NCERT Solutions for Class 12 Maths Chapter 3 Matrices

Q. Total number of possible matrices of order 3 × 3 with each entry 2 or 0 is :

  • (a) 9
  • (b) 27
  • (c) 81
  • (d) 512
Ans. (d) 512
 Explanation :

Number of entries in 3 × 3 matrix is 9. Since,
each entry has 2 choices, namely 2 or 0.
Therefore, number of possible matrices =
$$\underbrace{2×2×2......×2=2^9 }_{\text{9\space times}} = 512$$

$$\textbf{Q. The\space matrix\space P\space} = \begin{bmatrix} 0 & 0 & 4 \\ 0 & 4 & 0 \\ 4 & 0 & 0\end{bmatrix} \textbf{is\space not\space a\space :}$$
  • (a) square matrix
  • (b) diagonal matrix
  • (c) unit matrix
  • (d) none of these

Ans. (c) unit matrix

 Explanation :

If square matrix in which all diagonals elements are 1 and rest are 0, is called unit matrix.

Q. Which of the given values of x and y make the following pair of matrices equal : $$\begin{bmatrix} 3x+7 & 5 \\y+1 & 2-3x\end{bmatrix} = \begin{bmatrix} 0 & y-2 \\ 8 & 4 \end{bmatrix}$$

$$\text{(a)}\space x = \frac{-1}{3}, y = 7 \\ \text{(b) not possible to find}\\ \text{(c)}\space y\space =\space 7, \space \space= \space \frac{-2}{3} \\ \text{(d)} \space x=\space \frac{-1}{3}, y\space =\space \frac{-2}{3}$$

Ans. (c) Excretion

$$\textbf{Q. If}\begin{bmatrix}x+3 & z+4 & 2y-7 \\-6 & a-1 & 0 \\ b-3 & -21 & 0\end{bmatrix} = \begin{bmatrix} 0 & 6 & 3y-2 \\ -6 & -3 & 2c+2 \\ 2b+4 & -21 & 0\end{bmatrix} \textbf{find the values of a, b, c, x, y and z.} $$

Ans. As the given matrices are equal therefore, their corresponding elements must be equal.
Comparing the corresponding elements, we get
x + 3 = 0, z + 4 = 6, 2y – 7 = 3y – 2
a – 1 = – 3, 0 = 2c + 2, b – 3 = 2b + 4
Simplifying, we get
a = – 2, b = – 7, c = – 1
x = – 3, y = – 5 and z = 2.

$$\textbf{Q. Simplify : }cos\space \theta\begin{bmatrix}cos\space \theta & sin\space \theta \\-sin\space \theta & cos\space \theta\end{bmatrix} + sin \space \theta\begin{bmatrix}sin\space \theta & -cos\space \theta \\cos\space \theta & sin\space \theta\end{bmatrix} $$

Ans. We have,
$$cos\space \theta\begin{bmatrix}cos\space \theta & sin\space \theta \\-sin\space \theta & cos\space \theta\end{bmatrix} + sin \space \theta\begin{bmatrix}sin\space \theta & -cos\space \theta \\cos\space \theta & sin\space \theta\end{bmatrix}\\= \begin{bmatrix}cos^2\space \theta & sin\space \theta \space cos\space \theta \\-sin\space \theta \space cos\space \theta & cos^2\space \theta\end{bmatrix}+\begin{bmatrix}sin^2\space \theta & -sin\space \theta \space cos\space \theta \\sin\space \theta \space cos\space \theta & sin^2\space \theta\end{bmatrix} \\ = \begin{bmatrix}cos^2\space \theta +sin^2\space \theta & sin\space \theta \space cos\space \theta-sin\space \theta \space cos\space \theta \\-sin\space \theta \space cos\space \theta+sin\space \theta \space cos\space \theta & cos^2\space \theta +sin^2\space \theta\end{bmatrix} \\= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \space \space [\because sin^2\theta + cos^2 \theta = 1]$$

$$\textbf{Q. Construct a 3 × 2 matrix, whose elements are given by}\space a_{ij} \space= e^{ix}\space \textbf{sin jx.}$$

