NCERT Class 12 Maths Chapter 4 Notes, Determinants Class 12 Chapter 4 Notes

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NCERT Notes for Class 12 Chapter 4 Determinants

To every square matrix $\mathrm{A}=\left[a_{i j}\right]$ of order $n$, we can associate a number (real or complex) called the determinant of the matrix A, written as $\operatorname{det} \mathrm{A}$, where $a_{i j}$ is the $(i, j) t h$ element of A.
If $\mathrm{A}=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$, then determinant of A , denoted by $|\mathrm{A}|($ or $\operatorname{det} \mathrm{A})$, is given by

$|\mathrm{A}|=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|=a d-b c$

Let two equations $a_{1} x+b_{1}=0$ and $a_{2} x+b_{2}=0$ are satisfied by the same value of x.

$\Rightarrow x = \frac{-b{_1}}{a_{1}}= \frac{-b{_2}}{a_{2}}$

$\Rightarrow a_{1} b_{2}-a_{2} b_{1}=0$

The expression $\mathrm{a}_{1} \mathrm{~b}_{2}-\mathrm{a}_{2} \mathrm{~b}_{1}$ is called a determinant of the second-order and is denoted by :

$D = \left|\begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|$

Let's consider the system of equations:

$a_{1} x+b_{1} y+c_{1}=0$

$a_{2} x+b_{2} y+c_{2}=0$

$a_{3} x+b_{3} y+c_{3}=0$

If these equations are satisfied by the same x and y values, then eliminate x and y.

$a_{1} \left(b_{2} c_{3}-b_{3} c_{2}\right)+b_{1}\left(c_{2} a_{3}-c_{3} a_{2}\right)+c_{1}\left(a_{2} b_{3}-a_{3} b_{2}\right) =0$

The expression $a_{1} \left(b_{2} c_{3}-b_{3} c_{2}\right)+b_{1}\left(c_{2} a_{3}-c_{3} a_{2}\right)+c_{1}\left(a_{2} b_{3}-a_{3} b_{2}\right)$ is called a determinant of the third-order and is denoted by :

$D = \left|\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right|$

Note: Determinants consist of an equal number of rows and columns.

Value of a Determinant

Determinant of a matrix of order one

Let $\mathrm{A}=[a]$ be the matrix of order 1, then the determinant of A is defined to be equal to $a$.

Determinant of a matrix of order two

$\mathrm{A}=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]$

$\text { det }(\mathrm{A})=|\mathrm{Al}=\Delta=\left|\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right|$

$\text { det }(\mathrm{A})=a_{11} a_{22}-a_{21} a_{12}$

Determinant of a matrix of order three

$A=\left[\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right]$

$det(A) = D=\left|\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right|$

$D=a_{1}\left|\begin{array}{ll} b_{2} & c_{2} \\ b_{3} & c_{3} \end{array}\right|-b_{1}\left|\begin{array}{ll} a_{2} & c_{2} \\ a_{3} & c_{3} \end{array}\right|+c_{1}\left|\begin{array}{ll} a_{2} & b_{2} \\ a_{3} & b_{3} \end{array}\right|$

$D=a_{1}\left(b_{2} c_{3}-b_{3} c_{2}\right)-b_{1}\left(a_{2} c_{3}-a_{3} c_{2}\right)+c_{1}\left(a_{2} b_{3}-a_{3} b_{2}\right)$

