Class 12 Maths NCERT Solutions Chapter 3 Ex 3.3 – Transpose, Symmetric, Skew Symmetric & Invertible Matrices

The transpose of a matrix is obtained by interchanging its rows and columns.

(i) A column matrix transposes to a row matrix:

(ii)

(iii)

(i) A+B = \begin{bmatrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{bmatrix} \Rightarrow (A+B)' = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}

A' = \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix},\quad B' = \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{bmatrix} \Rightarrow A'+B' = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}

(ii) A-B = \begin{bmatrix} 3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \end{bmatrix} \Rightarrow (A-B)' = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}

A'-B' = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}

Taking the transpose of A' gives A = \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix}, and B' = \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix}.

(i) A+B = \begin{bmatrix} 2 & 1 & 1 \\ 5 & 4 & 4 \end{bmatrix} \Rightarrow (A+B)' = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}

A'+B' = \begin{bmatrix} 3-1 & 4+1 \\ -1+2 & 2+2 \\ 0+1 & 1+3 \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}

(ii) A-B = \begin{bmatrix} 4 & -3 & -1 \\ 3 & 0 & -2 \end{bmatrix} \Rightarrow (A-B)' = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}

A'-B' = \begin{bmatrix} 3+1 & 4-1 \\ -1-2 & 2-2 \\ 0-1 & 1-3 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}

Using (A+2B)' = A'+2B', where B' = \begin{bmatrix} -1 & 1 \\ 0 & 2 \end{bmatrix}:

A'+2B' = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix} + \begin{bmatrix} -2 & 2 \\ 0 & 4 \end{bmatrix}

(i) AB = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}\begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix} \Rightarrow (AB)' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}

A' = \begin{bmatrix} 1 & -4 & 3 \end{bmatrix},\ B' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \Rightarrow B'A' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}\begin{bmatrix} 1 & -4 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}

(ii) AB = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}\begin{bmatrix} 1 & 5 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{bmatrix} \Rightarrow (AB)' = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}

A' = \begin{bmatrix} 0 & 1 & 2 \end{bmatrix},\ B' = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix} \Rightarrow B'A' = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix}\begin{bmatrix} 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}

(i) A' = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}. Multiplying:

A'A = \begin{bmatrix} \cos^2\alpha+\sin^2\alpha & \cos\alpha\sin\alpha-\sin\alpha\cos\alpha \\ \sin\alpha\cos\alpha-\cos\alpha\sin\alpha & \sin^2\alpha+\cos^2\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I

(ii) A' = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}. Multiplying:

A'A = \begin{bmatrix} \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{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I

(i) Taking the transpose: A' = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix} = A.

(ii) Taking the transpose: A' = \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix} = -A.

A' = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}

(i) A+A' = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}. Its transpose is \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}, which equals A+A' itself.

(ii) A-A' = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}. Its transpose is \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = -(A-A').

Here A is already skew symmetric, so A' = \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix} = -A.

A+A' = O \Rightarrow \dfrac{1}{2}(A+A') = O

A-A' = 2A \Rightarrow \dfrac{1}{2}(A-A') = A

Every square matrix A can be written as A = P+Q, where P = \dfrac{1}{2}(A+A') is symmetric and Q = \dfrac{1}{2}(A-A') is skew symmetric.

(i) A' = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}. So P = \dfrac{1}{2}\begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} and Q = \dfrac{1}{2}\begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}.

(ii) Here A' = A already (A is already symmetric), so P = A and Q = O.

(iii) A' = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix}. So P = \dfrac{1}{2}\begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix} = \begin{bmatrix} 3 & \tfrac{1}{2} & -\tfrac{5}{2} \\ \tfrac{1}{2} & -2 & -2 \\ -\tfrac{5}{2} & -2 & 2 \end{bmatrix}

and Q = \dfrac{1}{2}\begin{bmatrix} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{bmatrix} = \begin{bmatrix} 0 & \tfrac{5}{2} & \tfrac{3}{2} \\ -\tfrac{5}{2} & 0 & 3 \\ -\tfrac{3}{2} & -3 & 0 \end{bmatrix}.

(iv) A' = \begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix}. So P = \dfrac{1}{2}\begin{bmatrix} 2 & 4 \\ 4 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix} and Q = \dfrac{1}{2}\begin{bmatrix} 0 & 6 \\ -6 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}.

Since A and B are symmetric, A' = A and B' = B. Taking the transpose of AB-BA:

(AB-BA)' = (AB)'-(BA)' = B'A'-A'B' = BA-AB = -(AB-BA)

A' = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix} \Rightarrow A+A' = \begin{bmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix}

Setting this equal to I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}: 2\cos\alpha = 1 \Rightarrow \cos\alpha = \dfrac{1}{2} \Rightarrow \alpha = \dfrac{\pi}{3}.

By definition, if A is a square matrix of order m and there exists another square matrix B of the same order such that AB = BA = I, then B is called the inverse of A (and A is the inverse of B).

It is not enough for just one of the two products to equal I or for the products to be equal to each other or to the zero matrix — both AB and BA must equal the identity matrix I.