc}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{array}\right] \)
So, \( A^{\prime}+\mathrm{B}^{\prime}=\left[\begin{array}{ccc}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{array}\right]+\left[\begin{array}{ccc}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{array}\right] \)
\( =A^{\prime}+\mathrm{B}^{\prime}=\left[\begin{array}{ccc}-1+(-4) & 5+1 & -2+1 \\ 2+1 & 7+2 & 1+3 \\ 3+(-5) & 9+0 & 1+1\end{array}\right] \)
\( A^{\prime}+\mathrm{B}^{\prime}=\left[\begin{array}{ccc}-5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2\end{array}\right]
\rightarrow \ldots (2)\)
From equation \( 1 \ \& \ 2 \) we verify that \( (\mathrm{A}+\mathrm{B})^{\prime}=\mathrm{A}^{\prime}+\mathrm{B}^{\prime} \).
Hence verified.

We will first calculate L.H.S i.e. (A+B)' and then consecutively we will calculate R.H.S and verify that both are equal.
So, \( A-B=\left[\begin{array}{ccc}-1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{array}\right]-\left[\begin{array}{ccc}-4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1\end{array}\right] \)
\( =\mathrm{A}-\mathrm{B}=\left[\begin{array}{ccc}-1-(-4) & 2-1 & 3-(-5) \\ 5-1 & 7-2 & 9-0 \\ -2-1 & 1-3 & 1-1\end{array}\right] \)
\( =A-B=\left[\begin{array}{ccc}3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0\end{array}\right] \)
Therefore, \( (A-B)=\left[\begin{array}{ccc}3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0\end{array}\right] \rightarrow \ldots (1)\)
Now, \( A^{\prime}=\left[\begin{array}{ccc}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{array}\right] \) and \( B^{\prime}=\left[\begin{array}{ccc}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{array}\right] \)
So, \( A^{\prime}-\mathrm{B}^{\prime}=\left[\begin{array}{ccc}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{array}\right]-\left[\begin{array}{ccc}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{array}\right] \)
\( A^{\prime}-\mathrm{B}^{\prime}=\left[\begin{array}{ccc}-1-(-4) & 5-1 & -2-1 \\ 2-1 & 7-2 & 1-3 \\ 3-(-5) & 9-0 & 1-1\end{array}\right] \)
\( =A^{\prime}-B^{\prime}=\left[\begin{array}{ccc}3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0\end{array}\right] \rightarrow \ldots (2)\)
From equation \( 1 \ \& \ 2 \) we verify that \( (A-B)^{\prime}=A^{\prime}-B^{\prime} \).
Hence verified.

\( A^{\prime}=\left[\begin{array}{cc}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right] \) and \( B^{\prime}=\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right] \)
\( (\mathrm{A}+\mathrm{B})^{\prime}=\mathrm{A}^{\prime}+\mathrm{B}^{\prime} \)
Explanation: Calculate matrix A by taking transpose of A' and also find the transpose of matrix B and the solve L.H.S and R.H.S separately and then compare the result.
So, \( \mathrm{A}= \) transpose of \( \mathrm{A}^{\prime} \)
\(
\begin{array}{l}
=\mathrm{A}=\left[\begin{array}{ccc}
3 & -1 & 0 \\
4 & 2 & 1
\end{array}\right] \\
\mathrm{B}^{\prime}=\operatorname{transpose} \text { of } \mathrm{B} \\
=\mathrm{B}^{\prime}=\left[\begin{array}{cc}
-1 & 1 \\
2 & 2 \\
1 & 3
\end{array}\right]
\end{array}
\)
Now, \( A+B=\left[\begin{array}{ccc}3 & -1 & 0 \\ 4 & 2 & 1\end{array}\right]+\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right] \)
\( =\mathrm{A}+\mathrm{B}=\left[\begin{array}{ccc}3+(-1) & -1+2 & 0+1 \\ 4+1 & 2+2 & 1+3\end{array}\right] \)
\( =A+B=\left[\begin{array}{lll}2 & 1 & 1 \\ 5 & 4 & 4\end{array}\right] \)
Therefore, \( (A+B)=\left[\begin{array}{ll}2 & 5 \\ 1 & 4 \\ 1 & 4\end{array}\right] \rightarrow \ldots (1)\)
Now, \( A^{\prime}+B^{\prime}=\left[\begin{array}{cc}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]+\left[\begin{array}{cc}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right] \)
\( =A^{\prime}+B^{\prime}=\left[\begin{array}{ll}2 & 5 \\ 1 & 4 \\ 1 & 4\end{array}\right] \rightarrow \ldots (2)\)
From equation \( 1 \ \& \ 2 \) we verify that \( (\mathrm{A}+\mathrm{B})^{\prime}=\mathrm{A}^{\prime}+\mathrm{B}^{\prime} \).
Hence verified.

