Unit – I – ALGEBRA

$\LARGE{\color {red}{CHAPTER\ 1.1:\ MATRICES\ AND\ DETERMINANTS (Text)}}$

Matrix and its applications are very important part of Mathematics. Also it is one of the most powerful tools of Mathematics.

Matrix notation and operation are used in electronic spread sheet programmes for personal computer, business budgeting, sales projection, cost estimation, analyzing the results of an experiment etc. Also many physical operations such as magnification, rotation and reflection through a plane can be represented mathematically by matrices. Also matrix used in Cryptography.

$\color {royalblue} {Definition\ of\ a\ Matrix}:\ \hspace{20cm}$

A matrix is a rectangular array of numbers arranged in rows and columns enclosed by brackets.

$\color {royalblue} {Ex:}\ 1)\ A =\begin{bmatrix} 3 & 6 \\ 1 & 2\\ \end{bmatrix}\ \hspace{2cm}\ 2)\ B =\begin{bmatrix} 1 & -1 & 2\\ 2 & -2 & 4\\ 3 & -3 & 6\\ \end{bmatrix}\ \hspace{10cm}$
$\color {royalblue} {Order\ of\ a\ Matrix}:\ \hspace{20cm}$

If there are  m rows and n columns in a matrix, then the order of  the matrix is  m × n.

$\color {royalblue} {Ex:}\ A =\begin{bmatrix} 1 & 2 & -1 & 3\\ 2 & 4 & -4 & 7\\ -1 & -2 & -2 & -2\\ \end{bmatrix}\ is\ a\ matrix\ of\ order\ 3×4 \hspace{10cm}$
$\color {royalblue} {Type\ of\ Matrix}:\ \hspace{20cm}$
$\color {green} {1.\ Row\ Matrix}:\ \hspace{20cm}$

A matrix having only one row and any number of columns is called a row matrix.

$\color {black} {Eg:}\ A =\begin{bmatrix} 1 & 2 & -3 \\ \end{bmatrix}\ is\ a\ matrix\ of\ order\ 1×3 \hspace{20cm}$
$\color {green} {2.\ Column\ Matrix}:\ \hspace{20cm}$

A matrix having only one column and any number of rows is called a column matrix.

$\color {black} {Eg:}\ A =\begin{bmatrix} 3 \\ -1\\ 5\\ \end{bmatrix}\ is\ a\ matrix\ of\ order\ 3×1 \hspace{20cm}$
$\color {green} {3.\ Null\ or\ zero\ Matrix}:\ \hspace{20cm}$

If all the elements of a matrix are zero, the matrix is called zero or null matrix.

$\color {black} {Eg:}\ 0 =\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}\ is\ a\ zero\ matrix\ of\ order\ 3×3 \hspace{20cm}$
$\color {green} {4.\ Square\ Matrix}:\ \hspace{20cm}$

A matrix which has equal number of rows and columns is called a square matrix.

$\color {black} {Eg:}\ 1)\ A =\begin{bmatrix} 3 & 6 \\ 1 & 4 \\ \end{bmatrix}\ is\ a\ square\ matrix\ of\ order\ 2×2 \hspace{20cm}$
$\color {black} {Eg:}\ 2)\ B =\begin{bmatrix} 1 & -1 & 2 \\ 2 & -2 & 4 \\ 3 & -3 & 6 \\ \end{bmatrix}\ is\ a\ square\ matrix\ of\ order\ 3×3 \hspace{20cm}$
$\color {green} {5.\ Triangular\ Matrix}:\ \hspace{20cm}$
$\color {black} {(a)\ Upper\ Triangular\ Matrix}:\ \hspace{10cm}$

In a square matrix if all the elements below the leading diagonal are zero is called Upper Triangular Matrix.

$\color {black} {Eg:}\ A =\begin{bmatrix} 1 & 4 & 2 \\ 0 & 3 & 4 \\ 0 & 0 & 1 \\ \end{bmatrix}\ \hspace{5cm}$
$\color {black} {(b)\ Lower\ Triangular\ Matrix}:\ \hspace{10cm}$

In a square matrix if all the elements above the leading diagonal are zero is called Lower Triangular Matrix.

