Maharashtra Board Class 12 Maths Part 1 Chapter 2 Matrices PDF Download

Matrices

Let's Study

2.1 Elementary Transformations

2.2 Inverse Of A Matrix

2.2.1 Elementary Transformation Method

2.2.2 Adjoint Method

2.3 Application Of Matrices

Solution Of A System Of Linear Equations

2.3.1 Method Of Inversion

2.3.2 Method Of Reduction

Let's Learn

A matrix of order m×m is a square arrangement of m² elements. The corresponding determinant of the same elements, after expansion is seen to be a value which is an element itself.

In standard XI, we have studied the types of matrices and algebra of matrices namely addition, subtraction, multiplication of two matrices.

The matrices are useful in almost every branch of science. Many problems in Statistics are expressed in terms of matrices. Matrices are also useful in Economics, Operation Research. It would not be an exaggeration to say that the matrices are the language of atomic Physics.

Hence, it is necessary to learn the uses of matrices with the help of elementary transformations and the inverse of a matrix.

2.1 Elementary Transformation

Let us first understand the meaning and applications of elementary transformations.

The elementary transformation of a matrix are the six operations, three of which are due to row and three are due to column.

They are as follows:

(a) Interchange Of Any Two Rows Or Any Two Columns

If we interchange the ith row and the jth row of a matrix then after this interchange the original matrix is transformed to a new matrix.

This transformation is symbolically denoted as \(R_i \leftrightarrow R_j\) or \(R_{ij}\).

The similar transformation can be due to two columns say \(C_k \leftrightarrow C_i\) or \(C_{ki}\).

(Recall that R and C symbolically represent the rows and columns of a matrix.)

For example, if \(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\) then \(R_1 \leftrightarrow R_2\) gives the new matrix \(\begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix}\) and \(C_1 \leftrightarrow C_2\) gives the new matrix \(\begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}\).

Note that \(A \neq \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix}\) and \(\neq \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}\) but we write \(A \sim \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix}\) and \(A \sim \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}\)

Note: The symbol ~ is read as equivalent to.

Teacher's Note

Elementary transformations help us change the shape of a matrix. Think of it like rearranging rows and columns in a classroom seating arrangement.

Exam Trick

Remember: When you swap rows or columns, use the ~ symbol to show the matrix is equivalent, not equal. Just like how rearranging students doesn't change who they are!

Points to Remember

Elementary transformations have three main types for rows and three for columns.
Swapping rows or columns is the first type of transformation.
The ~ symbol means the matrices are equivalent after transformation.
The original matrix element values stay the same, only their positions change.
These transformations are useful for solving systems of equations.

(b) Multiplication Of The Elements Of Any Row Or Column By A Non-Zero Scalar

If k is a non-zero scalar and the row Ri is to be multiplied by constant k then we multiply every element of Ri by the constant k and symbolically the transformation is denoted by kRi or \(R_i \to kR_i\).

For example, if \(A = \begin{bmatrix} 0 & 2 \\ 3 & 4 \end{bmatrix}\) then \(R_2 \to 4R_2\) gives \(A \sim \begin{bmatrix} 0 & 2 \\ 12 & 16 \end{bmatrix}\)

Similarly, if any column of a matrix is to be multiplied by a constant then we multiply every element of the column by the constant. It is denoted as kCl or \(C_l \to kC_l\).

For example, if \(A = \begin{bmatrix} 0 & 2 \\ 3 & 4 \end{bmatrix}\) then \(C_1 \to -3C_1\) gives \(A \sim \begin{bmatrix} 0 & 2 \\ -9 & 4 \end{bmatrix}\)

Can you say that \(A = \begin{bmatrix} 0 & 2 \\ 12 & 16 \end{bmatrix}\) or \(A = \begin{bmatrix} 0 & 2 \\ -9 & 4 \end{bmatrix}\)?

Teacher's Note

When you multiply a row or column by a number, every number in that row or column gets multiplied. This is like multiplying every student's score in a row by 2.

Exam Trick

Remember: Multiply ALL elements in the row or column by the same number k. If k is outside, put it in front of every element inside that row.

Points to Remember

k must be a non-zero number.
Every element in the chosen row or column is multiplied by k.
The matrix is still equivalent, not equal, to the original.
This transformation changes the numbers but keeps the matrix structure.
We write the transformation as \(R_i \to kR_i\) or \(C_i \to kC_i\).

(c) Adding The Scalar Multiples Of All The Elements Of Any Row (Column) To Corresponding Elements Of Any Other Row (Column)

If k is a non-zero scalar and the k-multiples of the elements of Ri (Ci) are to be added to the elements of Rj (Cj) then the transformation is symbolically denoted as \(R_j \to R_j + kR_i\), \(C_j \to C_j + kC_i\)

For example, if \(A = \begin{bmatrix} -1 & 4 \\ 2 & 5 \end{bmatrix}\) and k = 2 then \(R_1 \to R_1 + 2R_2\) gives \(A \sim \begin{bmatrix} -1 + 2(2) & 4 + 2(5) \\ 2 & 5 \end{bmatrix}\) i.e. \(A \sim \begin{bmatrix} 3 & 14 \\ 2 & 5 \end{bmatrix}\)

(Can you find the transformation of A using \(C_2 \to C_2 + (-3)C_1\)?)

Note (1): After the transformation, \(R_j \to R_j + kR_i\), Ri remains the same as in the original matrix. Similarly, with the transformation, \(C_j \to C_j + kC_i\), Ci remains the same as in the original matrix.

Note (2): After the elementary transformation, the matrix obtained is said to be equivalent to the original matrix.

Ex. 1: If \(A = \begin{bmatrix} 1 & 0 \\ -1 & 3 \end{bmatrix}\), apply the transformation \(R_1 \leftrightarrow R_2\) on A.

Solution:

As \(A = \begin{bmatrix} 1 & 0 \\ -1 & 3 \end{bmatrix}\)

\(R_1 \leftrightarrow R_2\) gives \(A \sim \begin{bmatrix} -1 & 3 \\ 1 & 0 \end{bmatrix}\)

Ex. 2: If \(A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 3 & 4 \end{bmatrix}\), apply the transformation \(C_1 \to C_1 + 2C_3\).

Solution:

\(A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 3 & 4 \end{bmatrix}\)

\(C_1 \to C_1 + 2C_3\) gives \(A \sim \begin{bmatrix} 1 + 2(2) & 0 & 2 \\ 2 + 2(4) & 3 & 4 \end{bmatrix}\) \(A \sim \begin{bmatrix} 5 & 0 & 2 \\ 10 & 3 & 4 \end{bmatrix}\)

Ex. 3: If \(A = \begin{bmatrix} 1 & 2 & -1 \\ -3 & -2 & 5 \end{bmatrix}\), apply \(R_1 \leftrightarrow R_2\) and then \(C_1 \to C_1 + 2C_3\) on A.

Solution:

\(A = \begin{bmatrix} 1 & 2 & -1 \\ -3 & -2 & 5 \end{bmatrix}\)

Teacher's Note

Adding a multiple of one row to another row is like combining information from different groups of students. We add information, not replace it.

Exam Trick

Remember: When you write \(R_j \to R_j + kR_i\), the original row i stays the same. Only row j changes. This is very important for exams!

Points to Remember

k can be any non-zero number, positive or negative.
The row or column you add to changes; the other stays the same.
You add k times ALL elements of one row to corresponding elements of another row.
The transformation changes one row while keeping the other unchanged.
Always use the arrow notation to show which row or column changed.

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