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Maths-

General

Easy

Question

If A = $open square brackets table row a b c row x y z row p q r end table close square brackets$ , B = $open square brackets table row q cell negative b end cell y row cell negative p end cell a cell negative x end cell row r cell negative c end cell z end table close square brackets$ then -

|A| = |B|
|A| = – |B|
|A| = 2|B|
A is invertible if and only if B is invertible

The correct answer is: |A| = – |B|

We have
|B| = $open vertical bar table row q cell negative b end cell y row cell negative p end cell a cell negative x end cell row r cell negative c end cell z end table close vertical bar$ = (–1) $open vertical bar table row q cell negative b end cell y row p cell negative a end cell x row r cell negative c end cell z end table close vertical bar$
= $open vertical bar table row q b y row p a x row r c z end table close vertical bar$ = – $open vertical bar table row p a x row q b y row r c z end table close vertical bar$
= $open vertical bar table row a p x row b q y row c r z end table close vertical bar$ = – $open vertical bar table row a x p row b y q row c z r end table close vertical bar$
= – |A| [using reflection property]
$therefore$ From here, we get |A| ≠ 0 if and only if |B| ≠ 0.

Related Questions to study

Let A, B, C be three square matrices of the same order, such that whenever AB = AC then B = C, if A is -

Let A, B, C be three square matrices of the same order, such that whenever AB = AC then B = C, if A is -

The inverse of a skew symmetric matrix (if it exists) is -

The inverse of a skew symmetric matrix (if it exists) is -

Let E( $alpha$ ) = $open square brackets table row cell cos to the power of 2 end exponent invisible function application alpha end cell cell cos invisible function application alpha sin invisible function application alpha end cell row cell cos invisible function application alpha sin invisible function application alpha end cell cell sin to the power of 2 end exponent invisible function application alpha end cell end table close square brackets$ . If $alpha$ and $beta$ differs by an odd multiple of $pi$ /2, then E( $alpha$ ) E( $beta$ ) is a -

Let E( $alpha$ ) = $open square brackets table row cell cos to the power of 2 end exponent invisible function application alpha end cell cell cos invisible function application alpha sin invisible function application alpha end cell row cell cos invisible function application alpha sin invisible function application alpha end cell cell sin to the power of 2 end exponent invisible function application alpha end cell end table close square brackets$ . If $alpha$ and $beta$ differs by an odd multiple of $pi$ /2, then E( $alpha$ ) E( $beta$ ) is a -

parallel

Let A and B be two 2 × 2 matrices. Consider the statements
i) AB = O $rightwards double arrow$ A = O or B = O
ii) AB = I₂ $rightwards double arrow$ A = B^–1
iii) (A + B)² = A² + 2AB + B²
Then

Let A and B be two 2 × 2 matrices. Consider the statements
i) AB = O $rightwards double arrow$ A = O or B = O
ii) AB = I₂ $rightwards double arrow$ A = B^–1
iii) (A + B)² = A² + 2AB + B²
Then

Statement-I :If $x equals cos space theta left parenthesis 1 plus sin space theta right parenthesis plus sin space theta left parenthesis 1 minus cos space theta right parenthesis$ and $y equals cos space theta left parenthesis 1 plus cos space theta right parenthesis minus sin space theta left parenthesis 1 minus sin space theta right parenthesis$ then $x plus y less or equal than 3$
Statement-II : $sum from straight r equals 1 to straight n minus 1 of cos squared space open parentheses fraction numerator straight r pi over denominator straight n end fraction close parentheses equals straight n over 2 minus 1$
Statement-III : $sin space open parentheses pi over n close parentheses plus sin space open parentheses fraction numerator 3 pi over denominator n end fraction close parentheses plus sin space open parentheses fraction numerator 5 pi over denominator n end fraction close parentheses plus midline horizontal ellipsis plus n text terms end text equals 0$

Statement-I :If $x equals cos space theta left parenthesis 1 plus sin space theta right parenthesis plus sin space theta left parenthesis 1 minus cos space theta right parenthesis$ and $y equals cos space theta left parenthesis 1 plus cos space theta right parenthesis minus sin space theta left parenthesis 1 minus sin space theta right parenthesis$ then $x plus y less or equal than 3$
Statement-II : $sum from straight r equals 1 to straight n minus 1 of cos squared space open parentheses fraction numerator straight r pi over denominator straight n end fraction close parentheses equals straight n over 2 minus 1$
Statement-III : $sin space open parentheses pi over n close parentheses plus sin space open parentheses fraction numerator 3 pi over denominator n end fraction close parentheses plus sin space open parentheses fraction numerator 5 pi over denominator n end fraction close parentheses plus midline horizontal ellipsis plus n text terms end text equals 0$

