`star` Determinants

`star` Determinant of a matrix of order one

`star` Determinant of a matrix of second one

`star` Determinant of a matrix of third

`star` Determinant of a matrix of order one

`star` Determinant of a matrix of second one

`star` Determinant of a matrix of third

`color{green} ✍️` To every square matrix `A = [a_(ij)]` of order `n,` we can associate a number (real or complex) called determinant of the square matrix A, where `a_(ij) = (i, j)^(th)` element of A.

`color{green} ✍️` If `M` is the set of square matrices, `K` is the set of numbers (real or complex) and `f : M → K` is defined by `f (A) = k,` where `A ∈ M` and `k ∈ K,` then f(A) is called the determinant of A. It is also denoted by `color{blue}{|A|}` or `color{blue}{"det" A}` or `Δ

`

If `A= [(a,b),(c,d)]` then determinant of A is written as `|A| = |(a,b),(c,d)| = det(A)`

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

(i) For matrix A, `|A|` is read as determinant of A and not modulus of A.

(ii) Only square matrices have determinants.

`color{green} ✍️` If `M` is the set of square matrices, `K` is the set of numbers (real or complex) and `f : M → K` is defined by `f (A) = k,` where `A ∈ M` and `k ∈ K,` then f(A) is called the determinant of A. It is also denoted by `color{blue}{|A|}` or `color{blue}{"det" A}` or `Δ

`

If `A= [(a,b),(c,d)]` then determinant of A is written as `|A| = |(a,b),(c,d)| = det(A)`

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

(i) For matrix A, `|A|` is read as determinant of A and not modulus of A.

(ii) Only square matrices have determinants.

`A = [a ]` be the matrix of order 1, then determinant of A is defined to be equal to `color{orange}{|A| = a}`

If `A = [(a_(11), a_(12)),(a_(21), a_(22))]` be a matrix of order 2 × 2,

then the determinant of A is defined as `color{orange}{|A| = |(a_(11), a_(12)),(a_(21), a_(22))|}`

then the determinant of A is defined as `color{orange}{|A| = |(a_(11), a_(12)),(a_(21), a_(22))|}`

Q 3134080852

Evaluate ` | ( 2,4 ), (-1,2 ) |`

Class 12 Chapter 4 Example 1

Class 12 Chapter 4 Example 1

We have ` | (2,4), (-1,2) | = 2 (2) - 4 (-1) = 4+4+` = 8 `

Q 3164080855

Evaluate ` | ( x, x+1 ), ( x-1 ,x ) |`

Class 12 Chapter 4 Example 2

Class 12 Chapter 4 Example 2

We have

` | ( x,x+1), ( x-1, x) | = x (x) – (x + 1) (x – 1) = x^2 – (x^2 – 1) = x^2 – x^2 + 1 = 1`

`color{green} ✍️` If `A = [(a_(11), a_(12),a_13),(a_(21), a_(22),a_23),(a_31,a_32,a_33)]` be a matrix of order `3 × 3, `

then the determinant of `A` is defined as `color{orange} |A| = |(a_(11), a_(12),a_13),(a_(21), a_(22),a_23),(a_31,a_32,a_33)| `

`color{red}{"Expansion of matrix "`

`color{green} ✍️` There are six ways of expanding a determinant of order 3 corresponding to each of three rows `(R_1 , R_2 and R_3)` and three columns `(C_1, C_2 and C3)` giving the same value as shown below

`|A| = |(a_11,a_12,a_13),(a_21,a_22,a_23),(a_31,a_32,a_33)|`

Expansion along first Row `(R_1)`

`color {green} text{Step 1 }` Multiply first element `a_(11) `of `R_1` by `(-1)^(1+1)` and with the second order determinant obtained by deleting the elements of first row `(R_1)` and first column `(C_1)` of `| A |` as `a_(11)` lies in `R_1` and `C_1`

i.e., `color {blue} {(-1)^(1+2) a_12 | (a_22, a_23),(a_32 ,a_33) |`

`color {green} text {Step 2 }` Multiply 2nd element `a_12` of `R_1` by `(–1)^(1 + 2)` `[(–1)^text(sum of suffixes in) a_12]` and the second order determinant obtained by deleting elements of first row `(R_1)` and 2nd column `(C_2)` of | A | as `a_12` lies in `R_1` and `C_2`,

