Mathematics Crash Course BITSAT>Chapter 17 - Matrices and Determinants>Exercise 1>Q 89

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Let $X = [\begin{array}{l} x \\ y \\ z \end{array}], D = [\begin{array}{l} 3 \\ 5 \\ 11 \end{array}]$ and $A = [\begin{array}{r} 1 & - 1 & - 2 \\ 2 & 1 & 1 \\ 4 & - 1 & - 2 \end{array}]$ , if $X = A^{- 1} D$ , then $X$ is equal to

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Important Questions on Matrices and Determinants

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IMPORTANT

Mathematics Crash Course BITSAT>Chapter 17 - Matrices and Determinants>Exercise 1>Q 90

A = [\begin{array}{l} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & - 2 & 4 \end{array}], I = [\begin{array}{l} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]

,

A^{- 1} = \frac{1}{6} (A^{2} + c A + d I)

where

c, d \in R,

then pair of values

(c, d)

is

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Mathematics Crash Course BITSAT>Chapter 17 - Matrices and Determinants>Exercise 1>Q 91

If

A

and

B

are square matrices of the same order and

A

is non-singular, then for a positive integer

n, {(A^{- 1} B A)}^{n}

is equal to

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Mathematics Crash Course BITSAT>Chapter 17 - Matrices and Determinants>Exercise 1>Q 92

Given

A = [\begin{matrix} 0 & - \tan α \\ \tan α & 0 \end{matrix}]

and

B (α) = [\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}],

then

(I + A) (I - A)^{- 1}

is equal to

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Mathematics Crash Course BITSAT>Chapter 17 - Matrices and Determinants>Exercise 1>Q 93

Let

M

and

N

be two

3 \times 3

skew-symmetric matrices such that

M N = N M .

If

P^{T}

denotes the transpose of

P

, then

M^{2} N^{2} {(M^{T} N)}^{- 1} {(M N^{- 1})}^{T}

is equal to

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Mathematics Crash Course BITSAT>Chapter 17 - Matrices and Determinants>Exercise 1>Q 94

If

A

and

a d j A

are non-singular square matrices of order

n

, then

a d j (A^{- 1}) =

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Mathematics Crash Course BITSAT>Chapter 17 - Matrices and Determinants>Exercise 1>Q 95

If

A

is a square matrix, then what is

adj (A^{- 1}) - {(adj A)}^{- 1}

equal to?

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Mathematics Crash Course BITSAT>Chapter 17 - Matrices and Determinants>Exercise 1>Q 96

If

[\begin{array}{lr} 1 & - \tan θ \\ \tan θ & 1 \end{array}] {[\begin{matrix} 1 & \tan θ \\ - \tan θ & 1 \end{matrix}]}^{- 1} = [\begin{array}{r} a & - b \\ b & a \end{array}]

, then

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Mathematics Crash Course BITSAT>Chapter 17 - Matrices and Determinants>Exercise 1>Q 97

The inverse of the matrix

A = [\begin{array}{l} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \end{array}]

is