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Assertion : There are only finitely many 2 × 2 matrices which commute with the matrix
$open square brackets table attributes columnalign center center columnspacing 1em end attributes row 1 2 row cell negative 1 end cell cell negative 1 end cell end table close square brackets$
Reason : If A is non-singular then it commutes with I, adj A and A^–1

If both (A) and (R) are true, and (R) is the correct explanation of (A).
If both (A) and (R) are true but (R) is not the correct explanation of (A).
If (A) is true but (R) is false.
If (A) is false but (R) is true.

The correct answer is: If (A) is false but (R) is true.

The reason R is true since
AI = IA, AA^–1 = A^–1A = I, A|adj A| = |adj. A|A
But a matrix can commute with general order matrices which may be infinite in number.
Let B = $open square brackets table attributes columnalign left left columnspacing 1em end attributes row cell a b end cell row cell c d end cell end table close square brackets$ be a matrix which commute with A then AB = BA
$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell not stretchy rightwards double arrow open square brackets table attributes columnalign center center columnspacing 1em end attributes row 1 2 row cell negative 1 end cell cell negative 1 end cell end table close square brackets open square brackets table attributes columnalign left left columnspacing 1em end attributes row a b row c d end table close square brackets equals open square brackets table attributes columnalign left left columnspacing 1em end attributes row a b row c d end table close square brackets open square brackets table attributes columnalign center center columnspacing 1em end attributes row 1 2 row cell negative 1 end cell cell negative 1 end cell end table close square brackets end cell row cell equals open square brackets table attributes columnalign center center columnspacing 1em end attributes row cell a plus 2 c end cell cell b plus 2 d end cell row cell negative a minus c end cell cell negative b minus d end cell end table close square brackets equals open square brackets table attributes columnalign left left columnspacing 1em end attributes row cell a minus b end cell cell 2 a minus b end cell row cell c minus d end cell cell 2 c minus d end cell end table close square brackets end cell row cell not stretchy rightwards double arrow a plus 2 c equals a minus b comma b plus 2 d equals 2 a minus b comma negative a minus c end cell row cell equals c minus d comma negative b minus d equals 2 c minus d end cell end table$
The above four relations are equivalent to only two independent relations
a – d = b, b + 2c = 0
If d = $lambda$ , then a = b + $lambda$ = –2c + $lambda$
Thus, $open square brackets table attributes columnalign center center columnspacing 1em end attributes row cell lambda minus 2 c end cell cell negative 2 c end cell row c lambda end table close square brackets$ are all possible 2 × 2 matrices which commute with given matrix A = $open square brackets table attributes columnalign center center columnspacing 1em end attributes row 1 2 row cell negative 1 end cell cell negative 1 end cell end table close square brackets$
$lambda$ and c being any arbitrary complex numbers. Thus assertion is therefore false.

Related Questions to study

Consider the system of equationsx – 2y + 3z = –1–x + y – 2z = kx – 3y + 4z = 1
Assertion : The system of equations has no solution for k $not equal to$ 3.and
Reason : The determinant $open vertical bar table row 1 3 cell negative 1 end cell row cell negative 1 end cell cell negative 2 end cell k row 1 4 1 end table close vertical bar$ $not equal to$ 0, for k $not equal to$ 3.

Consider the system of equationsx – 2y + 3z = –1–x + y – 2z = kx – 3y + 4z = 1
Assertion : The system of equations has no solution for k $not equal to$ 3.and
Reason : The determinant $open vertical bar table row 1 3 cell negative 1 end cell row cell negative 1 end cell cell negative 2 end cell k row 1 4 1 end table close vertical bar$ $not equal to$ 0, for k $not equal to$ 3.

Assertion : If a, b, c are distinct and x, y, z are not all zero given that ax + by + cz = 0, bx + cy + az = 0, cx + ay + bz = 0 then a + b + c $not equal to$ 0
Reason : a² + b² + c² > ab + bc + ca if a, b, c are distinct

Assertion : If a, b, c are distinct and x, y, z are not all zero given that ax + by + cz = 0, bx + cy + az = 0, cx + ay + bz = 0 then a + b + c $not equal to$ 0
Reason : a² + b² + c² > ab + bc + ca if a, b, c are distinct

Assertion : If A is a skew symmetric of order 3 then its determinant should be zero.
Reason : If A is square matrix then det A = det $A to the power of straight prime$ = det (– $A to the power of straight prime$ )

Assertion : If A is a skew symmetric of order 3 then its determinant should be zero.
Reason : If A is square matrix then det A = det $A to the power of straight prime$ = det (– $A to the power of straight prime$ )

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Assertion : There are only finitely many 2 ×2 matrices which commute with the matrix $open square brackets table row 1 2 row cell negative 1 end cell cell negative 1 end cell end table close square brackets$
Reason : If A is non-singular then it commutes with I, Adj A and A^–1.