i.e., `color {blue} {(-1)^(1 +2) a _12 | (a_21 , a _23),(a_31, a_33) |`

`color {green} text{ Step 3 }` Multiply third element `a_13` of `R_1` by `(–1)^(1 + 3)` `[(–1)^text(sum of suffixes in) a_13] ` and the second order determinant obtained by deleting elements of first row `(R_1)` and third column `(C_3)` of | A | as `a_13` lies in `R_1` and `C_3`,

i.e, `color {blue} {(-1)^(1+3) a_13 | (a_21, a_22), (a_31 ,a_32) |`

`color {green} text { Step 4 }` Now the expansion of determinant of A, that is, | A | written as sum of all three terms obtained in steps 1, 2 and 3 above is given by

`det A = |A| = (–1)^(1 + 1) a_11 | (a_22, a_23) ,( a_32 , a_33) | + (-1) ^(1+2) a_12 | (a_21, a_23 ), ( a_31 , a_33 ) | + (-1) ^(1 +3) a_13 | (a_21 ,a_22), (a_31 , a _32 ) |`

or ` |A| = a_11 (a_22 ,a_33 – a_32 a_23) – a_12 (a_21 a_33 – a_31 a_23) + a_13 (a_21 a_32 – a_31 a_22)`

`color {blue} { = a_11 a_22 a_33 – a_11 a_32 a_23 – a_12 a_21 a_33 + a_12 a_31 a_23 + a_13 a_21 a_32

– a_13 a_31 a_22}` .....................(1)

then the determinant of `A` is defined as `color{orange} |A| = |(a_(11), a_(12),a_13),(a_(21), a_(22),a_23),(a_31,a_32,a_33)| `

`color{red}{"Expansion of matrix "`

`color{green} ✍️` There are six ways of expanding a determinant of order 3 corresponding to each of three rows `(R_1 , R_2 and R_3)` and three columns `(C_1, C_2 and C3)` giving the same value as shown below

`|A| = |(a_11,a_12,a_13),(a_21,a_22,a_23),(a_31,a_32,a_33)|`

Expansion along first Row `(R_1)`

`color {green} text{Step 1 }` Multiply first element `a_(11) `of `R_1` by `(-1)^(1+1)` and with the second order determinant obtained by deleting the elements of first row `(R_1)` and first column `(C_1)` of `| A |` as `a_(11)` lies in `R_1` and `C_1`

i.e., `color {blue} {(-1)^(1+2) a_12 | (a_22, a_23),(a_32 ,a_33) |`

`color {green} text {Step 2 }` Multiply 2nd element `a_12` of `R_1` by `(–1)^(1 + 2)` `[(–1)^text(sum of suffixes in) a_12]` and the second order determinant obtained by deleting elements of first row `(R_1)` and 2nd column `(C_2)` of | A | as `a_12` lies in `R_1` and `C_2`,

i.e., `color {blue} {(-1)^(1 +2) a _12 | (a_21 , a _23),(a_31, a_33) |`

`color {green} text{ Step 3 }` Multiply third element `a_13` of `R_1` by `(–1)^(1 + 3)` `[(–1)^text(sum of suffixes in) a_13] ` and the second order determinant obtained by deleting elements of first row `(R_1)` and third column `(C_3)` of | A | as `a_13` lies in `R_1` and `C_3`,

i.e, `color {blue} {(-1)^(1+3) a_13 | (a_21, a_22), (a_31 ,a_32) |`

`color {green} text { Step 4 }` Now the expansion of determinant of A, that is, | A | written as sum of all three terms obtained in steps 1, 2 and 3 above is given by

`det A = |A| = (–1)^(1 + 1) a_11 | (a_22, a_23) ,( a_32 , a_33) | + (-1) ^(1+2) a_12 | (a_21, a_23 ), ( a_31 , a_33 ) | + (-1) ^(1 +3) a_13 | (a_21 ,a_22), (a_31 , a _32 ) |`

or ` |A| = a_11 (a_22 ,a_33 – a_32 a_23) – a_12 (a_21 a_33 – a_31 a_23) + a_13 (a_21 a_32 – a_31 a_22)`

