A- 5 5a a- 0 a 5a- 0 0 5 - If - A 2 - -25 then -a-

The correct option is A 5 A=[ [ 5 amp; 5a amp; a; 0 amp; a amp; 5a; 0 amp; 0 amp; 5 ]] #160; | A | =[ [ ...

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Byju's Answer

Standard XII

Mathematics

Determinant

A=[ 5 5a a; 0...

Question

$A = ⎡ ⎢ ⎣ \begin{matrix} 5 & 5 a & a 0 & a & 5 a 0 & 0 & 5 \end{matrix} ⎤ ⎥ ⎦$ If $∣ ∣ A^{2} ∣ ∣ = 25$ then $| a | =$

5

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5^{2}

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1

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\frac{1}{5}

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Solution

The correct option is A $5$
$A = ⎡ ⎢ ⎣ \begin{matrix} 5 & 5 a & a 0 & a & 5 a 0 & 0 & 5 \end{matrix} ⎤ ⎥ ⎦$
$| A | = ⎡ ⎢ ⎣ \begin{matrix} 5 & 5 a & a 0 & a & 5 a 0 & 0 & 5 \end{matrix} ⎤ ⎥ ⎦ = 25 a$
$| A^{2} | = | A |^{2} = 25$
$(25 a)^{2} = 25$
$a = \pm \frac{1}{5}$
$| a | = \frac{1}{5}$

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Similar questions

Q. Let

A = ⎡ ⎢ ⎣ \begin{matrix} 5 & 5 α & α 0 & α & 5 α 0 & 0 & 5 \end{matrix} ⎤ ⎥ ⎦

∣ ∣ A^{2} ∣ ∣ = 25,

then

| α |

equals to

Q. Let

A = ⎡ ⎢ ⎣ \begin{matrix} 5 & 5 α & α 0 & α & 5 α 0 & 0 & 5 \end{matrix} ⎤ ⎥ ⎦

∣ ∣ A^{2} ∣ ∣ = 25

, then

| α |

equals

If $a_{1} = \frac{1}{2 \times 5}, a_{2} = \frac{1}{5 \times 8}, a_{3} = \frac{1}{8 \times 11}, . . . . ., t h e n a_{1} + a_{2} + a_{3} + . . . . . + a_{100} i s$

Q. Simplify:

[\frac{1}{1 + a} + \frac{2 a}{1 - a^{2}}] \times \frac{(a^{2} + 4 a - 5)}{a^{2} + 10 a + 25}

Q. If

A = ⎛ ⎜ ⎝ \begin{matrix} 5 & 5 x & x 0 & x & 5 x 0 & 0 & 5 \end{matrix} ⎞ ⎟ ⎠

and

| A^{2} | = 25

, then

| x |

is equal to

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A=⎡⎢⎣55aa0a5a005⎤⎥⎦ If ∣∣A2∣∣=25 then |a|=

The correct option is A 5A=⎡⎢⎣55aa0a5a005⎤⎥⎦|A|=⎡⎢⎣55aa0a5a005⎤⎥⎦=25a|A2|=|A|2=25(25a)2=25a=±15|a|=15

$A = ⎡ ⎢ ⎣ \begin{matrix} 5 & 5 a & a 0 & a & 5 a 0 & 0 & 5 \end{matrix} ⎤ ⎥ ⎦$ If $∣ ∣ A^{2} ∣ ∣ = 25$ then $| a | =$