Properties of Determinants

• The value of a determinant doesn't change if the rows & columns are interchanged
  $D=\left|\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right|=\left|\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right|$
• If any two rows of a determinant are interchanged, the sign of the value of the determinant is changed
  Let $\mathrm{D}=\left|\begin{array}{lll} \mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ \mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2} \\ \mathrm{a}_{3} & \mathrm{~b}_{3} & \mathrm{c}_{3} \end{array}\right|$
  and $\mathrm{D'}=\left|\begin{array}{lll} \mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2} \\ \mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ \mathrm{a}_{3} & \mathrm{~b}_{3} & \mathrm{c}_{3} \end{array}\right|$
  then $\mathrm{D'}=-\mathrm{D}$
• If any two columns of a determinant are interchanged, the sign of the value of the determinant is changed
  Let $\mathrm{D}=\left|\begin{array}{lll} \mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ \mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2} \\ \mathrm{a}_{3} & \mathrm{~b}_{3} & \mathrm{c}_{3} \end{array}\right|$
  and$\mathrm{D'}=\left|\begin{array}{lll} \mathrm{b}_{1} & \mathrm{~a}_{1} & \mathrm{c}_{1} \\ \mathrm{b}_{2} & \mathrm{~a}_{2} & \mathrm{c}_{2} \\ \mathrm{b}_{3} & \mathrm{~a}_{3} & \mathrm{c}_{3} \end{array}\right|$
  then $\mathrm{D'}=-\mathrm{D}$
• If all the elements of a row are zero, then the value of the determinant is zero.
  $\mathrm{D}=\left|\begin{array}{lll} 0 &0&0 \\ \mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2} \\ \mathrm{a}_{3} & \mathrm{~b}_{3} & \mathrm{c}_{3} \end{array}\right|=0$
• If all the elements of a column are zero, then the value of the determinant is zero.
  $\mathrm{D}=\left|\begin{array}{lll}0 & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ 0 & \mathrm{~b}_{2} & \mathrm{c}_{2} \\ 0 & \mathrm{~b}_{3} & \mathrm{c}_{3} \end{array}\right|=0$
• If all the elements of any row are multiplied by the same number, then the determinant is multiplied by that number.
  Let $\mathrm{D}=\left|\begin{array}{lll} \mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ \mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2} \\ \mathrm{a}_{3} & \mathrm{~b}_{3} & \mathrm{c}_{3} \end{array}\right|$
  and $\mathrm{D'}=\left|\begin{array}{lll} k\mathrm{a}_{1} & k\mathrm{~b}_{1} & k\mathrm{c}_{1} \\ \mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2} \\ \mathrm{a}_{3} & \mathrm{~b}_{3} & \mathrm{c}_{3} \end{array}\right|$
  then $\mathrm{D'}=k\mathrm{D}$
• If all the elements of any column are multiplied by the same number, then the determinant is multiplied by that number.
  Let $\mathrm{D}=\left|\begin{array}{lll} \mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ \mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2} \\ \mathrm{a}_{3} & \mathrm{~b}_{3} & \mathrm{c}_{3} \end{array}\right|$
  and $\mathrm{D'}=\left|\begin{array}{lll} k\mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ k\mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2} \\ k\mathrm{a}_{3} & \mathrm{~b}_{3} & \mathrm{c}_{3} \end{array}\right|$
  then $\mathrm{D'}=k\mathrm{D}$
• If all the elements of a row are proportional to the element of any other row, then the determinant's value is zero.
  $\mathrm{D}=\left|\begin{array}{lll} \mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ k\mathrm{a}_{1} & k\mathrm{~b}_{1} & k\mathrm{c}_{1} \\ \mathrm{a}_{3} & \mathrm{~b}_{3} & \mathrm{c}_{3} \end{array}\right|=0$
• If all the elements of a column are proportional to the element of any other column, then the determinant's value is zero.
  $\mathrm{D}=\left|\begin{array}{lll} k\mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ k\mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2} \\ k\mathrm{a}_{3} & \mathrm{~b}_{3} & \mathrm{c}_{3} \end{array}\right|$
• If some or all elements of a row or column of a determinant are expressed as the sum of two (or more) terms, then the determinant can be expressed as the sum of two (or more) determinants.
  $D = \left|\begin{array}{ccc} a_{1}+ p & a_{2}+q & a_{3}+r \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right|=\left|\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right|+\left|\begin{array}{lll} p&q&r\\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right|$
• The value of the determinant remains the same if we apply the operation $R_{i }\rightarrow R_{i} + kR_{j} \ or \ C_{i} \rightarrow C_{i} + k C_{j}$

