QBManager

Matrix and Determinant

78972 If $A = [\begin{array}{ccc} 6 & - 7 & 8 \\ 1 & - 3 & 1 \\ 2 & 1 & - 4 \end{array}]$ , then $A^{- 1} =$

1

\frac{1}{80} [\begin{array}{ccc} 11 & - 20 & 17 \\ 6 & - 40 & 2 \\ 7 & - 20 & - 11 \end{array}]

2

\frac{1}{80} [\begin{array}{ccc} 11 & 6 & 7 \\ - 20 & - 40 & - 20 \\ 17 & 2 & - 11 \end{array}]

3

\frac{1}{80} [\begin{array}{ccc} - 11 & 20 & - 17 \\ - 6 & 40 & - 2 \\ - 7 & 20 & 11 \end{array}]

4

\frac{1}{80} [\begin{array}{ccc} 11 & 20 & 17 \\ 6 & 40 & - 2 \\ 7 & 20 & - 11 \end{array}]

Explanation:

(A) : We have,
$A = [\begin{array}{ccc} 6 & - 7 & 8 \\ 1 & - 3 & 1 \\ 2 & 1 & - 4 \end{array}]$
Here, $| A | = 6 (12 - 1) + 7 (- 4 - 2) + 8 (1 + 6) = 80 \neq 0$
$∵ | A | \neq 0$
$\Rightarrow A^{- 1}$ exists.
Now, $adj A = [\begin{array}{ccc} 11 & - 20 & 17 \\ 6 & - 40 & 2 \\ 7 & - 20 & - 11 \end{array}]$
We know that,
$A^{- 1} = \frac{1}{| A |} (adj A)$
$A^{- 1} = \frac{1}{80} [\begin{array}{ccc} 11 & - 20 & 17 \\ 6 & - 40 & 2 \\ 7 & - 20 & - 11 \end{array}]$

COMEDK-2020

Matrix and Determinant

78973 If $A = [\begin{array}{lll} 0 & - 1 & 2 \\ 2 & - 2 & 0 \end{array}], B = [\begin{array}{ll} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{array}]$ and $M = A B$ , then find $M^{- 1}$ .

1

[\begin{array}{cc} \frac{1}{6} & \frac{- 1}{3} \\ \frac{1}{3} & \frac{1}{3} \end{array}]

2

[\begin{array}{cc} \frac{1}{6} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{6} \end{array}]

3

[\begin{array}{cc} \frac{1}{3} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{3} \end{array}]

4

[\begin{array}{cc} \frac{1}{3} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6} \end{array}]

Explanation:

(A) : We have,
$A = [\begin{array}{lll} 0 & - 1 & 2 \\ 2 & - 2 & 0 \end{array}], B = [\begin{array}{ll} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{array}]$
It is given here, $M = A B$
$M = [\begin{array}{lll} 0 & - 1 & 2 \\ 2 & - 2 & 0 \end{array}] [\begin{array}{ll} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{array}] = [\begin{array}{cc} 1 & 2 \\ - 2 & 2 \end{array}]$
Here, $| M | = 6$ , and adj $M = [\begin{array}{cc} 1 & - 2 \\ 2 & 2 \end{array}]$
Now, $M^{- 1} = \frac{1}{6} [\begin{array}{cc} 1 & - 2 \\ 2 & 2 \end{array}]$
$M^{- 1} = [\begin{array}{cc} 1 / 6 & - 1 / 3 \\ 1 / 3 & 1 / 3 \end{array}]$

BITSAT-2006

Matrix and Determinant

78974 Inverse matrix of $[\begin{array}{cc} 2 & - 3 \\ - 4 & 2 \end{array}]$

1

- \frac{1}{8} [\begin{array}{ll} 2 & 3 \\ 4 & 2 \end{array}]

2

- \frac{1}{8} [\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}]

3

\frac{1}{8} [\begin{array}{ll} 2 & 3 \\ 4 & 2 \end{array}]

4

[\begin{array}{ll} 2 & 3 \\ 4 & 2 \end{array}]

Explanation:

(A) : Let the given matrix be A
i.e., $A = [\begin{array}{cc} 2 & - 3 \\ - 4 & 2 \end{array}]$
Here, $| A | = - 8$
And $adj A = {[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}]}^{T} = [\begin{array}{ll} 2 & 3 \\ 4 & 2 \end{array}]$
$∴ A^{- 1} = \frac{1}{| A |}$ adj $A$
$A^{- 1} = - \frac{1}{8} [\begin{array}{ll} 2 & 3 \\ 4 & 2 \end{array}]$

BITSAT-2005

Matrix and Determinant

78975 If $M (α) = [\begin{array}{ccc} \cos α & - \sin α & 0 \\ \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{array}]$ ;
$M (β) = [\begin{array}{ccc} \cos β & 0 & \sin β \\ 0 & 1 & 0 \\ - \sin β & 0 & \cos β \end{array}]$ then $[M (α) M (β)]^{- 1}$
is equal to-

1

M (β) M (α)

2

M (- α) M (- β)

3

M (- β) M (- α)

4

- M (β) M (α)

Explanation:

(C) : According to given summation,
$[M (α) M (β)]^{- 1} = M (β)^{- 1} M (α)^{- 1}]$
Now $M (α)^{- 1} = [\begin{array}{ccc} \cos α & \sin α & 0 \\ - \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{array}]$
$= [\begin{array}{ccc} \cos (- α) & - \sin (- α) & 0 \\ \sin (- α) & \cos (- α) & 0 \\ 0 & 0 & 1 \end{array}] = M (- α)$
$M (β)^{- 1} = [\begin{array}{ccc} \cos β & 0 & - \sin β \\ 0 & 1 & 0 \\ \sin β & 0 & \cos β \end{array}]$
$= [\begin{array}{ccc} \cos (- β) & 0 & \sin (- β) \\ 0 & 1 & 0 \\ - \sin (- β) & 0 & \cos (- β) \end{array}] = M (- β)$
$[M (α) M (β)]^{- 1} = M (- β) M (- α)$

BITSAT-2009

Matrix and Determinant

78972 If $A = [\begin{array}{ccc} 6 & - 7 & 8 \\ 1 & - 3 & 1 \\ 2 & 1 & - 4 \end{array}]$ , then $A^{- 1} =$

1

\frac{1}{80} [\begin{array}{ccc} 11 & - 20 & 17 \\ 6 & - 40 & 2 \\ 7 & - 20 & - 11 \end{array}]

2

\frac{1}{80} [\begin{array}{ccc} 11 & 6 & 7 \\ - 20 & - 40 & - 20 \\ 17 & 2 & - 11 \end{array}]

3

\frac{1}{80} [\begin{array}{ccc} - 11 & 20 & - 17 \\ - 6 & 40 & - 2 \\ - 7 & 20 & 11 \end{array}]

4

\frac{1}{80} [\begin{array}{ccc} 11 & 20 & 17 \\ 6 & 40 & - 2 \\ 7 & 20 & - 11 \end{array}]

Explanation:

(A) : We have,
$A = [\begin{array}{ccc} 6 & - 7 & 8 \\ 1 & - 3 & 1 \\ 2 & 1 & - 4 \end{array}]$
Here, $| A | = 6 (12 - 1) + 7 (- 4 - 2) + 8 (1 + 6) = 80 \neq 0$
$∵ | A | \neq 0$
$\Rightarrow A^{- 1}$ exists.
Now, $adj A = [\begin{array}{ccc} 11 & - 20 & 17 \\ 6 & - 40 & 2 \\ 7 & - 20 & - 11 \end{array}]$
We know that,
$A^{- 1} = \frac{1}{| A |} (adj A)$
$A^{- 1} = \frac{1}{80} [\begin{array}{ccc} 11 & - 20 & 17 \\ 6 & - 40 & 2 \\ 7 & - 20 & - 11 \end{array}]$

COMEDK-2020

Matrix and Determinant

78973 If $A = [\begin{array}{lll} 0 & - 1 & 2 \\ 2 & - 2 & 0 \end{array}], B = [\begin{array}{ll} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{array}]$ and $M = A B$ , then find $M^{- 1}$ .

