Mathematics Determinants , Determinant of a matrix of order one, Determinant of a matrix of second one and Determinant of a matrix of third For CBSE-NCERT
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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

Determinants

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.

Determinant of a matrix of order one

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

Determinant of a matrix of order two (2xx2)

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))|}
Q 3134080852

Evaluate  | ( 2,4 ), (-1,2 ) |
Class 12 Chapter 4 Example 1
Solution:

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
Solution:

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

Determinant of a matrix of order 3 × 3

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)

Expansion along second row (R_2)

|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)

Expansion along first Column (C_1)

|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."}
Q 3184080857

Evaluate the determinant Δ = | (1,2,4), ( -1,3, 0 ), ( 4,1, 0 ) |
Class 12 Chapter 4 Example 3
Solution:

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
Solution:

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
Solution:

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

Remarks

(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}`