Multiplication of Determinants

Let the two determinants of the third-order be

$
\Delta_1=\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right| \text { and } \Delta_2=\left|\begin{array}{ccc}
\alpha_1 & \beta_1 & \gamma_1 \\
\alpha_2 & \beta_2 & \gamma_2 \\
\alpha_3 & \beta_3 & \gamma_3
\end{array}\right|
$
We can multiply these row-by-row or column-by-column or row-by-column or column-by-row
Row-by-row multiplication of these two determinants is given by

$
\Delta_1 \times \Delta_2=\left|\begin{array}{lll}
a_1 \alpha_1+b_1 \beta_1+c_1 \gamma_1 & a_1 \alpha_2+b_1 \beta_2+c_1 \gamma_2 & a_1 \alpha_3+b_1 \beta_3+c_1 \gamma_3 \\
a_2 \alpha_1+b_2 \beta_1+c_2 \gamma_1 & a_2 \alpha_2+b_2 \beta_2+c_2 \gamma_2 & a_2 \alpha_3+b_2 \beta_3+c_2 \gamma_3 \\
a_3 \alpha_1+b_3 \beta_1+c_3 \gamma_1 & a_3 \alpha_2+b_3 \beta_2+c_3 \gamma_2 & a_3 \alpha_3+b_3 \beta_3+c_3 \gamma_3
\end{array}\right|
$
Multiplication can also be performed row by column; column by row or column by column as required in the problem.

To express a determinant as a product of two determinants, one requires lots of practice, and this can be done only by inspection and trial.

Property:

If $\mathrm{A}_1, \mathrm{~B}_1, \mathrm{C}_1, \ldots$...are respectively the cofactors of the elements $\mathrm{a}_1, \mathrm{~b}_1, \mathrm{c}_1,...$ of the determinant $\Delta=\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|, \Delta \neq 0$, then $\left|\begin{array}{lll}A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3\end{array}\right|=\Delta^2$

Proof:
given, $\quad \Delta=\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|$ and, $\mathrm{A}_1, \mathrm{~B}_1, \mathrm{C}_1, \ldots$. are respectively the cofactors of the elements $\mathrm{a}_1, \mathrm{~b}_1, \mathrm{c}_1 \ldots \ldots \ldots$. Hence,

$
\begin{aligned}
& \left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|\left|\begin{array}{lll}
A_1 & B_1 & C_1 \\
A_2 & B_2 & C_2 \\
A_3 & B_3 & C_3
\end{array}\right| \\
& =\left|\begin{array}{ccc}
a_1 A_1+b_1 B_1+c_1 C_1 & a_1 A_2+b_1 B_2+c_1 C_2 & a_1 A_3+b_1 B_3+c_1 C_3 \\
a_2 A_1+b_2 B_1+c_2 C_1 & a_2 A_2+b_2 B_2+c_2 C_2 & a_2 A_3+b_2 B_3+c_2 C_3 \\
a_3 A_1+b_3 B_1+c_3 C_1 & a_3 A_2+b_3 B_2+c_3 C_2 & a_3 A_3+b_3 B_3+c_3 C_3
\end{array}\right|
\end{aligned}
$

[row by row multiplication]

$
\begin{aligned}
& =\left|\begin{array}{ccc}
\Delta & 0 & 0 \\
0 & \Delta & 0 \\
0 & 0 & \Delta
\end{array}\right|=\Delta^3 \\
& \because \mathrm{a}_{\mathrm{i}} A_{\mathrm{j}}+\mathrm{b}_{\mathrm{i}} \mathrm{~B}_{\mathrm{j}}+\mathrm{c}_{\mathrm{i}} \mathrm{C}_{\mathrm{j}}=\left\{\begin{array}{cc}
\Delta, & i=j \\
0, & i \neq j
\end{array}\right. \\
& \Rightarrow \Delta\left|\begin{array}{lll}
A_1 & B_1 & C_1 \\
A_2 & B_2 & C_2 \\
A_3 & B_3 & C_3
\end{array}\right|=\Delta^3 \\
& \Rightarrow\left|\begin{array}{ccc}
A_1 & B_1 & C_1 \\
A_2 & B_2 & C_2 \\
A_3 & B_3 & C_3
\end{array}\right|=\Delta^2
\end{aligned}
$

