`star` Properties of Determinants

`color{red}{"The value of the determinant remains unchanged if its rows and columns are interchanged."}`

Verification : Let `Delta = |(a_1,a_2,a_3),(b_1,b_2,b_3),(c_1,c_2,c_3)|`

Expanding along first row, we get `Delta = a_1|(b_2,b_3),(c_2,c_3)| - a_2|(b_1,b_3),(c_1,c_3)| + a_3|(b_1,b_2),(c_1,c_2)|`

`=a_1(b_2 c_3– b_3 c_2) - a_2(b_1 c_3– b_3 c_1) + a_3(b_1 c_2 – b_2 c_1)`

● By interchanging the rows and columns of Δ, we get the determinant

` Δ_1 = | (a_1 ,b_1 ,c_1), ( a_2,b_2, c_2), ( a_3,b_3,c_3) |`

● Expanding `Δ_1` along first column, we get

`Δ_1 = a_1 (b_2 c_3 – c_2 b_3) – a_2 (b_1 c_3 – b_3 c_1) + a_3 (b_1 c_2 – b_2 c_1)`

Hence `Δ = Δ_1`

`color{blue}{"Remark :"}` If A is a square matrix, then `color{orange}{det (A) = det (A′),}` where A′ = transpose of A.

`color{blue}{"Note :"}` If `R_i = i^(th)` row and `C_i = i^(th)` column, then for interchange of row and columns, we will symbolically write `C_i↔ R_i`

Verification : Let `Delta = |(a_1,a_2,a_3),(b_1,b_2,b_3),(c_1,c_2,c_3)|`

Expanding along first row, we get `Delta = a_1|(b_2,b_3),(c_2,c_3)| - a_2|(b_1,b_3),(c_1,c_3)| + a_3|(b_1,b_2),(c_1,c_2)|`

`=a_1(b_2 c_3– b_3 c_2) - a_2(b_1 c_3– b_3 c_1) + a_3(b_1 c_2 – b_2 c_1)`

● By interchanging the rows and columns of Δ, we get the determinant

` Δ_1 = | (a_1 ,b_1 ,c_1), ( a_2,b_2, c_2), ( a_3,b_3,c_3) |`

● Expanding `Δ_1` along first column, we get

`Δ_1 = a_1 (b_2 c_3 – c_2 b_3) – a_2 (b_1 c_3 – b_3 c_1) + a_3 (b_1 c_2 – b_2 c_1)`

Hence `Δ = Δ_1`

`color{blue}{"Remark :"}` If A is a square matrix, then `color{orange}{det (A) = det (A′),}` where A′ = transpose of A.

`color{blue}{"Note :"}` If `R_i = i^(th)` row and `C_i = i^(th)` column, then for interchange of row and columns, we will symbolically write `C_i↔ R_i`

Q 3154180954

Verify Property 1 for `Δ = | ( 2,-3, 5), ( 6,0,4), ( 1,5, -7) |`

Class 12 Chapter 4 Example 6

Class 12 Chapter 4 Example 6

Expanding the determinant along first row, we have

` Δ = 2 | ( 0 ,4) , ( 5,-7) | - (-3) | (6,4), ( 1,-7) | + 5 | ( 6,0 ), (1,5) |`

= 2 (0 – 20) + 3 (– 42 – 4) + 5 (30 – 0)

= – 40 – 138 + 150 = – 28

By interchanging rows and columns, we get

`Δ_1 = | (2,6,1), (-3,0,5), (5,4 , -7) |` (Expanding along first column)

`=2 | ( 0 ,5 ), (4,-7) | - (-3) | ( 6,1), (4 , -7) | + 5 | ( 6,1), ( 0 ,5) |`

= 2 (0 – 20) + 3 (– 42 – 4) + 5 (30 – 0)

= – 40 – 138 + 150 = – 28

Clearly `Δ = Δ_1`

Hence, Property 1 is verified.