\( (A-B)^{\prime}=A^{\prime}-B^{\prime} \)
Explanation: Calculate matrix A by taking transpose of \( \mathrm{A}^{\prime} \) and also find the transpose of matrix B and the solve L.H.S and R.H.S separately and then compare the result.
\(
A-B=\left[\begin{array}{ccc}
3 & -1 & 0 \\
4 & 2 & 1
\end{array}\right]-\left[\begin{array}{ccc}
-1 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]
\)
\(
\begin{array}{l}
=A-B=\left[\begin{array}{ccc}
3-(-1) & -1-2 & 0-1 \\
4-1 & 2-2 & 1-3
\end{array}\right] \\
=A-B=\left[\begin{array}{ccc}
4 & -3 & -1 \\
3 & 0 & -2
\end{array}\right]
\end{array}
\)
Therefore, \( (A-B)=\left[\begin{array}{cc}4 & 3 \\ -3 & 0
\\ -1 & -2\end{array}\right] \rightarrow \ldots (1)\)
\( A^{\prime}-B^{\prime}=\left[\begin{array}{cc}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]-\left[\begin{array}{cc}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right] \)
\( =A^{\prime}-B^{\prime}=\left[\begin{array}{cc}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right] \rightarrow \ldots (2)\)
From equation \( 1 \ \& \ 2 \) we verify that
\( (A-B)^{\prime}=A^{\prime}-B^{\prime} \).
Hence verified.

\( A=\left[\begin{array}{cc}-2 & 3 \\ 1 & 2\end{array}\right] \) and \( B=\left[\begin{array}{cc}-1 & 0 \\ 1 & 2\end{array}\right] \)
Explanation: Whenever a constant term is multiplied with a matrix then it implies that every element in each rows and columns are to be multiplied with that constant term. So in order to solve this question we have to multiplied every element of the matrix B with constant term that is 2 .
\(
\begin{array}{l}
\text { So, } A+2 B=\left[\begin{array}{cc}
-2 & 1 \\
3 & 2
\end{array}\right]+2\left[\begin{array}{cc}
-1 & 0 \\
1 & 2
\end{array}\right] \\
=A+2 B=\left[\begin{array}{cc}
-2 & 1 \\
3 & 2
\end{array}\right]+\left[\begin{array}{cc}
-2 & 0 \\
2 & 4
\end{array}\right]
\end{array}
\)
\( =\mathrm{A}+2 \mathrm{~B}=\left[\begin{array}{cc}-4 & 1 \\ 5 & 6\end{array}\right] \)
Now, \( (A+2 B)^{\prime}= \) transpose of \( A+2 B \)
\( =(\mathrm{A}+2 \mathrm{~B})=\left[\begin{array}{cc}-4 & 5 \\ 1 & 6\end{array}\right] \)