Area of a Triangle:

The area of a triangle whose vertices are $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right) \text { and }\left(x_{3}, y_{3}\right)$ is given by:

$\Delta=\frac{1}{2}\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|$

Minors of Determinants

The minor of a given element of the determinant is the determinant obtained by deleting the row & the column in which the given element stands.

Let $D=\left|\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right|$

The minor of element a1 $M_{11}=\left|\begin{array}{ll} b_{2} & c_{2} \\ b_{3} & c_{3} \end{array}\right|$

The minor of element b1 =$M_{12}=\left|\begin{array}{ll} a_{2} & c_{2} \\ a_{3} & c_{3} \end{array}\right|$

The minor of element c1 = $M_{13} = \left|\begin{array}{ll} a_{2} & b_{2} \\ a_{3} & b_{3} \end{array}\right|$

Here Mij represents the minor of the element belonging to the ith row and jth column.

Cofactors of Determinants

The cofactor of element ith row and jth column is given by: $C_{ij} = (-1)^{i + j}M_{ij}$ where Mij represents the minor of the element belonging to ith row and jth column.

The cofactor of element a1 $C_{11}=\left|\begin{array}{ll} b_{2} & c_{2} \\ b_{3} & c_{3} \end{array}\right|$

The cofactor of element b1 =$C_{12}=(-1)\left|\begin{array}{ll} a_{2} & c_{2} \\ a_{3} & c_{3} \end{array}\right|$

The cofactor of element c1 = $C_{13} = \left|\begin{array}{ll} a_{2} & b_{2} \\ a_{3} & b_{3} \end{array}\right|$

Expansion of Determinants in terms of Elements of any Row or Column

Let $D=\left|\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right|$

The sum of the product of elements of any row (column) with their corresponding cofactors is always equal to the value of the determinant.

So the determinant D can be in 6 forms

$D = a_{1} C_{11}+b_{1} C_{12}+c_{1} C_{13}$

$D = a_{2} C_{21}+b_{2} C_{22}+c_{2} C_{23}$

$D = a_{3} C_{31}+b_{3} C_{32}+c_{3} C_{33}$

$D = a_{1} C_{11}+a_{2} C_{21}+a_{3} C_{31}$

$D = b_{1} C_{12}+b_{2} C_{22}+b_{3} C_{32}$$D = c_{1} C_{13}+c_{2} C_{23}+c_{3} C_{33}$

Adjoint of a Matrix

Let a given matrix $A=\left[\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right]$ then adjoin of matrix A is defined as

$\text { adj(A) = Transpose of }\left[\begin{array}{lll} \mathrm{C}_{11} & \mathrm{C}_{12} & \mathrm{C}_{13} \\ \mathrm{C}_{21} & \mathrm{C}_{22} & \mathrm{C}_{23} \\ \mathrm{C}_{31} & \mathrm{C}_{32} & \mathrm{C}_{33} \end{array}\right]=\left[\begin{array}{lll} \mathrm{C}_{11} & \mathrm{C}_{21} & \mathrm{C}_{31} \\ \mathrm{C}_{12} & \mathrm{C}_{22} & \mathrm{C}_{32} \\ \mathrm{C}_{13} & \mathrm{C}_{23} & \mathrm{C}_{33} \end{array}\right]$

where - Cij is the cofactor of the element aij

Note
For any given matrix of order n

$\mathrm{A}(a d j \mathrm{~A})=(a d j \mathrm{~A}) \mathrm{A}=|\mathrm{A}| \mathrm{I}$

where I - Identity matrix of order n

Singular matrix

A square matrix A of order n is said to be singular if $|A|=0$

Non-Singular matrix

A square matrix A of order n is said to be non-singular if $|A| \neq 0$

The inverse of a Matrix

A square matrix A is invertible if and only if A is a nonsingular matrix, or in other words, a square matrix A is invertible if and only if $|A| \neq 0$.