Ans. In general, let A be a matrix of order 3 × 2.
$$\therefore A = [a_{ij}]_{3×2} = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix} \space …..(i) \\ \text{Where, }\space\space a_{ij} = e^{ix}sin \space jx \\ \therefore \space \space a_{11} = e^{x}sin \space x \\ \space a_{12} = e^{x}sin\space 2\space x \\ \space a_{21} = e^{2x}sin \space x \\ \space a_{22} = e^{2x}sin \space 2x \\ \space a_{31} = e^{3x}sin \space x \\ \space a_{32} = e^{3x}sin \space 2x\\ \text{On putting all the values in equation (i), we get}\\ A\space = \space \begin{bmatrix} e^x sin\space x & e^x sin\space 2x \\ e^{2x} sin\space x & e^{2x} sin\space 2x \\ e^{3x} sin\space x & e^{3x} sin\space 2x \end{bmatrix} $$

$$\textbf{Q. Find the matrix X so that} X \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} $$

Ans. Here multiplier has 2 × 3 order matrix and product matrix has 2 × 3 order matrix.
$$\text{Let} X = \begin{bmatrix} x & y \\ a & b \end{bmatrix} \\ \text{We have}\begin{bmatrix} x & y \\ a & b \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} \\ \Rightarrow \begin{bmatrix} x+4y & 2x + 5y & 3x + 6y \\ a + 4b & 2a+5b & 3a+6b \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}\\ \text{Equating the corresponding elements, we get}\\ \text{x + 4y = – 7\space \space \space a + 4b = 2} \\ \text{2x + 5y = – 8\space and\space 2a + 5b = 4}\\ \text{3x + 6y = – 9\space\space\space 3a + 6b = 6}\\ \text{On solving these equations, we get }\\ \text{x = 1, y = – 2}\\ \text{a = 2, b = 0} \\ \Rightarrow \space \space X = \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix} $$

$$ \textbf{Q. If A = } \begin{bmatrix}0 & -tan \alpha/2 \\ tan\space \alpha/2 & 0 \end{bmatrix} \textbf{and I is a 2 × 2 unit matrix, then prove that } \\ \textbf{I + A = (I – A) } \begin{bmatrix}cos \space \alpha & -sin\space \alpha \\ sin\space \alpha & cos\space \alpha \end{bmatrix} $$

$$\textbf{Ans. }\text{L.H.S. = I + A}\\ = \begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix} + \begin{bmatrix}0 & -tan \alpha/2 \\ tan\space \alpha/2 & 0 \end{bmatrix}\\ = \begin{bmatrix}1 & -tan \alpha/2 \\ tan\space \alpha/2 & 1 \end{bmatrix} \\ \\ R.H.S. = (I -A)\begin{bmatrix}cos\space \alpha & -sin \space \alpha \\ sin \space \alpha & cos \space \alpha \end{bmatrix} \\ = \left(\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix}0 & -tan\space \alpha /2 \\ tan\space \alpha /2 & 0 \end{bmatrix}\right)\begin{bmatrix}cos\space \alpha & -sin \space \alpha \\ sin \space \alpha & cos \space \alpha \end{bmatrix} \\ = \begin{bmatrix}1 & tan\space \alpha /2 \\ -tan\space \alpha /2 & 1 \end{bmatrix} \begin{bmatrix}cos\space \alpha & -sin \space \alpha \\ sin \space \alpha & cos \space \alpha \end{bmatrix}\\ = \begin{bmatrix}cos\space \alpha + tan \space \alpha/2 sin\space \alpha & – sin\space \alpha + tan\space \alpha/2 cos\space \alpha \\ -tan\space \alpha/2 \space cos\space \alpha + sin\space \alpha & tan\space \alpha/2 \space sin\space \alpha + cos\space \alpha \end{bmatrix} \\ = \begin{bmatrix}cos\space \alpha + \frac{sin\space \alpha/2}{cos\space \alpha/2}sin\space \alpha & -sin\space \alpha + \frac{sin\space \alpha/2}{cos\space \alpha/2}cos\space \alpha \\ \frac{-sin\space \alpha/2}{cos\space \alpha/2}cos\space \alpha + sin\space \alpha & \frac{sin\space \alpha/2}{cos\space \alpha/2}sin\space \alpha + cos\space \alpha \end{bmatrix} \\ = \begin{bmatrix}\frac{cos\space \alpha\space cos\space \alpha/2+ sin\space \alpha/2\space sin\space \alpha}{cos\space \alpha /2} \\ \frac{-sin\space \alpha/2\space cos\space \alpha+ sin\space \alpha\space cos\space \alpha/2}{cos\space \alpha / 2} \end{bmatrix}$$