`color{red}{"If any two rows (or columns) of a determinant are interchanged,"}`

`color{red}{"then sign of determinant changes. "}`

`"Verification :"` Let Δ = ` | (a_1, a_2, a_3 ), ( b_1, b_2 , b_3 ), ( c_1, c_2, c_3) |`

Expanding along first row, we get

`Δ = a_1 (b_2 c_3 – b_3 c_2) – a_2 (b_1 c_3 – b_3 c_1) + a_3 (b_1 c_2 – b_2 c_1)`

`=>` Interchanging first and third rows, the new determinant obtained is given by

`Δ_1 = | (c_1, c_2, c_3 ), ( b_1, b_2 , b_3 ), ( a_1, a_2 , a_3) |`

`=>` Expanding along third row, we get

`Δ_1 = a_1 (c_2 b_3 – b_2 c_3) – a_2 (c_1 b_3 – c_3 b_1) + a_3 (b_2 c_1 – b_1 c_2)`

`= – [a_1 (b_2 c_3 – b_3 c_2) – a_2 (b_1 c_3 – b_3 c_1) + a_3 (b_1 c_2 – b_2 c_1)]`

Clearly `Δ_1 = – Δ`

`"Similarly, we can verify the result by interchanging any two columns."`

`color{orange}{" We can denote the interchange of rows by"\ \ R_i ↔ R_j \ \ "and interchange of

columns by" C_i ↔ C_j}`.

`color{red}{"then sign of determinant changes. "}`

`"Verification :"` Let Δ = ` | (a_1, a_2, a_3 ), ( b_1, b_2 , b_3 ), ( c_1, c_2, c_3) |`

Expanding along first row, we get

`Δ = a_1 (b_2 c_3 – b_3 c_2) – a_2 (b_1 c_3 – b_3 c_1) + a_3 (b_1 c_2 – b_2 c_1)`

`=>` Interchanging first and third rows, the new determinant obtained is given by

`Δ_1 = | (c_1, c_2, c_3 ), ( b_1, b_2 , b_3 ), ( a_1, a_2 , a_3) |`

`=>` Expanding along third row, we get

`Δ_1 = a_1 (c_2 b_3 – b_2 c_3) – a_2 (c_1 b_3 – c_3 b_1) + a_3 (b_2 c_1 – b_1 c_2)`

`= – [a_1 (b_2 c_3 – b_3 c_2) – a_2 (b_1 c_3 – b_3 c_1) + a_3 (b_1 c_2 – b_2 c_1)]`

Clearly `Δ_1 = – Δ`

`"Similarly, we can verify the result by interchanging any two columns."`

`color{orange}{" We can denote the interchange of rows by"\ \ R_i ↔ R_j \ \ "and interchange of

columns by" C_i ↔ C_j}`.

Q 3184180957

Verify Property 2 for `Δ = | (2,-3,5), ( 6,0,4), ( 1,5, -7) |`

Class 12 Chapter 4 Example 7

Class 12 Chapter 4 Example 7

`Δ = | (2,-3,5), ( 6,0,4), ( 1,5, -7) | = -28` (See Example 6)

Interchanging rows `R_2` and `R_3` i.e., `R_2 ↔ R_3`, we have

`Δ_1 = | ( 2,-3 ,5), ( 1,5, -7) , ( 6,0,4) |`

Expanding the determinant `Δ_1` along first row, we have

`Δ_1 = 2 | ( 5,-7), (0,4) | - ( -3) | (1,-7), (6,4) | + 5 | ( 1,5), (6 ,0) |`

= 2 (20 – 0) + 3 (4 + 42) + 5 (0 – 30)

= 40 + 138 – 150 = 28

Clearly `Δ_1 = – Δ`

Hence, Property 2 is verified

`color{red}{" If any two rows (or columns) of a determinant are identical "}`

`color{red}{"all corresponding elements are same), then value of determinant is zero."}`

`color{blue}{"Proof"}` If we interchange the identical rows (or columns) of the determinant Δ, then Δ

does not change. However, by Property 2, it follows that Δ has changed its sign

Therefore `Δ = – Δ`

or `Δ = 0`

`color{red}{"all corresponding elements are same), then value of determinant is zero."}`

`color{blue}{"Proof"}` If we interchange the identical rows (or columns) of the determinant Δ, then Δ

does not change. However, by Property 2, it follows that Δ has changed its sign

Therefore `Δ = – Δ`

or `Δ = 0`

Q 3154191054

Evaluate `Δ = | ( 3,2,3), ( 2,2,3), ( 3,2,3) |`

Class 12 Chapter 4 Example 8

Class 12 Chapter 4 Example 8

Expanding along first row, we get

Δ = 3 (6 – 6) – 2 (6 – 9) + 3 (4 – 6)

= 0 – 2 (–3) + 3 (–2) = 6 – 6 = 0

Here `R_1` and `R_3` are identical.