1

[\begin{array}{cc} \frac{1}{6} & \frac{- 1}{3} \\ \frac{1}{3} & \frac{1}{3} \end{array}]

2

[\begin{array}{cc} \frac{1}{6} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{6} \end{array}]

3

[\begin{array}{cc} \frac{1}{3} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{3} \end{array}]

4

[\begin{array}{cc} \frac{1}{3} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6} \end{array}]

Explanation:

(A) : We have,
$A = [\begin{array}{lll} 0 & - 1 & 2 \\ 2 & - 2 & 0 \end{array}], B = [\begin{array}{ll} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{array}]$
It is given here, $M = A B$
$M = [\begin{array}{lll} 0 & - 1 & 2 \\ 2 & - 2 & 0 \end{array}] [\begin{array}{ll} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{array}] = [\begin{array}{cc} 1 & 2 \\ - 2 & 2 \end{array}]$
Here, $| M | = 6$ , and adj $M = [\begin{array}{cc} 1 & - 2 \\ 2 & 2 \end{array}]$
Now, $M^{- 1} = \frac{1}{6} [\begin{array}{cc} 1 & - 2 \\ 2 & 2 \end{array}]$
$M^{- 1} = [\begin{array}{cc} 1 / 6 & - 1 / 3 \\ 1 / 3 & 1 / 3 \end{array}]$

BITSAT-2006

Matrix and Determinant

78974 Inverse matrix of $[\begin{array}{cc} 2 & - 3 \\ - 4 & 2 \end{array}]$

1

- \frac{1}{8} [\begin{array}{ll} 2 & 3 \\ 4 & 2 \end{array}]

2

- \frac{1}{8} [\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}]

3

\frac{1}{8} [\begin{array}{ll} 2 & 3 \\ 4 & 2 \end{array}]

4

[\begin{array}{ll} 2 & 3 \\ 4 & 2 \end{array}]

Explanation:

(A) : Let the given matrix be A
i.e., $A = [\begin{array}{cc} 2 & - 3 \\ - 4 & 2 \end{array}]$
Here, $| A | = - 8$
And $adj A = {[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}]}^{T} = [\begin{array}{ll} 2 & 3 \\ 4 & 2 \end{array}]$
$∴ A^{- 1} = \frac{1}{| A |}$ adj $A$
$A^{- 1} = - \frac{1}{8} [\begin{array}{ll} 2 & 3 \\ 4 & 2 \end{array}]$

BITSAT-2005

Matrix and Determinant

78975 If $M (α) = [\begin{array}{ccc} \cos α & - \sin α & 0 \\ \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{array}]$ ;
$M (β) = [\begin{array}{ccc} \cos β & 0 & \sin β \\ 0 & 1 & 0 \\ - \sin β & 0 & \cos β \end{array}]$ then $[M (α) M (β)]^{- 1}$
is equal to-

1

M (β) M (α)

2

M (- α) M (- β)

3

M (- β) M (- α)

4

- M (β) M (α)

Explanation:

(C) : According to given summation,
$[M (α) M (β)]^{- 1} = M (β)^{- 1} M (α)^{- 1}]$
Now $M (α)^{- 1} = [\begin{array}{ccc} \cos α & \sin α & 0 \\ - \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{array}]$
$= [\begin{array}{ccc} \cos (- α) & - \sin (- α) & 0 \\ \sin (- α) & \cos (- α) & 0 \\ 0 & 0 & 1 \end{array}] = M (- α)$
$M (β)^{- 1} = [\begin{array}{ccc} \cos β & 0 & - \sin β \\ 0 & 1 & 0 \\ \sin β & 0 & \cos β \end{array}]$
$= [\begin{array}{ccc} \cos (- β) & 0 & \sin (- β) \\ 0 & 1 & 0 \\ - \sin (- β) & 0 & \cos (- β) \end{array}] = M (- β)$
$[M (α) M (β)]^{- 1} = M (- β) M (- α)$