$\mathrm{A}(a d j \mathrm{~A})=(a d j \mathrm{~A}) \mathrm{A}=|\mathrm{A}| \mathrm{I}$

$\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} a d j \mathrm{~A}$

Solution of a System of Linear Equations using the Inverse of a Matrix

Let us give the system of linear equations

$\begin{aligned} &a_{1} x+b_{1} y+c_{1} z=d_{1} \\ &a_{2} x+b_{2} y+c_{2} z=d_{2} \\ &a_{3} x+b_{3} y+c_{3} z=d_{3} \end{aligned}$

Let $\mathrm{A}=\left[\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right]$

$\mathrm{X}=\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$

$\mathrm{B}=\left[\begin{array}{l} d_{1} \\ d_{2} \\ d_{3} \end{array}\right]$

Then the system of linear equations can be written as

AX = B

$\left[\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} d_{1} \\ d_{2} \\ d_{3} \end{array}\right]$

- If A is a non-singular matrix

$\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}$

- If A is a singular matrix

• If $adj(A). B \neq 0$ then the linear equation does not have solutions, and is an inconsistent linear equation
• If $adj(A).B = 0$ then linear equations are either consistent or inconsistent and the system has either infinitely many solutions or no solution.

Note 1:
If A and B are square matrices of the same order, then the determinant of the product of matrices is equal to the product of their respective determinants, i.e.

$|AB|=|A|.|B|$

Note 2:
If A is a square matrix of order n, then

$|adj(A)| = | A|^{ n -1}$

System of linear equations

(i) Consider the equations:

$\begin{aligned} & a_1 x+b_1 y+c_1 z=d_1 \\ & a_2 x+b_2 y+c_2 z=d_2 \\ & a_3 x+b_3 y+c_3 z=d_3,\end{aligned}$

In matrix form, these equations can be written as $\mathrm{AX}=\mathrm{B}$, where

$\mathrm{A}=\left[\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right], \mathrm{X}=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{l}d_1 \\ d_2 \\ d_3\end{array}\right]$

(ii) Unique solution of equation $\mathrm{AX}=\mathrm{B}$ is given by $\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}$, where $|\mathrm{A}| \neq 0$.

(iii) A system of equations is consistent or inconsistent according as its solution exists or not.
(iv) For a square matrix A in matrix equation $\mathrm{AX}=\mathrm{B}$
(a) If $|\mathrm{A}| \neq 0$, then there exists unique solution.
(b) If $|\mathrm{A}|=0$ and $(\operatorname{adj} \mathrm{A}) \mathrm{B} \neq 0$, then there exists no solution.
(c) If $|\mathrm{A}|=0$ and $(\operatorname{adj} \mathrm{A}) \mathrm{B}=0$, then system may or may not be consistent.

Cramer’s law

For the system of equations in two variables

Let $a_1 x+b_1 y=c_1$ and $a_2 x+b_2 y=c_2$, where

$
\frac{\mathrm{a}_1}{\mathrm{a}_2} \neq \frac{\mathrm{b}_1}{\mathrm{~b}_2}
$
On solving this equation by cross multiplication, we get