$\color {black} {Eg:}\ B =\begin{bmatrix} 1 & 0 & 0 \\ 2 & 8 & 0 \\ 4 & 9 & 7 \\ \end{bmatrix}\ \hspace{5cm}$
$\color {green} {6.\ Transpose\ of\ Matrix}:\ \hspace{20cm}$

Let A be a square matrix. The transpose of A is obtained by changing rows into columns and vise-versa and is denoted by  AT

$\color {black} {Eg:}\ 1)\ A =\begin{bmatrix} 3 & 6 \\ 1 & 4 \\ \end{bmatrix}\ \hspace{10cm}$$A^T =\begin{bmatrix} 3 & 6 \\ 1 & 4 \\ \end{bmatrix}\ \hspace{8cm}$
$\color {black} {Eg:}\ 2)\ A =\begin{bmatrix} 3 & 4 & 1 \\ 0 & -1 & 2 \\ 5 & -2 & 6 \\ \end{bmatrix}\ \hspace{10cm}$$A^T =\begin{bmatrix} 3 & 0 & 5 \\ 4 & -1 & -2 \\ 1 & 2 & 6 \\ \end{bmatrix}\ \hspace{8cm}$
$\color {green} {7.\ Unit\ Matrix}:\ \hspace{20cm}$

Unit matrix is a square matrix in which the diagonal elements are all ones and all the other elements are zeros.

$\color {black} {Eg:}\ 1)\ I =\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}\ is\ a\ unit\ matrix\ of\ order\ 2×2 \hspace{10cm}$
$2)\ I =\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}\ is\ a\ unit\ matrix\ of\ order\ 3×3 \hspace{8cm}$
$\color {purple} {Operation\ on\ Matrices:}\ \hspace{20cm}$

i) Addition and subtraction of matrices
ii) Multiplication of matrix by a scalar
iii) Multiplication of matrices

$\color {purple} {i)\ Addition\ and\ Subtraction\ of\ Matrices:}\ \hspace{20cm}$

Two Matrices can be added (or) subtracted if they have the same order. We can add (or) subtract two matrices by the corresponding element by element.

$\color {black} {Example\ 1:}\ If\ A =\begin{bmatrix} 1 & 2 & 7 \\ 0 & 4 & 5 \\ 3 & 1 & 6 \\ \end{bmatrix}\ ,\ B =\begin{bmatrix} 1 & 3 & 1 \\ 2 & 4 & 0 \\ 1 & 7 & 5 \\ \end{bmatrix}\ \hspace{5cm}$
$Find\ A + B\ \hspace{10cm}$
$A + B=\begin{bmatrix} 1 + 1 & 2 + 3 & 7 + 1 \\ 0 + 2 & 4 + 4 & 5 + 0 \\ 3 + 1 & 1 + 7 & 6 + 5 \\ \end{bmatrix}\ \hspace{6cm}$
$A + B=\begin{bmatrix} 2 & 5 & 8 \\ 2 & 8 & 5 \\ 4 & 8 & 11 \\ \end{bmatrix}\ \hspace{7cm}$
$\color {black} {Example\ 2:}\ If\ A =\begin{bmatrix} 1 & 3 & 5 \\ 2 & 0 & 7 \\ 1 & 5 & 2 \\ \end{bmatrix}\ ,\ B =\begin{bmatrix} 7 & 3 & 4 \\ 1 & -1 & 5 \\ 0 & 2 & 4 \\ \end{bmatrix}\ \hspace{5cm}$
$Find\ A – B\ \hspace{10cm}$
$A – B=\begin{bmatrix} 1 – 7 & 3 – 3 & 5 – 4 \\ 2 – 1 & 0 + 1 & 7 – 5 \\ 1 – 0 & 5 – 2 & 2 – 4 \\ \end{bmatrix}\ \hspace{6cm}$
$A – B=\begin{bmatrix} -6 & 0 & 1 \\ 1 & 1& 2 \\ 1 & 3 & -2 \\ \end{bmatrix}\ \hspace{7cm}$
$\color {purple} {ii)\ Matrix\ Multiplication\ by\ a\ scalar}\ \hspace{20cm}$

We can multiply the matrix by any non-zero scalar [(value) number] obtain we get the matrix whose all the elements are multiplied by that same scalar.