\( A=\left[\begin{array}{c}1 \\ -4 \\ 3\end{array}\right] \) and \( B=\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right] \)
Explanation: The product of two matrices is defined or possible only if the number of columns of the former matrix is equal to number of rows of the latter matrix.
\(
\begin{array}{l}
\mathrm{AB}=\left[\begin{array}{c}
1 \\
-4 \\
3
\end{array}\right] \times\left[\begin{array}{lll}
-1 & 2 & 1
\end{array}\right] \\
=\mathrm{AB}=\left[\begin{array}{ccc}
1 \times(-1) & 1 \times 2 & 1 \times 1 \\
-4 \times(-1) & -4 \times 2 & -4 \times 1 \\
3 \times(-1) & 3 \times 2 & 3 \times 1
\end{array}\right] \\
=\mathrm{AB}=\left[\begin{array}{ccc}
-1 & 2 & 1 \\
4 & -8 & -4 \\
-3 & 6 & 3
\end{array}\right]
\end{array}
\)
\(
\text { So, }(A B)=\left[\begin{array}{ccc}
-1 & 4 & -3 \\
2 & -8 & 6 \\
1 & -4 & 3
\end{array}\right] \rightarrow \ldots (1)\)
\( A^{\prime}=\left[\begin{array}{lll}1 & -4 & 3 \end{array}\right] \) and \( B^{\prime}=\left[\begin{array}{c}-1 \\ 2 \\ 1\end{array}\right] \)
Therefore, B'A' \( =\left[\begin{array}{c}-1 \\ 2 \\ 1\end{array}\right] \times\left[\begin{array}{lll}1 & -4 & 3\end{array}\right] \)
\( =\mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{ccc}-1 \times 1 & -1 \times(-4) & -1 \times 3 \\ 2 \times 1 & 2 \times(-4) & 2 \times 3 \\ 1 \times 1 & 1 \times(-4) & 1 \times 3\end{array}\right] \)
\( =B^{\prime} A^{\prime}=\left[\begin{array}{ccc}-1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3\end{array}\right]
\rightarrow \ldots (2)\)
From equation \( 1 \ \& \ 2 \) we verify that \( (\mathrm{AB})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime} \).
Hence verified.

\(
A=\left[\begin{array}{l}
0 \\
1 \\
2
\end{array}\right] \text { and } B=\left[\begin{array}{lll}
1 & 5 & 7
\end{array}\right]
\)
Explanation: The product of two matrices is defined or possible only if the number of columns of the former matrix is equal to number of rows of the latter matrix.
\(
\mathrm{AB}=\left[\begin{array}{l}
0 \\
1 \\
2
\end{array}\right] \times\left[\begin{array}{lll}
1 & 5 & 7
\end{array}\right]
\)
\(
\begin{array}{l}
=\mathrm{AB}=\left[\begin{array}{lll}
0 \times 1 & 0 \times 5 & 0 \times 7 \\
1 \times 1 & 1 \times 5 & 1 \times 7 \\
2 \times 1 & 2 \times 5 & 2 \times 7
\end{array}\right] \\
=\mathrm{AB}=\left[\begin{array}{ccc}
0 & 0 & 0 \\
1 & 5 & 7 \\
2 & 10 & 14
\end{array}\right]
\end{array}
\)
Therefore, \( (A B)=\left[\begin{array}{ccc}0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14\end{array}\right] \rightarrow \ldots (1)\)
Now, \( A^{\prime}=\left[\begin{array}{lll}0 & 1 & 2\end{array}\right] \) and \( B^{\prime}=\left[\begin{array}{l}1 \\ 5 \\ 7\end{array}\right] \)
Therefore, \( B^{\prime} A^{\prime}=\left[\begin{array}{l}1 \\ 5 \\ 7\end{array}\right] \times\left[\begin{array}{lll}0 & 1 & 2\end{array}\right] \)
\( =\mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{lll}1 \times 0 & 1 \times 1 & 1 \times 2 \\ 5 \times 0 & 5 \times 1 & 5 \times 2 \\ 7 \times 0 & 7 \times 1 & 7 \times 2\end{array}\right] \)
\( =B^{\prime} A^{\prime}=\left[\begin{array}{ccc}0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14\end{array}\right] \rightarrow \ldots (2)\)
From equation \( 1 \ \& \ 2 \) we verify that \( (\mathrm{AB})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime} \).
Hence verified.

We know A' can be calculated by taking the transpose of the given matrix A.
Therefore, \( \mathrm{A}^{\prime}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right] \)
Now multiply A and A'. So,
\(
\begin{array}{l}
\mathrm{AA}^{\prime}=\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right] \times\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right] \\
=\mathrm{AA}^{\prime}= \\
{\left[\begin{array}{cc}
(\cos \alpha \times \cos \alpha)+(\sin \alpha) \times(\sin \alpha) & \cos \alpha \times(-\sin \alpha)+(\sin \alpha) \times \cos \alpha \\
-\sin \alpha \times \cos \alpha+\cos \alpha \sin \alpha & -\sin \alpha \times(-\sin \alpha)+\cos \alpha \times \cos \alpha
\end{array}\right]} \\
=\mathrm{AA}^{\prime}=\left[\begin{array}{cc}
\cos ^{2} \alpha+\sin ^{2} \alpha & -\sin \alpha \cos \alpha+\cos \alpha \sin \alpha \\
-\sin \alpha \cos \alpha+\cos \alpha \sin \alpha & \sin^{2} \alpha+\cos ^{2} \alpha
\end{array}\right] \\
=\mathrm{AA}^{\prime}=\left[\begin{array}{cc}
1 & 0 \\
1 & 0 \end{array}\right] \rightarrow \ldots (1) \quad \left(\therefore \cos \alpha^{2}+\sin \alpha^{2}=1\right)
\end{array}
\)
And we know ' \( I \) ' represents an identity matrix
Therefore, \( I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end {array}\right] \rightarrow \ldots(2)\)
From equation \( 1 \ \& \ 2 \) we can say that
\(
\mathrm{AA}^{\prime}=\mathrm{I}
\)
\( A^{\prime}{ }^{\prime}=I \). Hence verified.