BITSAT-2009

Matrix and Determinant

78972 If $A = [\begin{array}{ccc} 6 & - 7 & 8 \\ 1 & - 3 & 1 \\ 2 & 1 & - 4 \end{array}]$ , then $A^{- 1} =$

1

\frac{1}{80} [\begin{array}{ccc} 11 & - 20 & 17 \\ 6 & - 40 & 2 \\ 7 & - 20 & - 11 \end{array}]

2

\frac{1}{80} [\begin{array}{ccc} 11 & 6 & 7 \\ - 20 & - 40 & - 20 \\ 17 & 2 & - 11 \end{array}]

3

\frac{1}{80} [\begin{array}{ccc} - 11 & 20 & - 17 \\ - 6 & 40 & - 2 \\ - 7 & 20 & 11 \end{array}]

4

\frac{1}{80} [\begin{array}{ccc} 11 & 20 & 17 \\ 6 & 40 & - 2 \\ 7 & 20 & - 11 \end{array}]

Explanation:

(A) : We have,
$A = [\begin{array}{ccc} 6 & - 7 & 8 \\ 1 & - 3 & 1 \\ 2 & 1 & - 4 \end{array}]$
Here, $| A | = 6 (12 - 1) + 7 (- 4 - 2) + 8 (1 + 6) = 80 \neq 0$
$∵ | A | \neq 0$
$\Rightarrow A^{- 1}$ exists.
Now, $adj A = [\begin{array}{ccc} 11 & - 20 & 17 \\ 6 & - 40 & 2 \\ 7 & - 20 & - 11 \end{array}]$
We know that,
$A^{- 1} = \frac{1}{| A |} (adj A)$
$A^{- 1} = \frac{1}{80} [\begin{array}{ccc} 11 & - 20 & 17 \\ 6 & - 40 & 2 \\ 7 & - 20 & - 11 \end{array}]$

COMEDK-2020

Matrix and Determinant

78973 If $A = [\begin{array}{lll} 0 & - 1 & 2 \\ 2 & - 2 & 0 \end{array}], B = [\begin{array}{ll} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{array}]$ and $M = A B$ , then find $M^{- 1}$ .

1

[\begin{array}{cc} \frac{1}{6} & \frac{- 1}{3} \\ \frac{1}{3} & \frac{1}{3} \end{array}]

2

[\begin{array}{cc} \frac{1}{6} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{6} \end{array}]

3

[\begin{array}{cc} \frac{1}{3} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{3} \end{array}]

4

[\begin{array}{cc} \frac{1}{3} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6} \end{array}]

Explanation:

(A) : We have,
$A = [\begin{array}{lll} 0 & - 1 & 2 \\ 2 & - 2 & 0 \end{array}], B = [\begin{array}{ll} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{array}]$
It is given here, $M = A B$
$M = [\begin{array}{lll} 0 & - 1 & 2 \\ 2 & - 2 & 0 \end{array}] [\begin{array}{ll} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{array}] = [\begin{array}{cc} 1 & 2 \\ - 2 & 2 \end{array}]$
Here, $| M | = 6$ , and adj $M = [\begin{array}{cc} 1 & - 2 \\ 2 & 2 \end{array}]$
Now, $M^{- 1} = \frac{1}{6} [\begin{array}{cc} 1 & - 2 \\ 2 & 2 \end{array}]$
$M^{- 1} = [\begin{array}{cc} 1 / 6 & - 1 / 3 \\ 1 / 3 & 1 / 3 \end{array}]$

BITSAT-2006

Matrix and Determinant

78974 Inverse matrix of $[\begin{array}{cc} 2 & - 3 \\ - 4 & 2 \end{array}]$

1

- \frac{1}{8} [\begin{array}{ll} 2 & 3 \\ 4 & 2 \end{array}]

2

- \frac{1}{8} [\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}]

3

\frac{1}{8} [\begin{array}{ll} 2 & 3 \\ 4 & 2 \end{array}]

4

[\begin{array}{ll} 2 & 3 \\ 4 & 2 \end{array}]

Explanation:

(A) : Let the given matrix be A
i.e., $A = [\begin{array}{cc} 2 & - 3 \\ - 4 & 2 \end{array}]$
Here, $| A | = - 8$
And $adj A = {[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}]}^{T} = [\begin{array}{ll} 2 & 3 \\ 4 & 2 \end{array}]$
$∴ A^{- 1} = \frac{1}{| A |}$ adj $A$
$A^{- 1} = - \frac{1}{8} [\begin{array}{ll} 2 & 3 \\ 4 & 2 \end{array}]$

BITSAT-2005

Matrix and Determinant

78975 If $M (α) = [\begin{array}{ccc} \cos α & - \sin α & 0 \\ \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{array}]$ ;
$M (β) = [\begin{array}{ccc} \cos β & 0 & \sin β \\ 0 & 1 & 0 \\ - \sin β & 0 & \cos β \end{array}]$ then $[M (α) M (β)]^{- 1}$
is equal to-

1

M (β) M (α)

2

M (- α) M (- β)

3

M (- β) M (- α)

4

- M (β) M (α)

Explanation:

(C) : According to given summation,
$[M (α) M (β)]^{- 1} = M (β)^{- 1} M (α)^{- 1}]$
Now $M (α)^{- 1} = [\begin{array}{ccc} \cos α & \sin α & 0 \\ - \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{array}]$
$= [\begin{array}{ccc} \cos (- α) & - \sin (- α) & 0 \\ \sin (- α) & \cos (- α) & 0 \\ 0 & 0 & 1 \end{array}] = M (- α)$
$M (β)^{- 1} = [\begin{array}{ccc} \cos β & 0 & - \sin β \\ 0 & 1 & 0 \\ \sin β & 0 & \cos β \end{array}]$
$= [\begin{array}{ccc} \cos (- β) & 0 & \sin (- β) \\ 0 & 1 & 0 \\ - \sin (- β) & 0 & \cos (- β) \end{array}] = M (- β)$
$[M (α) M (β)]^{- 1} = M (- β) M (- α)$

BITSAT-2009

Matrix and Determinant

78972 If $A = [\begin{array}{ccc} 6 & - 7 & 8 \\ 1 & - 3 & 1 \\ 2 & 1 & - 4 \end{array}]$ , then $A^{- 1} =$

1

\frac{1}{80} [\begin{array}{ccc} 11 & - 20 & 17 \\ 6 & - 40 & 2 \\ 7 & - 20 & - 11 \end{array}]

2

\frac{1}{80} [\begin{array}{ccc} 11 & 6 & 7 \\ - 20 & - 40 & - 20 \\ 17 & 2 & - 11 \end{array}]

3

\frac{1}{80} [\begin{array}{ccc} - 11 & 20 & - 17 \\ - 6 & 40 & - 2 \\ - 7 & 20 & 11 \end{array}]

4

\frac{1}{80} [\begin{array}{ccc} 11 & 20 & 17 \\ 6 & 40 & - 2 \\ 7 & 20 & - 11 \end{array}]

Explanation:

(A) : We have,
$A = [\begin{array}{ccc} 6 & - 7 & 8 \\ 1 & - 3 & 1 \\ 2 & 1 & - 4 \end{array}]$
Here, $| A | = 6 (12 - 1) + 7 (- 4 - 2) + 8 (1 + 6) = 80 \neq 0$
$∵ | A | \neq 0$
$\Rightarrow A^{- 1}$ exists.
Now, $adj A = [\begin{array}{ccc} 11 & - 20 & 17 \\ 6 & - 40 & 2 \\ 7 & - 20 & - 11 \end{array}]$
We know that,
$A^{- 1} = \frac{1}{| A |} (adj A)$
$A^{- 1} = \frac{1}{80} [\begin{array}{ccc} 11 & - 20 & 17 \\ 6 & - 40 & 2 \\ 7 & - 20 & - 11 \end{array}]$

COMEDK-2020

Matrix and Determinant

78973 If $A = [\begin{array}{lll} 0 & - 1 & 2 \\ 2 & - 2 & 0 \end{array}], B = [\begin{array}{ll} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{array}]$ and $M = A B$ , then find $M^{- 1}$ .