$\color {black} {Example\ 1:}\ If\ A =\begin{bmatrix} 4 & 3 & 2\\ 5 & 1 & 0 \\ 7 & 2 & 8 \\ \end{bmatrix}\ and\ B =\begin{bmatrix} -3 & 1 & 0 \\ 2 & 7 & 1 \\ 4 & 3 & 5\\ \end{bmatrix}\ \hspace{2cm}$
$Then\ Find\ 2A\ and\ 7B\ \hspace{5cm}$
$\color {black}{Solution:}\ \hspace{15cm}$
$2A = 2 \begin{bmatrix} 4 & 3 & 2\\ 5 & 1 & 0 \\ 7 & 2 & 8 \\ \end{bmatrix}\ \hspace{10cm}$
$= \begin{bmatrix} 2 × 4 & 2 × 3 & 2 × 2\\ 2 × 5 & 2 × 1 & 2 × 0 \\ 2 × 7 & 2 × 2 & 2 × 8 \\ \end{bmatrix}\ \hspace{10cm}$
$\therefore\ 2A= \begin{bmatrix} 8 & 6 & 4\\ 10 & 2 & 0 \\ 14 & 4 & 16 \\ \end{bmatrix}\ \hspace{12cm}$
$7B = 7 \begin{bmatrix} -3 & 1 & 0\\ 2 & 7 & 1 \\ 4 & 3 & 5 \\ \end{bmatrix}\ \hspace{10cm}$
$= \begin{bmatrix} 7 × -3 & 7 × 1 & 7 × 0\\ 7 × 2 & 7 × 7 & 7 × 1 \\ 7× 4 & 7 × 3 & 7 × 5 \\ \end{bmatrix}\ \hspace{10cm}$
$\therefore\ 7B= \begin{bmatrix} -21 & 7 & 0\\ 14 & 49 & 7 \\ 28 & 21 & 35 \\ \end{bmatrix}\ \hspace{12cm}$
$\color {black} {Example\ 2:}\ If\ A =\begin{bmatrix} 1 & 2 \\ 3 & 5\\ \end{bmatrix}\ and\ B =\begin{bmatrix} -5 & 7 \\ 0 & 4\\ \end{bmatrix}\ \hspace{15cm}$
$Then\ Find\ 4A\ -\ 2B\ \hspace{10cm}$
$\color {black}{Solution:}\ \hspace{18cm}$
$4A = 4\begin{bmatrix} 1 & 2 \\ 3 & 5\\ \end{bmatrix}\ \hspace{13cm}$
$4A = \begin{bmatrix} 4 & 8 \\ 12 & 20\\ \end{bmatrix}\ ———- (1)\ \hspace{8cm}$
$2B = 2\begin{bmatrix} -5 & 7 \\ 0 & 4\\ \end{bmatrix}\ \hspace{13cm}$
$2B = \begin{bmatrix} -10 & 14 \\ 0 & 8\\ \end{bmatrix}\ ———- (2)\ \hspace{8cm}$
$4A\ – 2B = \begin{bmatrix} 4 & 8 \\ 12 & 20\\ \end{bmatrix}\ -\ \begin{bmatrix} -10 & 14 \\ 0 & 8\\ \end{bmatrix}\ \hspace{8cm}$
$=\begin{bmatrix} 4 + 10 & 8 -14 \\ 12 – 0 & 20 -8\\ \end{bmatrix}\ \hspace{8cm}$
$4A\ – 2B = \begin{bmatrix} 14 & -6 \\ 12 & 12\\ \end{bmatrix}\ \hspace{10cm}$
$\color {purple} {iii)\ Multiplication\ of\ Matrices}\ \hspace{20cm}$

The condition of multiplication of two matrices A and B is the number of columns in A is equal to the number of rows in B.

$\color {black} {Eg:}\ If\ A =\begin{bmatrix} 1 & 2 \\ 3 & 4 \\ \end{bmatrix}\ ,\ B =\begin{bmatrix} 5 & 6 \\ 7 & 8 \\ \end{bmatrix}\ ,\ Find\ AB \hspace{12cm}$
$\color {black}{Solution:}\ \hspace{18cm}$
$AB =\begin{bmatrix} 1 & 2 \\ 3 & 4 \\ \end{bmatrix}\ \begin{bmatrix} 5 & 6 \\ 7 & 8 \\ \end{bmatrix}\ \hspace{12cm}$
$= \begin{bmatrix} 1 × 5 + 2 × 7 & 1 × 6 + 2 × 8 \\ 3 × 5 + 4 × 7 & 3 × 6 + 4 × 8\\ \end{bmatrix}\ \hspace{9cm}$
$= \begin{bmatrix} 5 + 14 & 6 + 16 \\ 15 + 28 & 18 + 32\\ \end{bmatrix}\ \hspace{10cm}$
$AB = \begin{bmatrix} 19 & 22 \\ 43 & 50\\ \end{bmatrix}\ \hspace{12cm}$
$\color {purple} {Reducing\ a\ Matrix\ into\ echelon\ form}\ \hspace{20cm}$

A matrix is in row echelon form (ref) when it satisfies the following conditions.