Again, \( \mathrm{A}^{\prime}=\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{array}\right] \)
\(
\begin{array}{l}
\mathrm{AA}^{\prime}=\left[\begin{array}{cc}
\sin \alpha & \cos \alpha \\
-\cos \alpha & \sin \alpha
\end{array}\right] \times\left[\begin{array}{cc}
\sin \alpha & -\cos \alpha \\
\cos \alpha & \sin \alpha
\end{array}\right] \\
=\mathrm{AA}^{\prime}= \\
{\left[\begin{array}{cc}
\sin \alpha \times \sin \alpha+\cos \alpha \times \cos \alpha & \sin \alpha \times(-\cos \alpha)+\cos \alpha \times \sin \alpha \\
-\cos \alpha \times \sin \alpha+\sin \alpha \times \cos \alpha & -\cos \alpha \times(-\cos \alpha)+\sin \alpha \times \sin \alpha
\end{array}\right]}
\end{array}
\)
\( =\mathrm{AA}^{\prime}=\left[\begin{array}{cc}\sin^{2} \alpha+\cos ^{2} \alpha & -\sin \alpha \cos \alpha+\cos \alpha \sin \alpha \\ -\cos \alpha \sin \alpha+\sin \alpha \cos \alpha & \cos ^{2} \alpha+\sin ^{2} \alpha\end{array}\right] \)
\( =\mathrm{AA}^{\prime}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \ldots(1) \quad \left(\therefore \cos ^{2} \alpha^{2}+\sin 2 \alpha=1\right) \)
And we know ' \( I \) ' represents an identity matrix
Therefore, \( I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end {array}\right] \rightarrow \ldots(2)\)
From equation \( 1 \ \& \ 2 \) we can say that
\( \mathrm{AA}^{\prime}=\mathrm{I} \)
\( A^{\prime}{ }^{\prime}=I \). Hence verified.

A matrix is aid to be symmetric only if the transpose of matrix and the matrix itself are equal or same. This means that \( \mathrm{A}=\mathrm{A}^{\prime} \).
\( A=\left[\begin{array}{ccc}1 & -1 & 5 \\ -1 & 2 & 1 \\
5 & 1 & 3\end{array}\right] \ldots (1)\)
Now, we know that the transpose of matrix A is
\( A^{\prime}=\left[\begin{array}{ccc}1 & -1 & 5 \\ -1 &
2 & 1 \\ 5 & 1 & 3\end{array}\right] \ldots (2) \)
So, from equation \( 1 \ \& \ 2 \) we get
\( \mathrm{A}=\mathrm{A}^{\prime} \), hence we can say that Matrix A is a symmetric matrix.
Hence proved.

A matrix is said to be skew symmetric if the transpose of the matrix is equal to the negative of the matrix. This means that \( A^{\prime}=-A \).
\( A=\left[\begin{array}{ccc}0 & 1 & -1 \\ -1 & 0 & 1 \\
1 & -1 & 0\end{array}\right] \ldots (1)\)
Now, we know that the transpose of matrix A is
\( \mathrm{A}^{\prime}=\left[\begin{array}{ccc}0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0\end{array}\right] \ldots (2)\)
Now if we carefully look at the equation 2 we can rewrite it as
\( A^{\prime}=(-1)\left[\begin{array}{ccc}0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0\end{array}\right] \) (by taking -1 common)
So, \( \mathrm{A}^{\prime}=(-1) \times \mathrm{A} \) (from equation 1\( ) \)
\( \Rightarrow A^{\prime}=-A \), hence we can say that Matrix \( A \) is a skew symmetric matrix.
Hence proved