`color{red}{" If each element of a row (or a column) of a determinant is"}`

`color{green}{"multiplied by a constant" k, "then its value gets multiplied by" k}`

`"Verification :"` Let `Δ = | (a_1, b_1, c_1), ( a_2,b_2, c_2), (a_3,b_3,c_3) |`

and `Δ_1` be the determinant obtained by multiplying the elements of the first row by `k.`

Then

`Δ_1 = | (k a_1, k b_1, k c_1 ), ( a_2, b_2, c_2 ), ( a_3, b_3, c_3) |`

`=>` Expanding along first row, we get

`Δ_1 = k a_1 (b_2 c_3 – b_3 c_2) – k b_1 (a_2 c_3 – c_2 a_3) + k c_1 (a_2 b_3 – b_2 a_3)`

`= k [ a_1 (b_2 c_3 – b_3 c_2) – b_1 (a_2 c_3 – c_2 a_3) + c_1 (a_2 b_3 – b_2 a_3)]`

`= k Δ`

`color{orange}{ | (k a_1, k b_1, k c_1 ), ( a_2, b_2, c_2), ( a_3, b_3, c_3) | = k | (a_1, b_1, c_1), ( a_2,b_2 ,c_2), (a_3, b_3, c_3) |}`

`color{blue}{"Remarks :"}`

(i) By this property, we can take out any common factor from any one row or any one column of a given determinant.

(ii) If corresponding elements of any two rows (or columns) of a determinant are proportional (in the same ratio), then its value is zero. For example

`Δ = | (a_1, a_2, a_3 ), (b_1, b_2, b_3 ), ( k a_1, k a_2, k a_3) |=0 ` (rows `R_1` and `R_2` are proportional)

`color{green}{"multiplied by a constant" k, "then its value gets multiplied by" k}`

`"Verification :"` Let `Δ = | (a_1, b_1, c_1), ( a_2,b_2, c_2), (a_3,b_3,c_3) |`

and `Δ_1` be the determinant obtained by multiplying the elements of the first row by `k.`

Then

`Δ_1 = | (k a_1, k b_1, k c_1 ), ( a_2, b_2, c_2 ), ( a_3, b_3, c_3) |`

`=>` Expanding along first row, we get

`Δ_1 = k a_1 (b_2 c_3 – b_3 c_2) – k b_1 (a_2 c_3 – c_2 a_3) + k c_1 (a_2 b_3 – b_2 a_3)`

`= k [ a_1 (b_2 c_3 – b_3 c_2) – b_1 (a_2 c_3 – c_2 a_3) + c_1 (a_2 b_3 – b_2 a_3)]`

`= k Δ`

`color{orange}{ | (k a_1, k b_1, k c_1 ), ( a_2, b_2, c_2), ( a_3, b_3, c_3) | = k | (a_1, b_1, c_1), ( a_2,b_2 ,c_2), (a_3, b_3, c_3) |}`

`color{blue}{"Remarks :"}`

(i) By this property, we can take out any common factor from any one row or any one column of a given determinant.

(ii) If corresponding elements of any two rows (or columns) of a determinant are proportional (in the same ratio), then its value is zero. For example

`Δ = | (a_1, a_2, a_3 ), (b_1, b_2, b_3 ), ( k a_1, k a_2, k a_3) |=0 ` (rows `R_1` and `R_2` are proportional)

Q 3104191058

Evaluate ` | (102, 18 ,36) , (1,3,4), ( 17,3,6) |`

Class 12 Chapter 4 Example 9

Class 12 Chapter 4 Example 9

Note that ` | (102,18 , 36), (1,3, 4 ), (17 ,3,6 ) | = | ( 6 (17) , 6 (3) , 6(6) ), ( 1,3,4), ( 17,3,6) | = 6 | (17 ,3,6 ), ( 1,3,4), ( 17,3,6) | = 0`

`color{red}{"If some or all elements of a row or column of a determinant are expressed as "}`

`color{red}{"sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants."}`

For example, ` | (a_1+ lambda_1 , a_2 + lambda_2, a_3 + lambda_3 ), ( b_1, b_2, b_3 ), ( c_1, c_2, c_3) | `

`= | (a_1, a_2, a_3 ), ( b_1, b_2 , b_3), (c_1, c_2 , c_3) | + | ( lambda_1, lambda_2, lambda_3 ) , ( b_1, b_2, b_3), (c_!, c_2, c_3) |`

Verification L.H.S.` = | (a_1 + lambda_1 , a_2 +lambda_2 , a_3+lambda_3 ), ( b_1, b_2, b_3), ( c_1, c_2, c_3) |`