$C l subscript 2 space i s space u s e d space i n space t h e space e x t r a c t i o n space o f colon$

$C l subscript 2 space i s space u s e d space i n space t h e space e x t r a c t i o n space o f colon$

chemistry-General

parallel

Barium adipate $not stretchy ⟶ with text Dry distillation end text on top left parenthesis straight A right parenthesis not stretchy ⟶ with MeCO subscript 3 straight H on top left parenthesis straight B right parenthesis$ The compound (A) and (B) , respectively, are:

Barium adipate $not stretchy ⟶ with text Dry distillation end text on top left parenthesis straight A right parenthesis not stretchy ⟶ with MeCO subscript 3 straight H on top left parenthesis straight B right parenthesis$ The compound (A) and (B) , respectively, are:

chemistry-General

There are two possible value of A in the solution of the matrix equation $open square brackets table row cell 2 A plus 1 end cell cell negative 5 end cell row cell negative 4 end cell A end table close square brackets to the power of negative 1 end exponent$ $open square brackets table row cell A minus 5 end cell B row cell 2 A minus 2 end cell C end table close square brackets$ = $open square brackets table row 14 D row E F end table close square brackets$ , where A, B, C, D, E, F are real numbers. The absolute value of the difference of these two solutions, is:

There are two possible value of A in the solution of the matrix equation $open square brackets table row cell 2 A plus 1 end cell cell negative 5 end cell row cell negative 4 end cell A end table close square brackets to the power of negative 1 end exponent$ $open square brackets table row cell A minus 5 end cell B row cell 2 A minus 2 end cell C end table close square brackets$ = $open square brackets table row 14 D row E F end table close square brackets$ , where A, B, C, D, E, F are real numbers. The absolute value of the difference of these two solutions, is:

For a given matrix A = $open square brackets table row cell cos invisible function application theta end cell cell negative sin invisible function application theta end cell row cell sin invisible function application theta end cell cell cos invisible function application theta end cell end table close square brackets$ which of the following statement holds good:

For a given matrix A = $open square brackets table row cell cos invisible function application theta end cell cell negative sin invisible function application theta end cell row cell sin invisible function application theta end cell cell cos invisible function application theta end cell end table close square brackets$ which of the following statement holds good:

parallel

A and B are two given matrices such that the order of A is 3 × 4, If A'B and BA' are both defined then:

A and B are two given matrices such that the order of A is 3 × 4, If A'B and BA' are both defined then:

If A is an involuntary matrix given byA = $open square brackets table row 0 1 cell negative 1 end cell row 4 cell negative 3 end cell 4 row 3 cell negative 3 end cell 4 end table close square brackets$ then the inverse of $fraction numerator A over denominator 2 end fraction$ will be

If A is an involuntary matrix given byA = $open square brackets table row 0 1 cell negative 1 end cell row 4 cell negative 3 end cell 4 row 3 cell negative 3 end cell 4 end table close square brackets$ then the inverse of $fraction numerator A over denominator 2 end fraction$ will be

Matrix A = $open square brackets table row x 3 2 row 1 y 4 row 2 2 z end table close square brackets$ , If xyz = 60 and 8x + 4y + 3z = 20, then A(adj A) is equal to:

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parallel

Let A = $open square brackets table row 1 0 0 row 2 1 0 row 3 2 1 end table close square brackets$ . If U₁, U₂, U₃ are column matrices satisfying AU₁ = $open square brackets table row 1 row 0 row 0 end table close square brackets$ , AU₂ = $open square brackets table row 2 row 3 row 0 end table close square brackets$ , AU₃ = $open square brackets table row 2 row 3 row 1 end table close square brackets$ and U is 3 × 3 matrix whose columns are U₁, U₂ and U₃. Then |U| =

Let A = $open square brackets table row 1 0 0 row 2 1 0 row 3 2 1 end table close square brackets$ . If U₁, U₂, U₃ are column matrices satisfying AU₁ = $open square brackets table row 1 row 0 row 0 end table close square brackets$ , AU₂ = $open square brackets table row 2 row 3 row 0 end table close square brackets$ , AU₃ = $open square brackets table row 2 row 3 row 1 end table close square brackets$ and U is 3 × 3 matrix whose columns are U₁, U₂ and U₃. Then |U| =

If the matrix A = $open square brackets table row 8 cell – 6 end cell 2 row cell – 6 end cell 7 cell – 4 end cell row 2 cell – 4 end cell lambda end table close square brackets$ is singular, then λ equal to -

If the matrix A = $open square brackets table row 8 cell – 6 end cell 2 row cell – 6 end cell 7 cell – 4 end cell row 2 cell – 4 end cell lambda end table close square brackets$ is singular, then λ equal to -

If AB = I and B = A' then :

If AB = I and B = A' then :

parallel

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