\( A=\left[\begin{array}{ll}1 & 5 \\ 6 & 7\end{array}\right] \)
\( \left(\mathrm{A}+\mathrm{A}^{\prime}\right) \) is a symmetric matrix.
So, \( A=\left[\begin{array}{ll}1 & 5 \\ 6 & 7\end{array}\right] \) and \( A^{\prime}=\left[\begin{array}{ll}1 & 6 \\ 5 & 7\end{array}\right] \)
On adding them we get,
\( A+A^{\prime}=\left[\begin{array}{ll}1 & 5 \\ 6 & 7\end{array}\right]+\left[\begin{array}{ll}1 & 6 \\ 5 & 7\end{array}\right] \)
\( =A+A^{\prime}=\left[\begin{array}{ll}1+1 & 5+6 \\ 6+5 & 7+7\end{array}\right] \)
\( =A+A^{\prime}=\left[\begin{array}{cc}2 & 11 \\ 11 &
14\end{array}\right] \rightarrow \ldots (1)\)
Explanation: Now to show that the matrix obtained i.e. \( \left(A+A^{\prime}\right) \) is symmetric we need to calculate its transpose and prove that the matrix \( \left(A^{\prime}+A^{\prime}\right) \) and its transpose are equal. This means that \( \left(A+A^{\prime}\right)=(A+ \) \( \left.A^{\prime}\right)^{\prime} \).
Therefore, \( \left(A+A^{\prime}\right)=\left[\begin{array}{cc}2 & 11 \\ 11 & 14\end{array}\right] \rightarrow \ldots (2)\)
So, from equation \( 1 \ \& \ 2 \) we get,
\( \left(\mathrm{A}+\mathrm{A}^{\prime}\right)=\left(\mathrm{A}+\mathrm{A}^{\prime}\right)^{\prime} \), hence we can say that \( \left(\mathrm{A}+\mathrm{A}^{\prime}\right) \) is a symmetric matrix.
Hence proved.

( \( \left.\mathrm{A}-\mathrm{A}^{\prime}\right) \) is a skew symmetric matrix.
So, \( A=\left[\begin{array}{ll}1 & 5 \\ 6 & 7\end{array}\right] \) and \( A^{\prime}=\left[\begin{array}{ll}1 & 6 \\ 5 & 7\end{array}\right] \)
subtracting \( \mathrm{A}^{\prime} \) from A , we get,
\(
\begin{array}{l}
A-A^{\prime}=\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]-\left[\begin{array}{ll}
1 & 6 \\
5 & 7
\end{array}\right] \\
=A-A^{\prime}=\left[\begin{array}{ll}
1-1 & 5-6 \\
6-5 & 7-7
\end{array}\right]
\end{array}
\)
Explanation: Now to show that the matrix obtained i.e. \( \left(\mathrm{A}+\mathrm{A}^{\prime}\right) \) is skew symmetric we need to calculate its transpose and prove that the matrix \( \left(A+A^{\prime}\right) \) is equal to the negative of its transpose are equal. This means that \( \left(A+A^{\prime}\right)=-\left(A+A^{\prime}\right)^{\prime} \).
Therefore, \( \left(A-A^{\prime}\right)=\left[\begin{array}
{cc}0 & 1 \\ -1 & 0\end{array}\right] \ldots (1)\)
We can rewrite above equation as
\(
\left(A-A^{\prime}\right)=(-1)\left[\begin{array}{cc}
0 & -1 \\
1 & 0 \end{array}\right] \ldots (2)\)
Also, \( \left(\mathrm{A}-\mathrm{A}^{\prime}\right)^{\prime}=(-1) \times\left(\mathrm{A}-\mathrm{A}^{\prime}\right)( \) from equation 1\( ) \)
\( \left(A-A^{\prime}\right)^{\prime}=-\left(A-A^{\prime}\right) \), hence we can say that Matrix \( A \) is a skew symmetric matrix.
Hence proved.

