The inverse of the matrix 1 0 0 3 3 0 5 2 - 1 is

- 1 3 - 3 0 0 3 - 1 0 - 9 - 2 3

- 1 3 3 0 0 3 - 1 0 - 9 - 2 - 3

- 1 3 - 3 0 0 - 3 - 1 0 - 9 - 2 3

HARD

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If $B_{0} = [\begin{matrix} - 4 & - 3 & - 3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{matrix}]$ , $B_{n} = a d j (B_{n - 1}), \forall n \in N$ and $I$ is an identity matrix of order $3$ , then $B_{1} + B_{3} + B_{5} + B_{7} + B_{9}$ is equal to

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Important Questions on Matrices and Determinants

EASY

Mathematics>Algebra>Matrices and Determinants>Inverse of a Matrix

A = [\begin{matrix} 2 & 2 \\ 9 & 4 \end{matrix}]

and

I = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}],

then

10 A^{- 1}

, is equal to.

MEDIUM

Mathematics>Algebra>Matrices and Determinants>Inverse of a Matrix

Inverse of the matrix

[\begin{matrix} \cos 2 θ & - \sin 2 θ \\ \sin 2 θ & \cos 2 θ \end{matrix}]

EASY

Mathematics>Algebra>Matrices and Determinants>Inverse of a Matrix

A

is a non-singular matrix and

(A + I) (A - I) = 0

then

A + A^{- 1} = \dots \dots

MEDIUM

Mathematics>Algebra>Matrices and Determinants>Inverse of a Matrix

A^{- 1} = \frac{1}{11} [\begin{matrix} - 3 & 4 \\ 5 & - 3 \end{matrix}]

, then

A =

EASY

Mathematics>Algebra>Matrices and Determinants>Inverse of a Matrix

A = [\begin{matrix} - 1 & 2 \\ 2 & - 1 \end{matrix}]

and

B = [\begin{matrix} 3 \\ 1 \end{matrix}], A X = B

, then

X

is equal to

EASY

Mathematics>Algebra>Matrices and Determinants>Inverse of a Matrix

The inverse of matrix is

[\begin{matrix} 2 & 5 & 0 \\ 0 & 1 & 1 \\ - 1 & 0 & 3 \end{matrix}]

EASY

Mathematics>Algebra>Matrices and Determinants>Inverse of a Matrix

B = [\begin{matrix} 5 & 2 α & 1 \\ 0 & 2 & 1 \\ α & 3 & - 1 \end{matrix}]

is the inverse of a

3 \times 3

matix

A,

then the sum of all values of

α

for which

d e t (A) + 1 = 0,

is:

MEDIUM

Mathematics>Algebra>Matrices and Determinants>Inverse of a Matrix

A = [\begin{matrix} 2 & - 1 \\ - 1 & 2 \end{matrix}],

such that

A^{2} - 4 A + 3 I = 0,

then

A^{- 1} =

HARD

Mathematics>Algebra>Matrices and Determinants>Inverse of a Matrix

The inverse of the matrix

[\begin{matrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & - 1 \end{matrix}]

EASY

Mathematics>Algebra>Matrices and Determinants>Inverse of a Matrix

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

,

then the sum of all the diagonal entries of

A^{- 1}

EASY

Mathematics>Algebra>Matrices and Determinants>Inverse of a Matrix

(\begin{array}{l} 2 & 1 \\ 3 & 2 \end{array}) A = (\begin{array}{l} 1 & 0 \\ 0 & 1 \end{array})

, then the matrix

A

EASY

Mathematics>Algebra>Matrices and Determinants>Inverse of a Matrix

Inverse of a diagonal non-singular matrix is

EASY

Mathematics>Algebra>Matrices and Determinants>Inverse of a Matrix

A = [\begin{array}{r} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{array}]

then

{(A^{2})}^{- 1} =

MEDIUM

Mathematics>Algebra>Matrices and Determinants>Inverse of a Matrix

Let

M = [\begin{matrix} \sin^{4} θ & - 1 - \sin^{2} θ \\ 1 + \cos^{2} θ & \cos^{4} θ \end{matrix}] = α I + β M^{- 1},

where

α = α (θ)

and

β = β (θ)

are real number, and

I

is the

2 \times 2

identity matrix. If

α^{*}

is the minimum of the set {

α (θ) : θ \in [0, 2 π)}

and

β^{*}

is the minimum of the set

\{β (θ) : θ \in [0, 2 π)\},

then the value of

α^{*} + β^{*}

MEDIUM

Mathematics>Algebra>Matrices and Determinants>Inverse of a Matrix

Which of the following matrix is invertible? $A_{1} = [\begin{array}{r} 4 & 2 \\ 2 & 1 \end{array}]$ , $A_{2} = [\begin{matrix} - 1 & - 2 & 3 \\ 4 & 5 & 7 \\ 2 & 4 & - 6 \end{matrix}]$ , $A_{3} = [\begin{matrix} 1 & 0 & 0 \\ 5 & 2 & 1 \\ 7 & 2 & 1 \end{matrix}]$ , $A_{4} = [\begin{array}{l} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{array}]$ .

MEDIUM

Mathematics>Algebra>Matrices and Determinants>Inverse of a Matrix

The values of

t

such that the matrix

[\begin{matrix} 1 & 3 & 2 \\ 2 & 5 & t \\ 4 & 7 - t & - 6 \end{matrix}]

has no inverse, are

MEDIUM

Mathematics>Algebra>Matrices and Determinants>Inverse of a Matrix

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

and

B = [\begin{matrix} 2 & - 3 \\ - 1 & 2 \end{matrix}],

then

{(B^{- 1} A^{- 1})}^{- 1} =

EASY

Mathematics>Algebra>Matrices and Determinants>Inverse of a Matrix

A = [\begin{matrix} 2 x & 0 \\ x & x \end{matrix}]

and

A^{- 1} = [\begin{matrix} 1 & 0 \\ - 1 & 2 \end{matrix}]

,

then

x =

EASY

Mathematics>Algebra>Matrices and Determinants>Inverse of a Matrix

The inverse of the matrix

[\begin{matrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{matrix}]

MEDIUM

Mathematics>Algebra>Matrices and Determinants>Inverse of a Matrix

A = [\begin{matrix} e^{t} & e^{- t} c o s t & e^{- t} \sin t \\ e^{t} & - e^{- t} \cos t - e^{- t} \sin t & - e^{- t} \sin t + e^{- t} \cos t \\ e^{t} & {2 e}^{- t} \sin t & - 2 e^{- t} \cos t \end{matrix}],

then

A

is:

If B0=-4-3-3 1 0 1 4 4 3, Bn=adjBn-1, ∀n∈N and I is an identity matrix of order 3, then B1+B3+B5+B7+B9 is equal to

Important Questions on Matrices and Determinants

Learn and Prepare for any exam you want !

If $B_{0} = [\begin{matrix} - 4 & - 3 & - 3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{matrix}]$ , $B_{n} = a d j (B_{n - 1}), \forall n \in N$ and $I$ is an identity matrix of order $3$ , then $B_{1} + B_{3} + B_{5} + B_{7} + B_{9}$ is equal to