HSC Science (General) 11th Standard - Maharashtra State Board Question Bank Solutions for Mathematics and Statistics

Find the polar co-ordinates of points whose Cartesian co-ordinates are:

`(- sqrt(3), 1)`

If A = `[(1, -3),(4, 2)], "B" = [(4, 1),(3, -2)]` show that AB ≠ BA.

If A = `[(-1, 1, 1),(2, 3, 0),(1, -3, 1)], "B" = [(2, 1, 4),(3, 0, 2),(1, 2, 1)]`. State whether AB = BA? Justify your answer.

Show that AB = BA where,

A = `[(-2, 3, -1),(-1, 2, -1),(-6, 9, -4)], "B" = [(1, 3, -1),(2, 2, -1),(3, 0, -1)]`

Show that AB = BA where, 

A = `[(costheta, - sintheta),(sintheta, costheta)], "B" = [(cosphi, -sinphi),(sinphi, cosphi)]`

If A = `[(4, 8),(-2, -4)]`, prove that A² = 0

Verify A(BC) = (AB)C of the following case:

A = `[(1, 0, 1),(2, 3, 0),(0, 4, 5)], "B" = [(2, -2),(-1, 1),(0, 3)] and "C" = [(3, 2, -1),(2, 0, -2)]`

Verify A(BC) = (AB)C of the following case:

A = `[(2, 4, 3),(-1, 3, 2)], "B" = [(2, -2),(3, 3),(-1, 1)], "C" = [(3, 1),(1, 3)]`

Verify that A(B + C) = AB + AC of the following matrix:

A = `[(4, -2),(2, 3)], "B" = [(-1, 1),(3, -2)] and "C" = [(4, 1),(2, -1)]`

Verify that A(B + C) = AB + AC of the following matrix:

A = `[(1, -1, 3),(2, 3, 2)], "B" = [(1, 0),(-2, 3),(4, 3)], "C" = [(1, 2),(-2, 0),(4, -3)]`

If A = `[(1, -2),(5, 6)], "B" = [(3, -1),(3, 7)]`, Find AB - 2I, where I is unit matrix of order 2.

If A = `[(4, 3, 2),(-1, 2, 0)], "B" = [(1, 2),(-1, 0),(1, -2)]` show that matrix AB is non singular

If A = `[(1, 2, 0),(5, 4, 2),(0, 7, -3)]`, find the product (A + I)(A − I)

A = `[(alpha, 0),(1, 1)], "B" = [(1, 0),(2, 1)]` find α, if A² = B.

If A = `[(8, 4),(10, 5)], "B" = [(5, -4),(10, -8)]` show that (A + B)² = A² + AB + B² 

If A = `[(3, 4),(-4, 3)] and "B" = [(2, 1),(-1, 2)]`, show that (A + B)(A – B) = A² – B² 

If A = `[(1, 2),(-1, -2)], "B" = [(2, "a"),(-1, "b")]` and if (A + B)² = A² + B² . find values of a and b

Find matrix X such that AX = B, where A = `[(1, -2),(-2, 1)]` and B = `[(-3),(-1)]`

Find k, if A= `[(3, -2),(4, -2)]` and if A² = kA – 2I

Find x, if `[(1, "x", 1)][(1, 2, 3),(4, 5, 6),(3, 2, 5)] [(1),(-2), (3)]` = 0