Arts (English Medium) कक्षा १२ - CBSE Question Bank Solutions for Mathematics

Using the property of determinants and without expanding, prove that:

`|(b+c, q+r, y+z),(c+a, r+p, z +x),(a+b, p+q, x + y )| = 2|(a,p,x),(b,q,y),(c, r,z)|`

By using properties of determinants, show that:

`|(0,a, -b),(-a,0, -c),(b, c,0)| = 0`

By using properties of determinants, show that:

`|(-a^2, ab, ac),(ba, -b^2, bc),(ca,cb, -c^2)| = 4a^2b^2c^2`

By using properties of determinants, show that:

`|(1,a,a^2),(1,b,b^2),(1,c,c^2)| = (a - b)(b-c)(c-a)`

By using properties of determinants, show that:

`|(1,1,1),(a,b,c),(a^3, b^3,c^3)|` = (a-b)(b-c)(c-a)(a+b+c)

By using properties of determinants, show that:

`|(x,x^2,yz),(y,y^2,zx),(z,z^2,xy)| = (x-y)(y-z)(z-x)(xy+yz+zx)`

By using properties of determinants, show that:

`|(x+4,2x,2x),(2x,x+4,2x),(2x , 2x, x+4)| = (5x + 4)(4-x)^2`

By using properties of determinants, show that:

`|(y+k,y, y),(y, y+k, y),(y, y, y+k)| = k^2(3y + k)`

By using properties of determinants, show that:

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

By using properties of determinants, show that:

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

By using properties of determinants, show that:

`|(1,x,x^2),(x^2,1,x),(x,x^2,1)| = (1-x^3)^2`

By using properties of determinants, show that:

`|(1+a^2-b^2, 2ab, -2b),(2ab, 1-a^+b^2, 2a),(2b, -2a, 1-a^2-b^2)| = (1+a^2+b^2)`

By using properties of determinants, show that:

`|(a^2+1, ab, ac),(ab, b^2+1, bc),(ca, cb, c^2+1)| = 1+a^2+b^2+c^2`

Without expanding the determinant, prove that

`|(a, a^2,bc),(b,b^2, ca),(c, c^2,ab)| = |(1, a^2, a^3),(1, b^2, b^3),(1, c^2, c^3)|`

Evaluate `|(x, y, x+y),(y, x+y, x),(x+y, x, y)|`

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

Using properties of determinants, prove that:

`|(alpha, alpha^2,beta+gamma),(beta, beta^2, gamma+alpha),(gamma, gamma^2, alpha+beta)|` =  (β – γ) (γ – α) (α – β) (α + β + γ)

Using properties of determinants, prove that:

`|(x, x^2, 1+px^3),(y, y^2, 1+py^3),(z, z^2, 1+pz^2)|` = (1 + pxyz) (x – y) (y – z) (z – x), where p is any scalar.

Using properties of determinants, prove that:

`|(3a, -a+b, -a+c),(-b+a, 3b, -b+c),(-c+a, -c+b, 3c)|`= 3(a + b + c) (ab + bc + ca)

Using properties of determinants, prove that:

`|(1, 1+p, 1+p+q),(2, 3+2p, 4+3p+2q),(3,6+3p,10+6p+3q)| =  1`