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Determinants

Multiple Choice Questions

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

1/5

straight A squared space equals space open square brackets table row 5 cell 5 straight alpha end cell straight alpha row 0 straight alpha cell 5 straight alpha end cell row 0 0 5 end table close square brackets open square brackets table row 5 cell 5 straight alpha end cell straight alpha row 0 straight alpha cell 5 straight alpha end cell row 0 0 5 end table close square brackets straight A squared space equals space open square brackets table row 25 cell 25 straight alpha plus 5 straight alpha squared end cell cell 5 straight alpha end cell row 0 cell straight alpha squared end cell cell 5 straight alpha squared plus 25 straight alpha end cell row 0 0 25 end table close square brackets 625 space straight alpha squared space equals space 25 rightwards double arrow space vertical line straight alpha vertical line space equals space 1 fifth

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

If det A = ± 1, then A^–1 exists but all its entries are not necessarily integers
If detA ≠ ± 1, then A^–1 exists and all its entries are non-integers
If detA = ± 1, then A^–1 exists and all its entries are integers
If detA = ± 1, then A–1 need not exist

If detA = ± 1, then A^–1 exists and all its entries are integers

Each entry of A is integer, so the cofactor of every entry is an integer and hence each entry in the adjoint of matrix A is integer. Now detA = ± 1 and A^–1 =(1/ det(A)) (adj A)

⇒ all entries in A^–1 are integers

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Let a, b, c be any real numbers. Suppose that there are real numbers x, y, z not all zero such that x = cy + bz, y = az + cx and z = bx + ay. Then a² + b² + c² + 2abc is equal to

2
-1
0
1

The system of equations x – cy – bz = 0, cx – y + az = 0 and bx + ay – z = 0 have non-trivial solution if

$open vertical bar table row 1 cell negative straight c end cell cell negative straight b end cell row straight c cell negative 1 end cell straight a row straight b straight a cell negative 1 end cell end table close vertical bar space equals space 0 space rightwards double arrow space 1 left parenthesis space 1 minus straight a squared right parenthesis space plus space straight c left parenthesis negative straight c minus ab right parenthesis minus straight b left parenthesis ca plus straight b right parenthesis space equals space 0 space rightwards double arrow space straight a squared space plus straight b squared space plus space straight c squared space plus space 2 abc space equals space 1$

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If D = $open vertical bar table row 1 1 1 row 1 cell 1 plus straight x end cell 1 row 1 1 cell 1 plus straight y end cell end table close vertical bar$ for x ≠ 0, y ≠ 0 then D is

divisible by neither x nor y
divisible by both x and y
divisible by x but not y
divisible by y but not x

divisible by both x and y

straight D space equals space open vertical bar table row 1 1 1 row 1 cell 1 plus straight x end cell 1 row 1 1 cell 1 plus straight y end cell end table close vertical bar

C₂ → C₂ – C₁ & C₃ → C₃ – C₁

open vertical bar table row 1 0 0 row 1 straight x 0 row 1 0 straight y end table close vertical bar space equals space xy

Hence D is divisible by both x and y.

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The system of equations

x + y + z = 0

2x + 3y + z = 0

and x + 2y = 0

has

a unique solution; x = 0, y = 0, z = 0
infinite solutions
no solution
finite number of non-zero solutions

infinite solutions

The given system of equations are

x + y + z = 0,

2x + 3y + z = 0,

and x + 2y = 0

$Here, |\begin{matrix} 1 & 1 & 1 \\ 2 & 3 & 1 \\ 1 & 2 & 0 \end{matrix}| = 1 (0 - 2) - 1 (0 - 1) + 1 (4 - 3) = - 2 + 1 + 1 = 0 ∴ This system has infinite solutions .$

If x = - 5 is a root of $|\begin{matrix} 2 x + 1 & 4 & 8 \\ 2 & 2 x & 2 \\ 7 & 6 & 2 x \end{matrix}|$ = 0, then the other roots are :

3, 3.5
1, 3.5
1, 7
2, 7

1, 3.5

$|\begin{matrix} 2 x + 1 & 4 & 8 \\ 2 & 2 x & 2 \\ 7 & 6 & 2 x \end{matrix}| = 0 \Rightarrow |\begin{matrix} 2 x + 10 & 2 x + 10 & 2 x + 10 \\ 2 & 2 x & 2 \\ 7 & 6 & 2 x \end{matrix}| = 0 (R_{1} \to R_{1} + R_{2} + R_{3}) \Rightarrow (2 x + 10) |\begin{matrix} 1 & 1 & 1 \\ 2 & 2 x & 2 \\ 7 & 6 & 2 x \end{matrix}| = 0 \Rightarrow (2 x + 10) |\begin{matrix} 1 & 0 & 0 \\ 2 & 2 x - 2 & 0 \\ 7 & - 1 & 2 x - 7 \end{matrix}| = 0 (C_{3} \to C_{3} - C_{1} and C_{2} \to C_{2} - C_{1}) \Rightarrow (2 x + 10) (2 x - 2) (2 x - 7) = 0 \Rightarrow x = - 5, 1, \frac{7}{2} Hence, other roots are 1 and \frac{7}{2} .$

