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Q5(i):
Construct a $3 \times 4$ matrix, whose elements are given by: (i) $a_{ij} = \frac{1}{2}|-3i + j|$
Solution :
Step 1: Theoretical Foundation of the Matrix
A $3 \times 4$ matrix $A$ consists of $3$ rows and $4$ columns. The general representation of this matrix is:
$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \end{bmatrix}$
Each element $a_{ij}$ is determined by the mapping function $a_{ij} = \frac{1}{2}|-3i + j|$, where $i \in \{1, 2, 3\}$ and $j \in \{1, 2, 3, 4\}$.
Step 2: Systematic Calculation of Elements
We evaluate the elements row by row using the defined modulus function [Definition of Absolute Value: $|x| = x$ if $x \ge 0$, and $-x$ if $x < 0$].
| Row ($i$) | $j=1$ | $j=2$ | $j=3$ | $j=4$ |
|---|---|---|---|---|
| $i=1$ | $\frac{1}{2}|-3(1)+1| = 1$ | $\frac{1}{2}|-3(1)+2| = 0.5$ | $\frac{1}{2}|-3(1)+3| = 0$ | $\frac{1}{2}|-3(1)+4| = 0.5$ |
| $i=2$ | $\frac{1}{2}|-3(2)+1| = 2.5$ | $\frac{1}{2}|-3(2)+2| = 2$ | $\frac{1}{2}|-3(2)+3| = 1.5$ | $\frac{1}{2}|-3(2)+4| = 1$ |
| $i=3$ | $\frac{1}{2}|-3(3)+1| = 4$ | $\frac{1}{2}|-3(3)+2| = 3.5$ | $\frac{1}{2}|-3(3)+3| = 3$ | $\frac{1}{2}|-3(3)+4| = 2.5$ |
Step 3: Matrix Assembly
Substituting the calculated values into the matrix structure defined in Step 1:
$A = \begin{bmatrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2} \end{bmatrix}$
Step 4: Geometric Representation of Matrix Indices
Final Solution: $A = \begin{bmatrix} 1 & 0.5 & 0 & 0.5 \\ 2.5 & 2 & 1.5 & 1 \\ 4 & 3.5 & 3 & 2.5 \end{bmatrix}$
More Questions from Class 12 Mathematics Matrices EXERCISE 3.1
- Q1(i): In the matrix $A = \begin{bmatrix} 2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17 \end{bmatrix}$, write: (i) The order of the matrix
- Q1(ii): In the matrix $A = \begin{bmatrix} 2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17 \end{bmatrix}$, write: (ii) The number of elements
- Q1(iii): In the matrix $A = \begin{bmatrix} 2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17 \end{bmatrix}$, write: (iii) Write the elements $a_{13}, a_{21}, a_{33}, a_{24}, a_{23}$.
- Q2: If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
- Q3: If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
- Q4(i): Construct a $2 \times 2$ matrix, $A = [a_{ij}]$, whose elements are given by: (i) $a_{ij} = \frac{(i+j)^2}{2}$
- Q4(ii): Construct a $2 \times 2$ matrix, $A = [a_{ij}]$, whose elements are given by: (ii) $a_{ij} = \frac{i}{j}$
- Q4(iii): Construct a $2 \times 2$ matrix, $A = [a_{ij}]$, whose elements are given by: (iii) $a_{ij} = \frac{(i+2j)^2}{2}$
- Q5(ii): Construct a $3 \times 4$ matrix, whose elements are given by: (ii) $a_{ij} = 2i-j$
- Q6(i): Find the values of $x, y$ and $z$ from the following equations: (i) $\begin{bmatrix} 4 & 3 \\ x & 5 \end{bmatrix} = \begin{bmatrix} y & z \\ 1 & 5 \end{bmatrix}$
- Q6(ii): Find the values of $x, y$ and $z$ from the following equations: (ii) $\begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}$
- Q6(iii): Find the values of $x, y$ and $z$ from the following equations: (iii) $\begin{bmatrix} x+y+z \\ x+z \\ y+z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}$
- Q7: Find the value of $a, b, c$ and $d$ from the equation: $\begin{bmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}$
- Q8: $A = [a_{ij}]_{m \times n}$ is a square matrix, if
- Q9: Which of the given values of $x$ and $y$ make the following pair of matrices equal $\begin{bmatrix} 3x+7 & 5 \\ y+1 & 2-3x \end{bmatrix}$, $\begin{bmatrix} 0 & y-2 \\ 8 & 4 \end{bmatrix}$
- Q10: The number of all possible matrices of order $3 \times 3$ with each entry 0 or 1 is:
CBSE Solutions for Class 12 Mathematics Matrices
Chapters in CBSE - Class 12 Mathematics
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