\(
\mathrm{A}=\left[\begin{array}{ccc}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{array}\right]
\)
(i) \( \frac{1}{2}\left(\mathrm{~A}+\mathrm{A}^{\prime}\right) \)
\( \mathrm{A}=\left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right] \) and \( \mathrm{A}^{\prime}=\left[\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right] \)
On adding A and A ', we get,
\( \mathrm{A}+\mathrm{A}^{\prime}=\left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]+\left[\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right] \)
\( =\mathrm{A}+\mathrm{A}^{\prime}=\left[\begin{array}{ccc}0+0 & a+(-a) & b+(-b) \\ -a+a & 0+0 & c+(-c) \\ -b+b & -c+c & 0+0\end{array}\right] \)
\( =A+A^{\prime}=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right] \)
\( =A+A^{\prime}=0 \) (as \( \left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right] \) a null or zero matrix is represented by ' 0 ')
Therefore, \( \left(\mathrm{A}^{\prime}+\mathrm{A}^{\prime}\right)=0 \)
Now, \( \frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}(0) \)
\( =\frac{1}{2}\left(\mathrm{~A}+\mathrm{A}^{\prime}\right)=0 \)
\( \frac{1}{2}\left(A+A^{\prime}\right)=0 \)
(ii) \( \frac{1}{2}\left(\mathrm{~A}-\mathrm{A}^{\prime}\right) \)
\( \mathrm{A}=\left[\begin{array}{ccc}0 & a & b \\ -a & 0 &
c \\ -b & -c & 0\end{array}\right] \ldots (1) \quad \) and \( \mathrm{A}^{\prime}=\left[\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right] \)
On adding A and A ', we get,
\( \mathrm{A}-\mathrm{A}^{\prime}=\left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]-\left[\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right] \)
\( =\mathrm{A}-\mathrm{A}^{\prime}=\left[\begin{array}{ccc}0-0 & a-(-a) & b-(-b) \\ -a-a & 0-0 & c-(-c) \\ -b-b & -c-c & 0-0\end{array}\right] \)
\( =\mathrm{A}-\mathrm{A}^{\prime}=\left[\begin{array}{ccc}0 & 2 a & 2 b \\ -2 a & 0 & 2 c \\ -2 b & -2 c & 0\end{array}\right] \)
We can rewrite the above equation as'
\( \mathrm{A}-\mathrm{A}^{\prime}=(2)\left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right] \)
Therefore, \( \left(A^{\prime}-A^{\prime}\right)=2(A)( \) from equation 1\( ) \)
Now, \( \frac{1}{2}\left(A-A^{\prime}\right)=\frac{1}{2}(2(A) \)
\( =\frac{1}{2}\left(\mathrm{~A}+\mathrm{A}^{\prime}\right)=\mathrm{A} \)
Ans. \( \frac{1}{2}\left(\mathrm{~A}+\mathrm{A}^{\prime}\right)=\mathrm{A} \).