The simultaneous equations Kx + 2y - z = 1, (K - I)y - 2z = 2 and (K + 2)z = 3 have only one solution when :

K = - 2
K = - 1
K = 0
K = 1

K = - 1

The system of given equations are

Kx + 2y - z = 1 ...(i)

(K - 1)y - 2z = 2 ...(ii)

and (K + 2)z = 3 ...(iii)

This system of equations has a unique solution, if

$|\begin{matrix} K & 2 & - 1 \\ 0 & K - 1 & - 2 \\ 0 & 0 & K + 2 \end{matrix}| \neq 0 \Rightarrow (K + 2) |\begin{matrix} K & 2 \\ 0 & K - 1 \end{matrix}| \neq 0 \Rightarrow (K + 2) (K) (K - 1) \neq 0 \Rightarrow K \neq - 2, 0, 1 i . e ., K = - 1, is a required answer .$

Let A be a 2 × 2 matrix with real entries. Let I be the 2 × 2 identity matrix. Denote by tr (A), the sum of diagonal entries of A. Assume that A₂= I.
Statement −1: If A ≠ I and A ≠ − I, then det A = − 1.
Statement −2: If A ≠ I and A ≠ − I, then tr (A) ≠ 0.

Statement −1 is false, Statement −2 is true
Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1
Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1.
Statement − 1 is true, Statement − 2 is false.

Statement − 1 is true, Statement − 2 is false.

Let space straight A space equals space open square brackets table row straight a straight b row straight c straight d end table close square brackets space so space that space straight A squared space equals space open square brackets table row cell straight a squared plus bc end cell cell ab space plus space bd end cell row cell ac space plus dc end cell cell bc plus straight d squared end cell end table close square brackets space equals space open square brackets table row 1 0 row 0 1 end table close square brackets rightwards double arrow space straight a squared space plus bc space equals space 1 space equals space bc plus straight d squared and space left parenthesis straight a plus straight d right parenthesis straight c space equals space 0 space left parenthesis straight a plus straight d right parenthesis straight b. Since space space straight A space not equal to space straight I comma space straight A space not equal to space 1 comma space straight a space equals space minus straight d space and space hence space det space straight A space equals space open vertical bar table row cell square root of 1 minus bc end root end cell straight b row straight c cell negative square root of 1 minus bc end root end cell end table close vertical bar space equals space minus 1 plus bc minus bc space equals space minus 1 Statement space 1 space is space true But space straight A space equals space 0 space space and space hence space statement space 2 space is space false.

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If a₁, a₂, a₃,…, a_n,… are in G.P., then the determinant

$space increment space equals space open vertical bar table row cell log space straight a subscript straight n end cell cell log space straight a subscript straight n plus 1 end subscript end cell cell log space straight a subscript straight n plus 2 end subscript end cell row cell log space straight a subscript straight n plus 3 end subscript end cell cell log space straight a subscript straight n plus 4 end subscript end cell cell log space straight a subscript straight n plus 5 end subscript end cell row cell log space straight a subscript straight n plus 6 end subscript end cell cell log space straight a subscript straight n plus 7 end subscript end cell cell log space straight a subscript straight n plus 8 end subscript end cell end table close vertical bar space is space equal space to$

C₁ – C₂, C₂ – C₃ two rows becomes identical Answer: 0

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10.

Let $straight A space equals space open parentheses table row 1 2 row 3 4 end table close parentheses space and space straight B space equals open parentheses table row straight a 0 row 0 straight b end table close parentheses$ a , b ∈ N. Then

there cannot exist any B such that AB = BA
there exist more than one but finite number of B’s such that AB = BA
there exists exactly one B such that AB = BA
there exist infinitely many B’s such that AB = BA

there exist infinitely many B’s such that AB = BA

straight A space equals open square brackets table row 1 2 row 3 4 end table close square brackets space space space space straight B space equals open square brackets table row straight a 0 row 0 straight b end table close square brackets AB space equals open square brackets table row straight a cell 2 straight b end cell row cell 3 straight a end cell cell 4 straight b end cell end table close square brackets BA space equals space open square brackets table row straight a 0 row 0 straight b end table close square brackets open square brackets table row 1 2 row 3 4 end table close square brackets space equals space open square brackets table row straight a cell 2 straight a end cell row cell 3 straight b end cell cell 4 straight b end cell end table close square brackets

AB = BA only when a = b

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