`color {blue} { = a_11 a_22 a_33 – a_11 a_32 a_23 – a_12 a_21 a_33 + a_12 a_31 a_23 + a_13 a_21 a_32

– a_13 a_31 a_22}` .....................(1)

`|A| = | (a_11 , a_12 ,a_13 ),( a_21 , a_22 , a_23), (a_31, a_32 , a_33)|`

Expanding along `R_2`, we get

`|A | = (-1)^(2 +1) a_21 | (a_12, a_13), ( a_32 , a_33) | + ( -1)^(2 +2) a_22 | ( a_11 , a_13),( a _31 , a _33)|` `+ (-1)^(2 +3) a_23 | (a_11, a_12 ), (a_31 ,a_32)|`

`=-a_21 ( a_12 a_33 - a_32 a_13 ) + a_22 ( a_11 a_33 - a_31 a_13) - a_23 ( a_11 a_32 - a_31 a_12 )`

`|A| = - a_21 a_12 a_33 + a_21 a_32 a_13 + a_22 a_11 a_33 - a_22 a_31 a_13 - a_23 a_11 a_32 + a_23 a_31 a_12`

`= a_11 a_22 a_33 - a_11 a_23 a_32 - a_12 a_21 a_33 + a_12 a_23 a_31 + a_13 a_21 a_32 - a_13 a_31 a_22`.....(2)

Expanding along `R_2`, we get

`|A | = (-1)^(2 +1) a_21 | (a_12, a_13), ( a_32 , a_33) | + ( -1)^(2 +2) a_22 | ( a_11 , a_13),( a _31 , a _33)|` `+ (-1)^(2 +3) a_23 | (a_11, a_12 ), (a_31 ,a_32)|`

`=-a_21 ( a_12 a_33 - a_32 a_13 ) + a_22 ( a_11 a_33 - a_31 a_13) - a_23 ( a_11 a_32 - a_31 a_12 )`

`|A| = - a_21 a_12 a_33 + a_21 a_32 a_13 + a_22 a_11 a_33 - a_22 a_31 a_13 - a_23 a_11 a_32 + a_23 a_31 a_12`

`= a_11 a_22 a_33 - a_11 a_23 a_32 - a_12 a_21 a_33 + a_12 a_23 a_31 + a_13 a_21 a_32 - a_13 a_31 a_22`.....(2)

`|A| = | (a_11 , a_12 , a_13 ) , (a _21 , a_22 , a_23 ) , ( a_31 , a_32 , a_33 ) |`

`=>` By expanding along `C_1,` we get

`=> | A | = a_11(-1)^(1+1) | (a_22, a_23),(a_32 ,a_33) | + a _21 (-1)^(2 +1) | (a_12 , a_13 ), (a_32 , a_33 ) |` `+ a_31(-1)^(3+1) | (a_12, a_13 ), (a_22 , a_23 ) |`

`= a_11 (a _22 a_33 - a_23 a_32) -a_21 ( a_12 a_33 - a_13 a_32 ) + a_31 ( a_12 a_23 -a_13 a_22 )`

`|A| = a _11 a_22 a_33 - a_11 a_23 a_32 - a_21 a_12 a_33 + a_21 a_13 a_32 + a_31 a_12 a_23 - a_31 a_13 a_22`

`= a_11 a_22 a_33 - a_11 a_23 a_32 - a_12 a_21 a_33 + a_12 a_23 a_31 + a_13 a_21 a_32 - a_13 a_31 a_22` .....(3)

`=>` Clearly, all values of |A| in (1), (2) and (3) are equal. It is left as an exercise to the reader to verify that the values of |A| by expanding along `R_3, C_2` and `C_3` are equal to the value of |A| obtained in (1), (2) or (3).