`=>` Expanding the determinants along the first row, we get

`Δ = (a_1 + λ_1) (b_2 c_3 – c_2 b_3) – (a_2 + λ_2) (b_1 c_3 – b_3 c_1)`

`+ (a_3 + λ_3) (b_1 c_2 – b_2 c_1)`

`= a_1 (b_2 c_3 – c_2 b_3) – a_2 (b_1 c_3 – b_3 c_1) + a_3 (b_1 c_2 – b_2 c_1)`

`+ λ_1 (b_2 c_3 – c_2 b_3) – λ_2 (b_1 c_3 – b_3 c_1) + λ_3 (b_1 c_2 – b_2 c_1)`

(by rearranging terms)

`= | (a_1, a_2, a_3), ( b_1, b_2, b_3), ( c_1, c_2, c_3) | + | (λ_1 , λ_2, λ_3 ), ( b_1 ,b_2, b_3 ), ( c_1, c_2, c_3) | = R.H.S.`

`=>` Similarly, we may verify Property 5 for other rows or columns.

`color{red}{"sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants."}`

For example, ` | (a_1+ lambda_1 , a_2 + lambda_2, a_3 + lambda_3 ), ( b_1, b_2, b_3 ), ( c_1, c_2, c_3) | `

`= | (a_1, a_2, a_3 ), ( b_1, b_2 , b_3), (c_1, c_2 , c_3) | + | ( lambda_1, lambda_2, lambda_3 ) , ( b_1, b_2, b_3), (c_!, c_2, c_3) |`

Verification L.H.S.` = | (a_1 + lambda_1 , a_2 +lambda_2 , a_3+lambda_3 ), ( b_1, b_2, b_3), ( c_1, c_2, c_3) |`

`=>` Expanding the determinants along the first row, we get

`Δ = (a_1 + λ_1) (b_2 c_3 – c_2 b_3) – (a_2 + λ_2) (b_1 c_3 – b_3 c_1)`

`+ (a_3 + λ_3) (b_1 c_2 – b_2 c_1)`

`= a_1 (b_2 c_3 – c_2 b_3) – a_2 (b_1 c_3 – b_3 c_1) + a_3 (b_1 c_2 – b_2 c_1)`

`+ λ_1 (b_2 c_3 – c_2 b_3) – λ_2 (b_1 c_3 – b_3 c_1) + λ_3 (b_1 c_2 – b_2 c_1)`

(by rearranging terms)

`= | (a_1, a_2, a_3), ( b_1, b_2, b_3), ( c_1, c_2, c_3) | + | (λ_1 , λ_2, λ_3 ), ( b_1 ,b_2, b_3 ), ( c_1, c_2, c_3) | = R.H.S.`

`=>` Similarly, we may verify Property 5 for other rows or columns.

Q 3124291151

Show that ` | (a,b,c ), ( a+2x , b +2y , c +2z), ( x,y,z ) | = 0 `

Class 12 Chapter 4 Example 10

Class 12 Chapter 4 Example 10

We have` | ( a,b,c), ( a+2x , b +2y ,c+ 2z), ( x,y,z) | = | ( a,b,c ), ( a,b,c), ( x,y,z) | + | (a,b,c), ( 2x +2y +2z), ( x,y,z ) |`

`= 0 +0 += 0 `

`color{green} ✍️` If to each element of any row or column of a determinant, the equimultiples of corresponding elements of other row (or column) are added, then value of determinant remains the same, i.e., the value of determinant remain same if we apply the operation

`color{red}{R_i → R_i + k R_j` or `C_i → C_i + k C_j}` .

`"Verification : "` Let ` Δ = | (a_1, a_2, a_3), ( b_1, b_2, b_3), (c_1, c_2, c_3) | `

and ` Δ_1 = | (a_1+k c_1 , a_2 + kc_2 , a_3 + k c_3), ( b_1, b_2, b_3 ), ( c_1, c_2, c_3) |` ,

`=>` where `Δ_1` is obtained by the operation `R_1 → R_1 + k R_3` .

`=>` Here, we have multiplied the elements of the third row `(R_3)` by a constant k and added them to the corresponding elements of the first row `(R_1)`.

Now, again

` Δ_1 = | (a_1 , a_2, a_3), ( b_1, b_2, b_3), ( c_!, c_2, c_3) | + | (k c_1, k c_2, k c_3), (b_1, b_2, b_3), ( c_1, c_2, c_3) | ` (Using Property 5)

`= Δ + 0` (since `R_1` and `R_3` are proportional)

Hence `Δ = Δ_1`

Remarks

(i) If `Δ_1` is the determinant obtained by applying `R_i → k R_i` or `C_i → k C_i` to the determinant Δ, then `Δ_1 = kΔ`.