• The first non-zero element in each row, called the leading entry, is 1.
• Each leading entry is in a column to the right of the leading entry in the previous row.
• Number of zeros before the non zero element increasing with row number.
• If there are any zero rows, they should be below the non-zero row.
$\color {black} {Eg:}\ Reduce\ the\ matrix\ A =\begin{bmatrix} 1 & -2 & 3 \\ 2 & 1 & 2 \\ 5 & -5 & 11 \\ \end{bmatrix}\ to\ echelon\ form\ \hspace{10cm}$
$\color {black}{Solution:}\ A =\begin{bmatrix} 1 & -2 & 3 \\ 2 & 1 & 2 \\ 5 & -5 & 11 \\ \end{bmatrix}\ \hspace{18cm}$
$R_2\ \rightarrow\ R_2\ -\ 2R_1\ \\ R_3\ \rightarrow\ R_3\ -\ 5R_1$
$A =\begin{bmatrix} 1 & -2 & 3 \\ 0 & 5 & -4 \\ 0 & 5 & -4 \\ \end{bmatrix}\ \hspace{16cm}$
$R_3\ \rightarrow\ R_3\ -\ R_2$
$A =\begin{bmatrix} 1 & -2 & 3 \\ 0 & 5 & -4 \\ 0 & 0 & 0 \\ \end{bmatrix}\ is\ an\ echelon\ form\ \hspace{17cm}$
$\color {purple} { DETERMINANTS}\ \hspace{20cm}$
$\color {green} {Definition\ of\ Determinant}\ \hspace{20cm}$
$A\ system\ of\ Linear\ equations\ like\ \hspace{15cm}$
$a_1x + b_1y = c_1\ \hspace{15cm}$
$a_2x + b_2y = c_2\ \hspace{15cm}$

Determinant is a square arrangement of numbers within two vertical lines.

$\color {black} {Eg:}\ \Delta =\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \\ \end{vmatrix}\ \hspace{15cm}$
$\color {green} {Order\ and\ Value\ of\ the\ Determinant}\ \hspace{20cm}$
$\color {green} {Determinant\ of\ Second\ order:}\ \hspace{20cm}$
$A =\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \\ \end{vmatrix}\ \hspace{15cm}$

consisting of two rows and two columns is called  a determinant of second order.

Value of the Determinant  is

$\Delta =\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \\ \end{vmatrix}\ = a_1b_2\ -\ a_2b_1\ \hspace{15cm}$
$\color {black} {Example\ 1:}\ Find\ the\ determinant\ of\ A =\begin{vmatrix} 2 & -5 \\ 1 & 3 \\ \end{vmatrix}\ \hspace{15cm}$
$\color {black}{Solution:}\ A = \begin{vmatrix} 2 & -5 \\ 1 & 3 \\ \end{vmatrix}\ \hspace{15cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ = 2(3) – (-5)(1) \hspace{13cm}$
$= 6 + 5 \hspace{12cm}$
$\Delta= 11 \hspace{12cm}$
$\color {black} {Example\ 2:}\ Solve\ \begin{vmatrix} x & 2 \\ 2 & x \\ \end{vmatrix}\ = 0 \hspace{15cm}$
$\color {black}{Solution:}\ \begin{vmatrix} x & 2 \\ 2 & x \\ \end{vmatrix}\ = 0 \hspace{15cm}$
$x(x) – 2 (2) = 0\ \hspace{13cm}$
$x^2 – 4 = 0\ \hspace{13cm}$
$x^2 = 4\ \hspace{13cm}$
$x = \pm\sqrt{4}\ \hspace{13cm}$
$x = \pm 2\ \hspace{13cm}$
$\color {green} {Determinant\ of\ Third\ order:}\ \hspace{20cm}$
$The\ expression\ \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \\ \end{vmatrix}\ \hspace{20cm}$

consisting of three rows and three columns is called  a determinant of third order.

