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Matrices - Order of a matrix

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A matrix is an ordered rectangular array of numbers or functions. The numbers or functions are called the elements or the entries of the matrix.

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The horizontal lines of elements are said to constitute the rows of the matrix, denoted by mm.

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The vertical lines of elements are said to constitute the columns of the matrix, denoted by nn.

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A matrix having mm rows and nn columns is called a matrix of order m×nm \times n (read as mm by nn).

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In general, an m×nm \times n matrix has the following rectangular array: A=[a11a12…a1na21a22…a2n⋮⋮⋱⋮am1am2…amn]A = \begin{bmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \dots & a_{mn} \end{bmatrix}

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The element aija_{ij} represents the entry located in the ithi^{th} row and jthj^{th} column.

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The total number of elements in a matrix of order m×nm \times n is the product mnmn.

📐Formulae

Order of a Matrix=Number of rows×Number of columns\text{Order of a Matrix} = \text{Number of rows} \times \text{Number of columns}

Total number of elements=m⋅n\text{Total number of elements} = m \cdot n

A=[aij]m×n, where 1≤i≤m and 1≤j≤nA = [a_{ij}]_{m \times n}, \text{ where } 1 \leq i \leq m \text{ and } 1 \leq j \leq n

💡Examples

Problem 1:

Consider the matrix A=[35−1204]A = \begin{bmatrix} 3 & 5 & -1 \\ 2 & 0 & 4 \end{bmatrix}. State the order of the matrix and the total number of elements.

Solution:

The matrix AA has 22 rows and 33 columns. Therefore, its order is 2×32 \times 3. The total number of elements is 2×3=62 \times 3 = 6.

Explanation:

We count the horizontal rows first (2) and then the vertical columns (3) to determine the order m×nm \times n.

Problem 2:

If a matrix has 88 elements, what are the possible orders it can have?

Solution:

The possible orders are the pairs of natural numbers (m,n)(m, n) such that m×n=8m \times n = 8. These pairs are (1,8),(8,1),(2,4), and (4,2)(1, 8), (8, 1), (2, 4), \text{ and } (4, 2). Thus, the possible orders are 1×8,8×1,2×4, and 4×21 \times 8, 8 \times 1, 2 \times 4, \text{ and } 4 \times 2.

Explanation:

To find all possible orders, we look for all possible factor pairs of the total number of elements.

Problem 3:

Construct a 2×22 \times 2 matrix A=[aij]A = [a_{ij}], where the elements are given by aij=(i+j)22a_{ij} = \frac{(i+j)^2}{2}.

Solution:

For a 2×22 \times 2 matrix, i∈{1,2}i \in \{1, 2\} and j∈{1,2}j \in \{1, 2\}. a11=(1+1)22=42=2a_{11} = \frac{(1+1)^2}{2} = \frac{4}{2} = 2 a12=(1+2)22=92a_{12} = \frac{(1+2)^2}{2} = \frac{9}{2} a21=(2+1)22=92a_{21} = \frac{(2+1)^2}{2} = \frac{9}{2} a22=(2+2)22=162=8a_{22} = \frac{(2+2)^2}{2} = \frac{16}{2} = 8 Thus, A=[292928]A = \begin{bmatrix} 2 & \frac{9}{2} \\ \frac{9}{2} & 8 \end{bmatrix}.

Explanation:

Each element aija_{ij} is calculated by substituting its row index ii and column index jj into the given formula.