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Text Solution

Solution :

(i) `|[x,-7],[x,5x+1]| = [x(5x+1)+7x] = x(5x+1+7)= x(5x+8)`<br>
<br>
(ii)`|[cos theta,-sintheta],[sintheta,cos theta]| = (cos^2theta+sin^2theta) = 1`<br>
<br>
(iii) `|[cos15^@,sin15^@],[sin75^@,cos75^@]|`<br>
`= cos15^@cos75^@-sin15^@sin75^@`<br>
`=cos(15^@+75^@) = cos90^@ = 0`<br>
<br>
(iv) `|[a+ib,c+id],[-c+id,a-ib]|`
`=(a+ib)(a-ib) - (-c+id)(c+id)`<br>
`=a^2+b^2-(-c^2-d^2)`<br>
`=a^2+b^2+c^2+d^2.`<br>
**Definition and Determinant of order 1 order 2**

**Determinant of order 3**

**Determinant calculation by sarrus diagram**

**Determinant of matrix order >=4**

**Definition and example**

**Definition**

**Let A be a square matrix of order n; then the sum of the product of elements of any row (column) with their cofactors is always equal to `|A|`**

**Let A be a square matrix of order n; then the sum of the product of elements of any row (column) with the cofactors of the corresponding elements of some other row (column) is 0.**

**Let A be a square matrix of order n then `|A| = |A^T|`**

**Let `A=[a_(ij)]` be a square matrix of order `n>=2` and let B be a matrix obtained from A by interchanging any two rows (columns) of A then `|B|=-|A|`**