Value of the Determinant  is :

$\Delta =a_1\begin{vmatrix} b_2 & b_3 \\ c_2 & c_3 \\ \end{vmatrix}\ -\ a_2\begin{vmatrix} b_1 & b_3 \\ c_1 & c_3 \\ \end{vmatrix}\ +\ a_3\begin{vmatrix} b_1 & b_2 \\ c_1 & c_2 \\ \end{vmatrix}\ \hspace{10cm}$
$\Delta =a_1(b_2c_3\ -\ b_3c_2)\ – a_2 (b_1c_3\ -\ b_3c_1) + a_3(b_1c_2\ -\ b_2c_1)\ \hspace{10cm}$
$\color {black} {Example\ :}\ Find\ the\ determinant\ of\ A =\begin{bmatrix} 3 & 1 & -1 \\ 2 & -1 & 2 \\ 2 & 1 & -2 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {black}{Solution:}\ A =\begin{bmatrix} 3 & 1 & -1 \\ 2 & -1 & 2 \\ 2 & 1 & -2 \\ \end{bmatrix}\ \hspace{15cm}$
$\Delta =3\begin{vmatrix} -1 & 2 \\ 1 & -2 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & 2 \\ 2 & -2 \\ \end{vmatrix}\ +\ -1\begin{vmatrix} 2 & -1 \\ 2 & 1 \\ \end{vmatrix}\ \hspace{4cm}$
$=3(2\ -\ 2)\ – 1 (-4\ -\ 4) – 1(2\ +\ 2)\ \hspace{10cm}$
$=3(0)\ – 1 (-8) – 1(4)\ \hspace{14cm}$
$=0\ +8 – 4\ \hspace{15cm}$
$\Delta =4\ \hspace{16cm}$
$\color {skyblue} {Properties\ of\ Determinants:}\ \hspace{20cm}$
$\color {purple} {Property:\ 1}\ \hspace{20cm}$
$The\ value\ of\ the\ determinant\ is\ unaltered\ by\ changing\ rows\ into\ columns\ and\ vice\ versa.$
$i.e\ \begin{vmatrix} A \\ \end{vmatrix}\ =\ \begin{vmatrix} A^T \\ \end{vmatrix}\ \hspace{15cm}$
$\color {black}{Example:}\ A =\begin{bmatrix} 3 & 1 & -1 \\ 2 & -1 & 2 \\ 2 & 1 & -2 \\ \end{bmatrix}\ \hspace{15cm}$
$\color {black}{Solution:}\ A =\begin{bmatrix} 3 & 1 & -1 \\ 2 & -1 & 2 \\ 2 & 1 & -2 \\ \end{bmatrix}\ \hspace{15cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ =3\begin{vmatrix} -1 & 2 \\ 1 & -2 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & 2 \\ 2 & -2 \\ \end{vmatrix}\ +\ -1\begin{vmatrix} 2 & -1 \\ 2 & 1 \\ \end{vmatrix}\ \hspace{4cm}$
$=3(2\ -\ 2)\ – 1 (-4\ -\ 4) – 1(2\ +\ 2)\ \hspace{10cm}$
$=3(0)\ – 1 (-8) – 1(4)\ \hspace{14cm}$
$=0\ +8 – 4\ \hspace{15cm}$
$\begin{vmatrix} A \\ \end{vmatrix}\ =4\ ——- (1) \hspace{16cm}$
$A^T =\begin{bmatrix} 3 & 2 & 2 \\ 1 & -1 & 1 \\ -1 & 2 & -2 \\ \end{bmatrix}\ \hspace{10cm}$
$\begin{vmatrix} A^T \\ \end{vmatrix}\ =3\begin{vmatrix} -1 & 1 \\ 2 & -2 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 1 & 1 \\ -1 & -2 \\ \end{vmatrix}\ +\ 2\begin{vmatrix} 1 & -1 \\ -1 & 2 \\ \end{vmatrix}\ \hspace{4cm}$
$=3(2\ -\ 2)\ – 2 (-2\ +\ 1) + 2(2\ -\ 1)\ \hspace{10cm}$
$=3(0)\ – 2 (-1) + 2(1)\ \hspace{14cm}$
$=0\ +2 + 2\ \hspace{15cm}$
$\begin{vmatrix} A^T \\ \end{vmatrix}\ =4\ —– (2) \hspace{16cm}$
$from\ (1)\ and (2)\ \begin{vmatrix} A \\ \end{vmatrix}\ =\ \begin{vmatrix} A^T \\ \end{vmatrix}\ \hspace{15cm}$