As per Theorem 2 "Any square matrix can be expressed as the sum of a symmetric and skew symmetric matrix." So in order to prove this we will be using Theorem 1 which states that "For any square matrix A with real number entries, \( \mathrm{A}+\mathrm{A} \) ' is a symmetric matrix and \( \mathrm{A}-\mathrm{A} \) ' is a skew symmetric matrix."
Now, let \( A=\left[\begin{array}{cc}3 & 5 \\ 1 & -1\end{array}\right] \)
Therefore, \( A^{\prime}=\left[\begin{array}{cc}3 & 1 \\ 5 & -1\end{array}\right] \)
Now, on adding A and A' we will get,
\( A+A^{\prime}=\left[\begin{array}{cc}3 & 5 \\ 1 & -1\end{array}\right]+\left[\begin{array}{cc}3 & 1 \\ 5 & -1\end{array}\right] \)
\( =A+A^{\prime}=\left[\begin{array}{cc}3+3 & 5+1 \\ 1+5 & -1+(-1)\end{array}\right] \)
\( =\mathrm{A}+\mathrm{A}^{\prime}=\left[\begin{array}{cc}6 & 6 \\ 6 & -2\end{array}\right] \)
Now, let \( \mathrm{M}=\frac{1}{2}\left(\mathrm{~A}+\mathrm{A}^{\prime}\right) \)
Therefore, \( M=\frac{1}{2}\left[\begin{array}{cc}6 & 6 \\ 6 & -2\end{array}\right] \)
Now, \( \mathrm{M}^{\prime}=\left[\begin{array}{cc}3 & 3 \\ 3 & -1\end{array}\right] \)
\( =M^{\prime}=M \) thus \( M=\frac{1}{2}\left(A+A^{\prime}\right) \) is a symmetric matrix as \( M^{\prime}=M N \), on subtracting A' from A we will get,
\( A-A^{\prime}=\left[\begin{array}{cc}3 & 5 \\ 1 & -1\end{array}\right]-\left[\begin{array}{cc}3 & 1 \\ 5 & -1\end{array}\right] \)
\( =A-A^{\prime}=\left[\begin{array}{cc}0 & 4 \\ -4 & 0\end{array}\right] \)
Now, let \( \mathrm{N}=\frac{1}{2}\left(\mathrm{~A}-\mathrm{A}^{\prime}\right) \)
Therefore, \( \mathrm{N}=\frac{1}{2}\left[\begin{array}{cc}0 & 4 \\ -4 & 0\end{array}\right] \)
\( =\mathrm{N}=\left[\begin{array}{cc}0 & 2 \\ -2 & 0\end{array}\right] \)
Now, \( \mathrm{N}^{\prime}=\left[\begin{array}{cc}0 & -2 \\ 2 & 0\end{array}\right] \)
\(
\begin{array}{l}
=N^{\prime}=(-1)\left[\begin{array}{cc}
0 & 2 \\
-2 & 0
\end{array}\right] \\
=N^{\prime}=-N
\end{array}
\)
Thus, \( \mathrm{M}=\frac{1}{2}\left(\mathrm{~A}^{+}+\mathrm{A}^{\prime}\right) \) is a skew symmetric matrix as \( \mathrm{N}^{\prime}=-\mathrm{N} \)
Now, Add M and N, we get,
\(
\begin{array}{l}
M+N=\left[\begin{array}{cc}
3 & 3 \\
3 & -1
\end{array}\right]+\left[\begin{array}{cc}
0 & 2 \\
-2 & 0
\end{array}\right] \\
=M+N=\left[\begin{array}{cc}
3+0 & 3+2 \\
3+(-2) & -1+0
\end{array}\right] \\
=M+N=\left[\begin{array}{cc}
3 & 5 \\
1 & -1
\end{array}\right]
\end{array}
\)
So, we see here, \( M+N=\left[\begin{array}{cc}3 & 5 \\ 1 & -1\end{array}\right]=A \)
us, A is represented as the sum of a symmetric matrix M and a skew symmetric matrix N .
Hence proved

As per Theorem 2 "Any square matrix can be expressed as the sum of a symmetric and skew symmetric matrix." So in order to prove this we will be using Theorem 1 which states that "For any square matrix A with real number entries, \( \mathrm{A}+\mathrm{A} \) ' is a symmetric matrix and \( \mathrm{A}-\mathrm{A} \) ' is a skew symmetric matrix."
Now, let \( \mathrm{A}=\left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right] \)
Therefore, \( A^{\prime}=\left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right] \)
Now, on adding A and A' we will get,
\(
\begin{array}{l}
A+A^{\prime}=\left[\begin{array}{ccc}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]+\left[\begin{array}{ccc}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right] \\
=A+A^{\prime}=\left[\begin{array}{ccc}
6+6 & -2+(-2) & 2+2 \\
-2+(-2) & 3+3 & -1+(-1) \\
2+2 & -1+(-1) & 3+3
\end{array}\right] \\
=A+A^{\prime}=\left[\begin{array}{ccc}
12 & -4 & 4 \\
-4 & 6 & -2 \\
2 & -2 & 6
\end{array}\right]
\end{array}
\)
Therefore, \( M=\frac{1}{2}\left[\begin{array}{ccc}12 & -4 & 4 \\ -4 & 6 & -2 \\ 2 & -2 & 6\end{array}\right] \)
\( =M=\left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & 1 & 3\end{array}\right] \)
Now, \( M^{\prime}=\left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & 1 & 3\end{array}\right] \)
\( =\mathrm{M}^{\prime}=\mathrm{M} \)
Thus, \( \mathrm{M}=\frac{1}{2}\left(\mathrm{A}+\mathrm{A}^{\prime}\right) \) is a symmetric matrix as \( \mathrm{M}^{\prime}=\mathrm{M} \)
Now, on subtracting A' from A we will get,
\(
\begin{array}{l}
A-A^{\prime}=\left[\begin{array}{ccc}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]-\left[\begin{array}{ccc}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right] \\
=A-A^{\prime}=\left[\begin{array}{ccc}
6-6 & -2-(-2) & 2-2 \\
-2-(-2) & 3-3 & -1-(-1) \\
2-2 & -1-(-1) & 3-3
\end{array}\right] \\
=A-A^{\prime}=\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]
\end{array}
\)
Now, let \( \mathrm{N}=\frac{1}{2}\left(-\mathrm{A}^{\prime}\right) \)
Therefore, \( N=\frac{1}{2}\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right] \)
\(
\begin{array}{l}
\mathrm{N}=\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right] \\
=\mathrm{N}^{\prime}=(-1)\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right] \\
=\mathrm{N}=-\mathrm{N}
\end{array}
\)
Thus \( \mathrm{M}=\frac{1}{2}\left(\mathrm{~A}+\mathrm{A}^{\prime}\right) \) is a skew symmetric matrix as \( \mathrm{N}^{\prime}=-\mathrm{N} \)
Now, Add M and N, we get,
\(
\begin{aligned}
M+N & =\left[\begin{array}{ccc}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]+\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right] \\
M+N & =\left[\begin{array}{ccc}
6+0 & -2+0 & 2+0 \\
-2+0 & 3+0 & -1+0 \\
2+0 & -1+0 & 3+0
\end{array}\right]
\end{aligned}
\)
\(
=M+N=\left[\begin{array}{ccc}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]
\)
So, we see here, \( M+N=\left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]=A \)
Thus, A is represented as the sum of a symmetric matrix M and a skew symmetric matrix N .

