Class 12 Maths NCERT Solutions Chapter 3 Miscellaneous Exercise – Matrices

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)

Case 1: A is symmetric (A' = A). Then

(B'AB)' = B'A'(B')' = B'A'B = B'AB

Case 2: A is skew symmetric (A' = -A). Then

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

A' = \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix}

Multiplying row by column, every off-diagonal entry of A'A cancels out (for example, entry (1,2) = 0(2y)+x(y)+x(-y) = 0), while the diagonal entries are:

(A'A)_{11} = 0^2+x^2+x^2 = 2x^2,\quad (A'A)_{22} = (2y)^2+y^2+(-y)^2 = 6y^2,\quad (A'A)_{33} = z^2+(-z)^2+z^2 = 3z^2

So A'A = \begin{bmatrix} 2x^2 & 0 & 0 \\ 0 & 6y^2 & 0 \\ 0 & 0 & 3z^2 \end{bmatrix}. Setting this equal to I:

2x^2=1 \Rightarrow x=\pm\dfrac{1}{\sqrt{2}},\qquad 6y^2=1 \Rightarrow y=\pm\dfrac{1}{\sqrt{6}},\qquad 3z^2=1 \Rightarrow z=\pm\dfrac{1}{\sqrt{3}}

First multiply the matrix by the column vector: \begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = \begin{bmatrix} 0+4+0 \\ 0+0+x \\ 0+0+2x \end{bmatrix} = \begin{bmatrix} 4 \\ x \\ 2x \end{bmatrix}

Now multiply by the row vector: \begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\begin{bmatrix} 4 \\ x \\ 2x \end{bmatrix} = 4+2x+2x = 4+4x

Setting this equal to 0: 4+4x=0 \Rightarrow x=-1

A^2 = A\cdot A = \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}

5A = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix},\qquad 7I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}

A^2-5A+7I = \begin{bmatrix} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

First multiply the matrix by the column vector: \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}\begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = \begin{bmatrix} x+2 \\ 9 \\ 2x+3 \end{bmatrix}

Now multiply by the row vector: \begin{bmatrix} x & -5 & -1 \end{bmatrix}\begin{bmatrix} x+2 \\ 9 \\ 2x+3 \end{bmatrix} = x(x+2)-5(9)-1(2x+3) = x^2+2x-45-2x-3 = x^2-48

Setting this equal to 0: x^2-48=0 \Rightarrow x^2=48 \Rightarrow x=\pm4\sqrt{3}

(a) Revenue. Writing the sales as a matrix and prices as a column matrix and multiplying:

\begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix}\begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix} = \begin{bmatrix} 25000+3000+18000 \\ 15000+30000+8000 \end{bmatrix} = \begin{bmatrix} 46000 \\ 53000 \end{bmatrix}

(b) Gross profit. Total cost in each market, using the same sales matrix with the cost column instead:

\begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix}\begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix} = \begin{bmatrix} 20000+2000+9000 \\ 12000+20000+4000 \end{bmatrix} = \begin{bmatrix} 31000 \\ 36000 \end{bmatrix}

Gross profit = Revenue - Cost:

Market I: 46000-31000=15000; Market II: 53000-36000=17000

Since the given matrix is 2\times3 and the product is 2\times3, X must be a 2\times2 matrix. Let X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}. Then

X\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} a+4b & 2a+5b & 3a+6b \\ c+4d & 2c+5d & 3c+6d \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}

From the first row: a+4b=-7 and a+2b=-3 (dividing the third entry by 3). Subtracting gives 2b=-4 \Rightarrow b=-2, so a=-3-2(-2)=1.

From the second row: c+4d=2 and c+2d=2 (dividing the third entry by 3). Subtracting gives 2d=0 \Rightarrow d=0, so c=2.

A^2 = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}\begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} = \begin{bmatrix} \alpha^2+\beta\gamma & \alpha\beta-\alpha\beta \\ \gamma\alpha-\alpha\gamma & \gamma\beta+\alpha^2 \end{bmatrix} = \begin{bmatrix} \alpha^2+\beta\gamma & 0 \\ 0 & \alpha^2+\beta\gamma \end{bmatrix}

Setting A^2 = I: \alpha^2+\beta\gamma=1 \Rightarrow 1-\alpha^2-\beta\gamma=0

If A is symmetric, A' = A. If A is also skew symmetric, A' = -A. Combining these: A = -A \Rightarrow 2A = O \Rightarrow A = O.

Expanding (I+A)^3 using the binomial expansion (valid since I commutes with every matrix):

(I+A)^3 = I+3A+3A^2+A^3

Since A^2=A, it follows that A^3 = A^2\cdot A = A\cdot A = A^2 = A. Substituting:

(I+A)^3 = I+3A+3A+A = I+7A

So (I+A)^3-7A = I+7A-7A = I