$$\textbf{ Q. \space If}\space A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{bmatrix}, \text{prove that:} \\ A^n = \begin{bmatrix}3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{bmatrix}, n\space \epsilon\space N\space \text{by using mathematical induction.} $$

$$\textbf{Ans. }\text{Here}, \space\space A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{bmatrix} \\ \text{Let\space\space P(n):} A^n = \begin{bmatrix}3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{bmatrix} \\ \text{On putting n = 1, we get} \\ \text{ P(1):} A^1 = \begin{bmatrix}3^{1-1} & 3^{1-1} & 3^{1-1} \\ 3^{1-1} & 3^{1-1} & 3^{1-1} \\ 3^{1-1} & 3^{1-1} & 3^{1-1} \end{bmatrix} \\= \begin{bmatrix}3^{0} & 3^{0} & 3^{0} \\ 3^{0} & 3^{0} & 3^{0} \\ 3^{0} & 3^{0} & 3^{0} \end{bmatrix}\\ = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}\space\space ….(i) \\ \text{which is true for n = 1.} \\ \text{Let the result be true for n = K.} \\ \text{Then \space \space P(K) :} A^k = \begin{bmatrix} 3^{K-1} & 3^{K-1} & 3^{K-1} \\ 3^{K-1} & 3^{K-1} & 3^{K-1} \\ 3^{K-1} & 3^{K-1} & 3^{K-1} \end{bmatrix} \space \space …(ii) \\ \text{On putting n = K + 1, we get} \\ P(K + 1): A^{K+1} = \begin{bmatrix} 3^K & 3^K & 3^K \\ 3^K & 3^K & 3^K \\3^K & 3^K & 3^K \end{bmatrix} \\ \text{Now,} \space \space L.H.S = A^{K+1} = A^K. A^1 \\ = \begin{bmatrix} 3^{K-1} & 3^{K-1} & 3^{K-1} \\ 3^{K-1} & 3^{K-1} & 3^{K-1} \\ 3^{K-1} & 3^{K-1} & 3^{K-1} \end{bmatrix}\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} \\ \text{[using equations (i) and (ii)]} \\ = \begin{bmatrix} 3^{K-1}+3^{K-1}+3^{K-1} & 3^{K-1}+3^{K-1}+3^{K-1} & 3^{K-1}+3^{K-1}+3^{K-1} \\ 3^{K-1}+3^{K-1}+3^{K-1} & 3^{K-1}+3^{K-1}+3^{K-1} & 3^{K-1}+3^{K-1}+3^{K-1} \\3^{K-1}+3^{K-1}+3^{K-1} & 3^{K-1}+3^{K-1}+3^{K-1} & 3^{K-1}+3^{K-1}+3^{K-1} \end{bmatrix} \\ = \begin{bmatrix} 3×3^{K-1} & 3×3^{K-1} & 3×3^{K-1}\\ 3×3^{K-1} & 3×3^{K-1} & 3×3^{K-1} \\ 3×3^{K-1} & 3×3^{K-1} & 3×3^{K-1} \end{bmatrix} \\ = \begin{bmatrix} 3^K & 3^K & 3^K \\ 3^K & 3^K & 3^K \\ 3^K & 3^K & 3^K \end{bmatrix} = R.H.S \\ \text{Therefore, the result is true for n = K + 1, whenever it is true for n = K. So, by principle of mathematical induction, it is true for all n ∈ N.}$$