$
\begin{aligned}
& \frac{x}{b_2 c_1-b_1 c_2}=\frac{y}{a_1 c_2-a_2 c_1}=\frac{1}{a_1 b_2-a_2 b_1} \\
& \text { or } \frac{\mathrm{x}}{\left|\begin{array}{ll}
c_1 & b_1 \\
c_2 & b_2
\end{array}\right|}=\frac{\mathrm{y}}{\left|\begin{array}{ll}
a_1 & c_1 \\
a_2 & c_2
\end{array}\right|}=\frac{1}{\left|\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right|} \\
& \text { or } \mathrm{x}=\frac{\left|\begin{array}{ll}
c_1 & b_1 \\
c_2 & b_2
\end{array}\right|}{\left|\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right|}, \mathrm{y}=\frac{\left|\begin{array}{ll}
a_1 & c_1 \\
a_2 & c_2
\end{array}\right|}{\left|\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right|}
\end{aligned}
$

We can observe that the first column in the numerator of x is of constants, and 2nd column in the numerator of y is of constants, and the denominator is of the coefficient of variables.

We can follow this analogy for the system of equations of 3 variables, where the third column in the numerator of the value of z will be constant, and the denominator will be formed by the value of the coefficients of the variables.

For the system of equations in three variables

Let us consider the system of equations

$
\begin{aligned}
& \mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1 \mathrm{z}=\mathrm{d}_1 \\
& \mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2 \mathrm{z}=\mathrm{d}_2 \\
& \mathrm{a}_3 \mathrm{x}+\mathrm{b}_3 \mathrm{y}+\mathrm{c}_3 \mathrm{z}=\mathrm{d}_3
\end{aligned}
$

then $\Delta$, which will be determinant of the coefficient of variables, will be

$
\Delta=\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|
$

$\Delta_1$ numerator of $x$ is :
$\Delta_1=\left|\begin{array}{lll}d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3\end{array}\right|$
Similarly $\Delta_2=\left|\begin{array}{lll}a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3\end{array}\right|$ and $\Delta_3=\left|\begin{array}{lll}a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3\end{array}\right|$
i) If $\Delta \neq 0$, then the system of equations has a unique finite solution and so equations are consistent, and solutions are $\mathrm{x}=\frac{\Delta_1}{\Delta}, \mathrm{y}=\frac{\Delta_2}{\Delta}, \mathrm{z}=\frac{\Delta_3}{\Delta}$
ii) If $\Delta=0$, and any of $\Delta_1 \neq 0$ or $\Delta_2 \neq 0$ or $\Delta_3 \neq 0$

Then the system of equations is inconsistent, and hence no solution exists.
iii) If all $\Delta=\Delta_1=\Delta_2=\Delta_3=0$ then

The system of equations is consistent, and it has an infinite number of solutions (except when all three equations represent parallel planes, in which case there is no solution)

Determinants: Previous Year Question and Answer

Question 1:
The value of the determinant $\left|\begin{array}{ccc} x & x+y & x+2 y \\ x+2 y & x & x+y \\ x+y & x+2 y & x \end{array}\right|$ is:

Solution:
$
\begin{aligned}
& \left|\begin{array}{ccc}
x & x+y & x+2 y \\
x+2 y & x & x+y \\
x+y & x+2 y & x
\end{array}\right| \\
& =\mathrm{x}\left|\begin{array}{cc}
\mathrm{x} & \mathrm{x}+\mathrm{y} \\
\mathrm{x}+2 \mathrm{y} & \mathrm{x}
\end{array}\right|-(\mathrm{x}+\mathrm{y})\left|\begin{array}{cc}
\mathrm{x}+2 \mathrm{y} & \mathrm{x}+\mathrm{y} \\
\mathrm{x}+\mathrm{y} & \mathrm{x}
\end{array}\right|+(\mathrm{x}+2 \mathrm{y})\left|\begin{array}{cc}
\mathrm{x}+2 \mathrm{y} & \mathrm{x} \\
\mathrm{x}+\mathrm{y} & \mathrm{x}+2 \mathrm{y}
\end{array}\right| \\
& =x\left[x^2-(x+y)(x+2 y)\right]-(x+y)\left[(x+2 y)(x)-(x+y)^2\right]+(x+2 y)[(x+ \\
& \left.2 y)^2-x(x+y)\right] \\
& =x\left[x^2-x^2-3 x y-2 y^2\right]-(x+y)\left[x^2+2 x y-x^2-2 x y-y^2\right]+(x+2 y)\left[x^2+\right. \\
& \left.4 x y+4 y^2-x^2-x y\right] \\
& =x\left[-3 x y-2 y^2\right]-(x+y)\left[-y^2\right]+(x+2 y)\left[3 x y+4 y^2\right] \\
& =-3 x^2 y-2 x y^2+x y^2+y^3+3 x^2 y+4 x y^2+6 x y^2+8 y^3 \\
& =9 y^3+9 x y^2 \\
& =9 y^2(x+y)
\end{aligned}
$
Hence, the correct answer is $9 y^2(x+y)$.