(ii) If more than one operation like `R_i→ R_i + k R_j` is done in one step, care should be taken to see that a row that is affected in one operation should not be used in another operation. `A` similar remark applies to column operations.

`color{red}{R_i → R_i + k R_j` or `C_i → C_i + k C_j}` .

`"Verification : "` Let ` Δ = | (a_1, a_2, a_3), ( b_1, b_2, b_3), (c_1, c_2, c_3) | `

and ` Δ_1 = | (a_1+k c_1 , a_2 + kc_2 , a_3 + k c_3), ( b_1, b_2, b_3 ), ( c_1, c_2, c_3) |` ,

`=>` where `Δ_1` is obtained by the operation `R_1 → R_1 + k R_3` .

`=>` Here, we have multiplied the elements of the third row `(R_3)` by a constant k and added them to the corresponding elements of the first row `(R_1)`.

Now, again

` Δ_1 = | (a_1 , a_2, a_3), ( b_1, b_2, b_3), ( c_!, c_2, c_3) | + | (k c_1, k c_2, k c_3), (b_1, b_2, b_3), ( c_1, c_2, c_3) | ` (Using Property 5)

`= Δ + 0` (since `R_1` and `R_3` are proportional)

Hence `Δ = Δ_1`

Remarks

(i) If `Δ_1` is the determinant obtained by applying `R_i → k R_i` or `C_i → k C_i` to the determinant Δ, then `Δ_1 = kΔ`.

(ii) If more than one operation like `R_i→ R_i + k R_j` is done in one step, care should be taken to see that a row that is affected in one operation should not be used in another operation. `A` similar remark applies to column operations.

Q 3144291153

Prove that ` | ( a,a+b , a+b +c ), ( 2a , 3a +2b , 4a +3b +2c ), ( 3a , 6a +3b , 10a + 6b + 3c) | = a^3`

Class 12 Chapter 4 Example 11

Class 12 Chapter 4 Example 11

Applying operations `R_2 → R_2 – 2R_1` and `R_3 → R_3 – 3R_1` to the given

determinant Δ, we have

`Δ = | (a, a+b , a+b +c ), ( 0,a, a + b ), ( 0 ,3a , 7a +3b) |`

Now applying `R_3 → R_3 – 3R_2` , we get

`Δ = | ( a, a+b , a+b +c ), ( 0,a, 2a +b), ( 0,0, a) |`

Expanding along `C_1`, we obtain

`Δ = a | (a, 2a + b ) , ( 0 ,a) | + 0 + 0`

` = a (a^2 -0 ) = a (a^2 ) = a^3`

Q 3154291154

Without expanding, prove that

`Δ = | ( x+y , y+z, z+x ), ( z,x,y), ( 1,1,1) | = 0`

Class 12 Chapter 4 Example 12

`Δ = | ( x+y , y+z, z+x ), ( z,x,y), ( 1,1,1) | = 0`

Class 12 Chapter 4 Example 12

Applying `R_1 → R_1 + R_2` to Δ, we get

`Δ = | ( x+y +z, x+y +z , x+ y + z ), ( z ,x,y), ( 1,1,1) |`

Since the elements of `R_1` and `R_3` are proportional, Δ = 0.

Q 3164291155

Evaluate

` Δ = | ( 1,a, bc ), ( 1,b, ca), ( 1,c, ab ) |`

Class 12 Chapter 4 Example 13

` Δ = | ( 1,a, bc ), ( 1,b, ca), ( 1,c, ab ) |`

Class 12 Chapter 4 Example 13

Applying `R_2 → R_2 – R_1` and `R_3 → R_3 – R_1`, we get

`Δ = | (1,a, bc ), ( 0 , b-a , c (a-b) ) ,( 0 , c-a , b (a-c) ) |`

Taking factors (b – a) and (c – a) common from `R_2` and `R_3`, respectively, we get

`Δ = (b-a) (c-a) | ( 1,a, bc) , ( 0,1,-c), ( 0,1, -b) |`

= (b – a) (c – a) [(– b + c)] (Expanding along first column)

= (a – b) (b – c) (c – a)