1

[\begin{array}{cc} \frac{1}{6} & \frac{- 1}{3} \\ \frac{1}{3} & \frac{1}{3} \end{array}]

2

[\begin{array}{cc} \frac{1}{6} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{6} \end{array}]

3

[\begin{array}{cc} \frac{1}{3} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{3} \end{array}]

4

[\begin{array}{cc} \frac{1}{3} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6} \end{array}]

Explanation:

(A) : We have,
$A = [\begin{array}{lll} 0 & - 1 & 2 \\ 2 & - 2 & 0 \end{array}], B = [\begin{array}{ll} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{array}]$
It is given here, $M = A B$
$M = [\begin{array}{lll} 0 & - 1 & 2 \\ 2 & - 2 & 0 \end{array}] [\begin{array}{ll} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{array}] = [\begin{array}{cc} 1 & 2 \\ - 2 & 2 \end{array}]$
Here, $| M | = 6$ , and adj $M = [\begin{array}{cc} 1 & - 2 \\ 2 & 2 \end{array}]$
Now, $M^{- 1} = \frac{1}{6} [\begin{array}{cc} 1 & - 2 \\ 2 & 2 \end{array}]$
$M^{- 1} = [\begin{array}{cc} 1 / 6 & - 1 / 3 \\ 1 / 3 & 1 / 3 \end{array}]$

BITSAT-2006

Matrix and Determinant

78974 Inverse matrix of $[\begin{array}{cc} 2 & - 3 \\ - 4 & 2 \end{array}]$

1

- \frac{1}{8} [\begin{array}{ll} 2 & 3 \\ 4 & 2 \end{array}]

2

- \frac{1}{8} [\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}]

3

\frac{1}{8} [\begin{array}{ll} 2 & 3 \\ 4 & 2 \end{array}]

4

[\begin{array}{ll} 2 & 3 \\ 4 & 2 \end{array}]

Explanation:

(A) : Let the given matrix be A
i.e., $A = [\begin{array}{cc} 2 & - 3 \\ - 4 & 2 \end{array}]$
Here, $| A | = - 8$
And $adj A = {[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}]}^{T} = [\begin{array}{ll} 2 & 3 \\ 4 & 2 \end{array}]$
$∴ A^{- 1} = \frac{1}{| A |}$ adj $A$
$A^{- 1} = - \frac{1}{8} [\begin{array}{ll} 2 & 3 \\ 4 & 2 \end{array}]$

BITSAT-2005

Matrix and Determinant

78975 If $M (α) = [\begin{array}{ccc} \cos α & - \sin α & 0 \\ \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{array}]$ ;
$M (β) = [\begin{array}{ccc} \cos β & 0 & \sin β \\ 0 & 1 & 0 \\ - \sin β & 0 & \cos β \end{array}]$ then $[M (α) M (β)]^{- 1}$
is equal to-

1

M (β) M (α)

2

M (- α) M (- β)

3

M (- β) M (- α)

4

- M (β) M (α)

Explanation:

(C) : According to given summation,
$[M (α) M (β)]^{- 1} = M (β)^{- 1} M (α)^{- 1}]$
Now $M (α)^{- 1} = [\begin{array}{ccc} \cos α & \sin α & 0 \\ - \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{array}]$
$= [\begin{array}{ccc} \cos (- α) & - \sin (- α) & 0 \\ \sin (- α) & \cos (- α) & 0 \\ 0 & 0 & 1 \end{array}] = M (- α)$
$M (β)^{- 1} = [\begin{array}{ccc} \cos β & 0 & - \sin β \\ 0 & 1 & 0 \\ \sin β & 0 & \cos β \end{array}]$
$= [\begin{array}{ccc} \cos (- β) & 0 & \sin (- β) \\ 0 & 1 & 0 \\ - \sin (- β) & 0 & \cos (- β) \end{array}] = M (- β)$
$[M (α) M (β)]^{- 1} = M (- β) M (- α)$

BITSAT-2009