Question 2:
If A, B and C are angles of a triangle, then the determinant $\begin{array}{|ccc|} -1 & \cos \mathrm{C} & \cos \mathrm{B} \\ \cos \mathrm{C} & -1 & \cos \mathrm{A} \\ \cos \mathrm{B} & \cos \mathrm{A} & -1 \end{array} \mid$ is equal to:

Solution:
$
\left|\begin{array}{ccc}
-1 & \cos \mathrm{C} & \cos \mathrm{~B} \\
\cos \mathrm{C} & -1 & \cos \mathrm{~A} \\
\cos \mathrm{~B} & \cos \mathrm{~A} & -1
\end{array}\right|
$
Expand along Column 1

$
\begin{aligned}
& |A|=a_{11}(-1)^{1+1}\left|\begin{array}{cc}
a_{22} & a_{23} \\
a_{32} & a_{33}
\end{array}\right|+\mathrm{a}_{21}(-1)^{2+1}\left|\begin{array}{cc}
a_{12} & a_{13} \\
a_{32} & a_{33}
\end{array}\right|+a_{31}(-1)^{3+1}\left|\begin{array}{lc}
a_{12} & a_{13} \\
a_{22} & a_{23}
\end{array}\right| \\
& \Delta=(-1)\left|\begin{array}{cc}
-1 & \cos A \\
\cos A & -1
\end{array}\right|-\cos C\left|\begin{array}{cc}
\cos C & \cos B \\
\cos A & -1
\end{array}\right|+\cos B\left|\begin{array}{cc}
\cos C & \cos B \\
-1 & \cos A
\end{array}\right| \\
& =\left[(-1)\left\{1-\cos ^2 A\right\}-\cos C\{-\cos C-\cos A \cos B\}+\cos B\{\cos A \cos C+\right. \\
& \cos B\}]
\end{aligned}
$
$
=-1+\cos ^2 A+\cos ^2 C+\cos A \cos B \cos C+\cos A \cos B \cos C+\cos ^2 B
$
$
=-1+\cos ^2 A+\cos ^2 B+\cos ^2 C+2 \cos A \cos B \cos C
$
Using the formula

$
1+\cos 2 A=2 \cos ^2 A
$

$\\ \begin{aligned} &=-1+\frac{1+\cos 2 \mathrm{~A}}{2}+\frac{1+\cos 2 \mathrm{~B}}{2}+\frac{1+\cos 2 \mathrm{C}}{2}+2 \cos \mathrm{A} \cos \mathrm{B} \cos \mathrm{C}\\ &\text { Taking L.C.M, we get }\\ &=\frac{-2+1+\cos 2 A+1+\cos 2 B+1+\cos 2 C+4 \cos A \cos B \cos C}{2}\\ &=\frac{1+(\cos 2 A+\cos 2 B)+\cos 2 C+4 \cos C \cos A \cos B}{2}\\ &\text { Now use: } \cos (A+B) \cos (A-B)=2 \cos A \cos B\\ &so, \cos 2 A+\cos 2 B=2 \cos (A+B) \cos (A-B) \end{aligned}$