Q 3114291159

Prove that ` | ( b+c ,a,a ), ( b, c+a, b ), ( c,c , a+b ) | = 4 abc`

Class 12 Chapter 4 Example 14

Class 12 Chapter 4 Example 14

Let `Δ = | ( b+c, a,a), ( b, c+a, b), ( c,c, a+b ) | `

Applying `R_1 → R_1 – R_2 – R_3` to Δ, we get

`Δ = | ( 0 , -2c , -2b ), ( b, c+a , b ), ( c,c, a+b ) |`

Expanding along `R_1`, we obtain

` Δ = 0 | ( c+a , b ), ( c, a+b ) | - (-2c) | ( b,b), ( c, a+b) | + (-2b ) | ( b , c+a ), ( c,c) |`

`= 2 c (a b + b^2 – bc) – 2 b (b c – c^2 – ac)`

`= 2 a b c + 2 cb^2 – 2 bc^2 – 2 b^2 c + 2 bc^2 + 2 abc`

`= 4 abc`

Q 3114491350

If x, y, z are different and ` Δ = | (x,x^2 , 1+x^3 ), ( y , y^2 , 1+ y^3 ), ( z , z^2 , 1+z^3) | = 0` , then

show that 1 + xyz = 0

Class 12 Chapter 4 Example 15

show that 1 + xyz = 0

Class 12 Chapter 4 Example 15

We have

` Δ = | ( x, x^2 , 1+x^3 ), ( y, y^2 , 1+ y^3 ), ( z, z^2 ,1+z^3 ) |`

`= | ( x, x^2 ,1 ), ( y , y^2 ,1), ( z, z^2 ,1) | + | ( x, x^2 , x^3 ), ( y , y^2 , y^3 ), (z, z^2 , z^3) | ` (Using Property 5)

`= (-1)^2 | (1, x, x^2 ), ( 1, y , y^2 ), (1,z, z^2 ) | + xyz | ( 1,x, x^2 ), ( 1,y, y^2 ), ( 1,z, z^2) |` (Using `C_3↔C_2` and then `C_1 ↔ C_2` )

` = | (1,x,x^2 ), (1,y, y^2 ) , ( 1,z, z^2 ) | (1+ xyz)`

`= (1+ xyz) | (1,x, x^2 ), ( 0 , y-x , y^2 - x^2 ), ( 0, z-x , z^2 - x^2) |` (Using `R_2→R_2–R_1` and `R_3 → R_3–R_1`)

Taking out common factor (y – x) from `R_2` and (z – x) from `R_3`, we get

` Δ = (1+xyz) (y-x) (z-x) | ( 1,x, x^2 ) , ( 0,1, y+x), ( 0,1, z+x) |`

`= (1 + xyz) (y – x) (z – x) (z – y)` (on expanding along `C_1`)

Since Δ = 0 and x, y, z are all different, i.e., x – y ≠ 0, y – z ≠ 0, z – x ≠ 0, we get

1 + xyz = 0

Q 3184491357

Show that

` | (1+a ,1,1 ), ( 1,1+b ,1 ), ( 1,1, 1+c) | = abc (1+ 1/a +1/b +1/c) = abc + bc + ca +ab`

Class 12 Chapter 4 Example 16

` | (1+a ,1,1 ), ( 1,1+b ,1 ), ( 1,1, 1+c) | = abc (1+ 1/a +1/b +1/c) = abc + bc + ca +ab`

Class 12 Chapter 4 Example 16

Taking out factors a,b,c common from `R_1, R_2` and `R_3` , we get

L.H.S. `= abc | (1/a +1, 1/a ,1/a), ( 1/b , 1/b+1, 1/b ), ( 1/c , 1/c , 1/c +1 ) |`

Applying `R_1→ R_1 + R_2 + R_3`, we have

` Δ = abc | (1 +1/a +1/b +1/c , 1 +1/a +1/b +1/c ,1+1/a + 1/b +1/c ), (1/b , 1/b+1 ,1/b), ( 1/c , 1/c ,1/c+1) |`

`= abc (1 + 1/a +1/b +1/c) | (1,1,1), (1/b , 1/b+1 , 1/b), ( 1/c ,1/c ,1/c+1) |`

Now applying `C_2 → C_2 – C_1, C_3 → C_3 – C_1`, we get

` Δ = abc (1 +1/a +1/b +1/c) | (1,0,0 ), (1/b , 1, 0 ), ( 1/c , 0 ,1 ) |`

`= abc (1 + 1/a +1/b +1/c) [ 1 (1-0 ) ]`

`= abc (1+ 1/a + 1/b +1/c) = abc + bc + ca + ab =RHS`