Given A and B are symmetric matrix of same order
\(
\Rightarrow \mathrm{A}=\mathrm{A}^{\prime} \rightarrow \ldots(1)\)
\(\Rightarrow \mathrm{B}=\mathrm{B}^{\prime} \rightarrow \ldots(2)\)
So, \( \mathrm{AB}-\mathrm{BA}=\mathrm{A}^{\prime} \mathrm{B}^{\prime}-\mathrm{B}^{\prime} \mathrm{A}^{\prime}( \) from \( 1 \& 2) \)
\(\Rightarrow \mathrm{AB}-\mathrm{BA}=(\mathrm{BA})^{\prime}-(\mathrm{AB})^{\prime}\left(\therefore(\mathrm{AB})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime}\right)\)
\(\Rightarrow \mathrm{AB}-\mathrm{BA}=(-1)\left((\mathrm{AB})^{\prime}-(\mathrm{BA})^{\prime}\right) \text { (taking }-1 \text { common) }\)
\(\Rightarrow \mathrm{AB}-\mathrm{BA}=-(\mathrm{AB}-\mathrm{BA})^{\prime}\left(\therefore(\mathrm{A}-\mathrm{B})^{\prime}=\mathrm{A}^{\prime}-\mathrm{B}^{\prime}\right)
\)
Here we see that the relation between \( (A B-B A) \) and its transpose i.e. \( (A B-B A)^{\prime} \) is \( (A B-B A)=-(A B-B A)^{\prime} \), this implies that \( (A B-B A) \) is a skew symmetric matrix.
Hence proved.

Given \( \mathrm{A}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right] \)
Therefore, \( A^{\prime}=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right] \)
Also given that \( \mathrm{A}+\mathrm{A}^{\prime}=\mathrm{I} \)
(Putting the values in the above equation)
\(
\begin{array}{l}
{\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]+\left[\begin{array}{cc}
\cos \alpha & \text { si } n \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]} \\
=\left[\begin{array}{cc}
\cos \alpha+\cos \alpha & -\sin \alpha+\sin \alpha \\
\sin \alpha-\sin \alpha & \cos \alpha+\cos \alpha
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
=\left[\begin{array}{cc}
2 \cos \alpha & 0 \\
0 & 2 \cos \alpha
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right]
\end{array}
\)
We know when two matrices are equal only when all their corresponding elements or entries are equal i.e. if \( \mathrm{A}=\mathrm{B} \), then \( aij =\mathrm{bij} \) for all i and j .
This implies,
\(2 \cos \alpha=1\)
\(=\cos \alpha=\frac{1}{2}\)
\( =\cos \alpha=\cos \frac{\pi}{3}\left(\therefore \cos \frac{\pi}{3}=\frac{1}{2}\right)\)
\(=\alpha=\frac{\pi}{3} \)