$\\ \begin{aligned} &=\frac{1+\cos 2 C+\{2 \cos (A+B) \cos (A-B)\}+4 \cos A \cos B \cos C}{2}\\ &=\frac{1+2 \cos ^{2} C-1+\{2 \cos (A+B) \cos (A-B)\}+4 \cos A \cos B \cos C}{2}\\ &.=\frac{2 \cos ^{2} \mathrm{C}+[2 \cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})\}+4 \cos \mathrm{A} \cos \mathrm{B} \cos C}{2} \ldots (\mathrm{i})\\ &\text { We know that } A, B, C \text { are angles of triangle }\\ &\Rightarrow A+B+C=\pi\\ &\Rightarrow A+B=\pi-C \end{aligned}$

$\\ =\frac{2 \cos ^{2} C+\{2 \cos (\pi-C) \cos (A-B)\}+4 \cos A \cos B \cos C}{2} \\ =\frac{2 \cos ^{2} C+\{-2 \cos C \cos (A-B)\}+4 \cos A \cos B \cos C}{2} $
$\\ [\because \cos (\pi-x)=-\cos x]$
$=\frac{-2 \cos C\{\cos (A-B)-\cos C\}+4 \cos A \cos B \cos C}{2}$

$\begin{align*} &= -\cos C \{ \cos(A - B) - \cos C \} + 2\cos A \cos B \cos C\\ &= -\cos C[\cos(A - B) - \cos \{ \pi - (A + B) \}] + 2\cos A \cos B \cos C\\ &= -\cos C[\cos(A - B) + \cos(A + B)] + 2\cos A \cos B \cos C\\ &= -\cos C[2\cos A \cos B] + 2\cos A \cos B \cos C\\ &= 0 \end{align*}$

Hence, the correct answer is 0.

Question 3:
The value of determinant $\left|\begin{array}{lll} a-b & b+c & a \\ b-c & c+a & b \\ c-a & a+b & c \end{array}\right|$ is:

Solution:
$
\left|\begin{array}{lll}
a-b & b+c & a \\
b-c & c+a & b \\
c-a & a+b & c
\end{array}\right|
$
Apply $\mathrm{C} 2 \rightarrow \mathrm{C} 2+\mathrm{C} 3$

$
=\left|\begin{array}{lll}
a-b & a+b+c & a \\
b-c & c+a+b & b \\
c-a & a+b+c & c
\end{array}\right|
$
Take $(a+b+c)$ common from Column 2

$
=(a+b+c)\left|\begin{array}{lll}
a-b & 1 & a \\
b-c & 1 & b \\
c-a & 1 & c
\end{array}\right|
$
Apply $\mathrm{C}_1 \rightarrow \mathrm{C}_1-\mathrm{C}_3$

$
\begin{aligned}
& =(a+b+c)\left|\begin{array}{llll}
a-b-a & 1 & a \\
b-c-b & 1 & b \\
c-a-c & 1 & c
\end{array}\right| \\
& =(a+b+c)\left|\begin{array}{lll}
-b & 1 & a \\
-c & 1 & b \\
-a & 1 & c
\end{array}\right|
\end{aligned}
$

Expand along Row 1

$
\begin{aligned}
& =(a+b+c)\left[(-b)\{c-b\}-(1)\left\{-c^2-(-a b)\right\}+a\{-c-(-a)\}\right] \\
& =(a+b+c)\left(-b c+b^2+c^2-a b-a c+a^2\right) \\
& =a\left(-b c+b^2+c^2-a b-a c+a^2\right)+b\left(-b c+b^2+c^2-a b-a c+a^2\right)+c(-b c+ \\
& \left.b^2+c^2-a b-a c+a^2\right) \\
& =-a b c+a b^2+a c^2-a^2 b-a^2 c+a^3-b^2 c+b^3+b c^2-a b^2-a b c+a^2 b-b c^2+ \\
& b^2 c+c^3-a b c-a c^2+a^2 c \\
& =a^3+b^3+c^3-3 a b c
\end{aligned}
$

Hence, the correct answer is $a^3+b^3+c^3-3 a b c$.

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