`color{orange}{"Hence, expanding a determinant along any row or column gives same value."}`

`=>` By expanding along `C_1,` we get

`=> | A | = a_11(-1)^(1+1) | (a_22, a_23),(a_32 ,a_33) | + a _21 (-1)^(2 +1) | (a_12 , a_13 ), (a_32 , a_33 ) |` `+ a_31(-1)^(3+1) | (a_12, a_13 ), (a_22 , a_23 ) |`

`= a_11 (a _22 a_33 - a_23 a_32) -a_21 ( a_12 a_33 - a_13 a_32 ) + a_31 ( a_12 a_23 -a_13 a_22 )`

`|A| = a _11 a_22 a_33 - a_11 a_23 a_32 - a_21 a_12 a_33 + a_21 a_13 a_32 + a_31 a_12 a_23 - a_31 a_13 a_22`

`= a_11 a_22 a_33 - a_11 a_23 a_32 - a_12 a_21 a_33 + a_12 a_23 a_31 + a_13 a_21 a_32 - a_13 a_31 a_22` .....(3)

`=>` Clearly, all values of |A| in (1), (2) and (3) are equal. It is left as an exercise to the reader to verify that the values of |A| by expanding along `R_3, C_2` and `C_3` are equal to the value of |A| obtained in (1), (2) or (3).

`color{orange}{"Hence, expanding a determinant along any row or column gives same value."}`

Q 3184080857

Evaluate the determinant `Δ = | (1,2,4), ( -1,3, 0 ), ( 4,1, 0 ) |`

Class 12 Chapter 4 Example 3

Class 12 Chapter 4 Example 3

Note that in the third column, two entries are zero. So expanding along third

column `(C_3)`, we get

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

`= 4 (-1 -12) -0 + 0 = -52`

Q 3114080859

Evaluate `Δ = | ( 0 , sin alpha , -cos alpha ), ( - sin alpha , 0 , sin beta ), ( cos alpha , - sin beta , 0 ) |`

Class 12 Chapter 4 Example 4

Class 12 Chapter 4 Example 4

Expanding along `R_1`, we get

`Δ = 0 | ( 0 , sin beta ), ( - sin beta , 0 ) | - sin alpha | ( - sin alpha , sin beta ), ( cos alpha , 0 ) | - cos alpha | ( - sin alpha , 0 ), ( cos alpha , - sin beta) |`

= 0 – sin α (0 – sin β cos α) – cos α (sin α sin β – 0)

= sin α sin β cos α – cos α sin α sin β = 0

Q 3134180952

Find values of x for which ` | (3,x), (x,1) | = | ( 3,2), (4,1) |`

Class 12 Chapter 4 Example 5

Class 12 Chapter 4 Example 5

We have ` | ( 3,x), ( x,1) | = | ( 3,2), (4,1) |`

i.e. `3 – x^2 = 3 – 8`

i.e. `x^2 = 8`

Hence `x = ± 2 sqrt 2`

(i) For easier calculations, we shall expand the determinant along that row or column which contains maximum number of zeros.

(ii) While expanding, instead of multiplying by `(–1)^(i + j)`, we can multiply by +1 or –1 according as (i + j) is even or odd.

(iii) Let `A = [(2,2) , (4,0) ] ` and `B = [(1,1),(2, 0)]` . Then, it is easy to verify that A = 2B. Also |A| = 0 – 8 = – 8 and |B| = 0 – 2 = – 2.

Observe that, `|A| = 4 (– 2) = 2^2 |B|` or `|A| = 2^n |B|`, where n = 2 is the order of square matrices A and B.

`color{orange}{"In general, if A = kB where A and B are square matrices of order n, then" | A| = k^n | B |, where \ \ n = 1, 2, 3}`

(ii) While expanding, instead of multiplying by `(–1)^(i + j)`, we can multiply by +1 or –1 according as (i + j) is even or odd.

(iii) Let `A = [(2,2) , (4,0) ] ` and `B = [(1,1),(2, 0)]` . Then, it is easy to verify that A = 2B. Also |A| = 0 – 8 = – 8 and |B| = 0 – 2 = – 2.

Observe that, `|A| = 4 (– 2) = 2^2 |B|` or `|A| = 2^n |B|`, where n = 2 is the order of square matrices A and B.

`color{orange}{"In general, if A = kB where A and B are square matrices of order n, then" | A| = k^n | B |, where \ \ n = 1, 2, 3}`