$\color {purple} {Property:\ 2}\ \hspace{20cm}$
$If\ any\ two\ rows\ / \ columns\ of\ a\ determinant\ are\ interchanged,\ then\ the\ value\ of\ the$
$determinant\ changes\ in\ sign\ but\ its\ absolute\ value\ remains\ unaltered.$
$\color {purple} {Property:\ 3}\ \hspace{20cm}$
$If\ any\ two\ rows\ (or)\ two\ columns\ of\ a\ matrix\ are\ identical,$$then\ the\ value\ of\ the\ determinant\ is\ zero.$
$\color {purple} {Property:\ 4}\ \hspace{20cm}$
$If\ each\ element\ of\ a\ row\ (or\ column)\ is\ multiplied\ by\ any\ scalar\ K,$$then\ the\ value\ of\ the\ determinant\ is\ also\ multiplied\ by\ the\ same\ scalar\ K$
$\color {purple} {Property:\ 5}\ \hspace{20cm}$
$If\ each\ element\ of\ a\ row\ (or\ column)\ of\ a\ determinant\ is\ expressed\ as\ sum\ of\ two\ or\ more\ terms$$then\ the\ whole\ determinant\ can\ be\ expressed\ as\ the\ sum\ of\ two\ (or)\ more\ determinants\ of\ the\ same\ order.$
$\color {purple} {Property:\ 6}\ \hspace{20cm}$
$A\ determinant\ is\ unaltered\ when\ to\ each\ element\ of\ any\ row\ (or\ column)\ are\ added\ those\ of$
$several\ other\ rows\ or\ columns\ multiplied\ respectively\ by\ constant\ factors.$
$\color {purple} {Problems\ using\ Properties\ of\ Determinants}\ \hspace{20cm}$
$1.\ Evaluate:\ \begin{vmatrix} 1 & a & b + c \\ 1 & b & c + a \\ 1 & c & a + b \\ \end{vmatrix}\ \hspace{20cm}$
$\color {black}{Solution:}\ \hspace{20cm}$
$Let\ \Delta = \begin{vmatrix} 1 & a & b + c \\ 1 & b & c + a \\ 1 & c & a + b \\ \end{vmatrix}\ \hspace{20cm}$
$C_2\ =\ C_2\ + C_3$
$= \begin{vmatrix} 1 & a + b + c & b + c \\ 1 & b + c + a & c + a \\ 1 & c + a + b & a + b \\ \end{vmatrix}\ \hspace{20cm}$
$= (a + b + c) \begin{vmatrix} 1 & 1 & b + c \\ 1 & 1 & c + a \\ 1 & 1 & a + b \\ \end{vmatrix}\ \hspace{20cm}$
$= (a + b + c) (0)\ \because\ C_1 = C_2\ \hspace{20cm}$
$\begin{vmatrix} 1 & a & b + c \\ 1 & b & c + a \\ 1 & c & a + b \\ \end{vmatrix}\ = 0\ \hspace{15cm}$
$2.\ Prove\ that\ \begin{vmatrix} 2a + b & a & b \\ 2b + c & b & c \\ 2c + a & c & a\\ \end{vmatrix}\ =\ 0\ \hspace{20cm}$
$\color {black}{Solution:}\ \hspace{20cm}$
$LHS\ =\ \begin{vmatrix} 2a + b & a & b \\ 2b + c & b & c \\ 2c + a & c & a\\ \end{vmatrix}\ \hspace{15cm}$
$=\begin{vmatrix} 2a & a & b \\ 2b & b & c \\ 2c & c & a\\ \end{vmatrix}\ + \begin{vmatrix} b & a & b \\ c & b & c \\ a & c & a\\ \end{vmatrix}\ \hspace{15cm}$
$=2\begin{vmatrix} a & a & b \\ b & b & c \\ c & c & a\\ \end{vmatrix}\ + 0\ \because\ C_1 = C_3\ \hspace{15cm}$
$= 2 (0)\ \because\ C_1 = C_2\ \hspace{15cm}$
$\begin{vmatrix} 2a + b & a & b \\ 2b + c & b & c \\ 2c + a & c & a\\ \end{vmatrix}\ = 0\ \hspace{15cm}$