Assertion : There are only finitely many 2 ×2 matrices which commute with the matrix $open square brackets table row 1 2 row cell negative 1 end cell cell negative 1 end cell end table close square brackets$
Reason : If A is non-singular then it commutes with I, Adj A and A^–1.

Let a matrix A = $open square brackets table row 1 1 row 0 1 end table close square brackets$ & P = $open square brackets table row cell fraction numerator square root of 3 over denominator 2 end fraction end cell cell fraction numerator 1 over denominator 2 end fraction end cell row cell negative fraction numerator 1 over denominator 2 end fraction end cell cell fraction numerator square root of 3 over denominator 2 end fraction end cell end table close square brackets$ Q = PAP^T where P^T is transpose of matrix P. Find PT Q2005 P is

Let a matrix A = $open square brackets table row 1 1 row 0 1 end table close square brackets$ & P = $open square brackets table row cell fraction numerator square root of 3 over denominator 2 end fraction end cell cell fraction numerator 1 over denominator 2 end fraction end cell row cell negative fraction numerator 1 over denominator 2 end fraction end cell cell fraction numerator square root of 3 over denominator 2 end fraction end cell end table close square brackets$ Q = PAP^T where P^T is transpose of matrix P. Find PT Q2005 P is

Let A = $open square brackets table row 1 0 0 row 0 1 1 row 0 cell negative 2 end cell 4 end table close square brackets$ & I = $open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets$ and A–1 = $fraction numerator 1 over denominator 6 end fraction$ [A²+ cA + dI], find ordered pair (c, d) ?]

Let A = $open square brackets table row 1 0 0 row 0 1 1 row 0 cell negative 2 end cell 4 end table close square brackets$ & I = $open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets$ and A–1 = $fraction numerator 1 over denominator 6 end fraction$ [A²+ cA + dI], find ordered pair (c, d) ?]

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$open square brackets table row alpha 2 row 2 alpha end table close square brackets$ = A & | A³ | = 125, then $alpha$ is -

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If A = $open square brackets table row alpha 0 row 1 1 end table close square brackets$ , B = $open square brackets table row 1 0 row 5 1 end table close square brackets$ and A² = B, then

Let A be a square matrix all of whose entries are integers. Then which one of the following is true ?

Let A be a square matrix all of whose entries are integers. Then which one of the following is true ?

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Let A = $open square brackets table row 5 cell 5 alpha end cell alpha row 0 alpha cell 5 alpha end cell row 0 0 5 end table close square brackets$ If |A2| = 25, then | $alpha$ | equals

Let A = $open square brackets table row 5 cell 5 alpha end cell alpha row 0 alpha cell 5 alpha end cell row 0 0 5 end table close square brackets$ If |A2| = 25, then | $alpha$ | equals

Let A = $open parentheses table row 1 2 row 3 4 end table close parentheses$ and B = $open parentheses table row a 0 row 0 b end table close parentheses$ , a, b $element of$ N. Then

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If A and B are square matrices of size n × n such that A² – B² = (A – B) (A + B), then which of the following will be always true –

If A and B are square matrices of size n × n such that A² – B² = (A – B) (A + B), then which of the following will be always true –

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If A = $open square brackets table row 1 0 row 1 1 end table close square brackets$ and I = $open square brackets table row 1 0 row 0 1 end table close square brackets$ , then which one of the following holds for all n $greater or equal than$ 1, by the principle of mathematical induction -

If A = $open square brackets table row 1 0 row 1 1 end table close square brackets$ and I = $open square brackets table row 1 0 row 0 1 end table close square brackets$ , then which one of the following holds for all n $greater or equal than$ 1, by the principle of mathematical induction -

If A² – A + I = 0, then the inverse of A is

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Let A = $open parentheses table row 1 cell negative 1 end cell 1 row 2 1 cell negative 3 end cell row 1 1 1 end table close parentheses$ and (10) B = $open parentheses table row 4 2 2 row cell negative 5 end cell 0 alpha row 1 cell negative 2 end cell 3 end table close parentheses$ . If B is the inverse of matrix A, then $alpha$ is

Let A = $open parentheses table row 1 cell negative 1 end cell 1 row 2 1 cell negative 3 end cell row 1 1 1 end table close parentheses$ and (10) B = $open parentheses table row 4 2 2 row cell negative 5 end cell 0 alpha row 1 cell negative 2 end cell 3 end table close parentheses$ . If B is the inverse of matrix A, then $alpha$ is

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