A- - b2c2 bc b-c- c2a2 ca c-a- a2b2 ab a-b - then - A - -

The correct option is D 0 |A|=b^2c^2 amp; bc amp; b+c c^2a^2 amp; ac amp; c+a a^2b^2 amp; ab amp; a+b =b^2c^2[a^2c+a ...

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Byju's Answer

Standard XII

Mathematics

Determinant

A= [ b2c2 ...

Question

A= $⎡ ⎢ ⎣ \begin{matrix} b^{2} c^{2} & b c & b + c c^{2} a^{2} & c a & c + a a^{2} b^{2} & a b & a + b \end{matrix} ⎤ ⎥ ⎦$ then $| A |$ =?

a b c

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a b c - 1

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a b c + 1

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0

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Solution

The correct option is D $0$
$| A | = ∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} c^{2} & b c & b + c c^{2} a^{2} & a c & c + a a^{2} b^{2} & a b & a + b \end{matrix} ∣ ∣ ∣ ∣ ∣$

$= b^{2} c^{2} [a^{2} c + a b c - a b c - a^{2} b] - b c [a^{3} c^{2} + a^{2} b c^{2} - a^{2} b^{2} c - a^{3} b^{2}] + (b + c) [a^{3} b c^{2} - a^{3} b^{2} c]$

$= a^{2} b^{2} c^{3} - a^{2} b^{3} c^{2} - a^{3} b c^{3} - a^{2} b^{2} c^{3} + a^{2} b^{3} c^{2} + a^{3} b^{3} c + a^{3} b^{2} c^{2} + a^{3} b c^{3} - a^{3} b^{3} c - a^{3} b^{3} c - a^{3} b^{2} c^{2}$
$= 0$

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Similar questions

⎡ ⎢ ⎣ \begin{matrix} b^{2} c^{2} & b c & b + c c^{2} a^{2} & c a & c + a a^{2} b^{2} & a b & a + b \end{matrix} ⎤ ⎥ ⎦

is equal to

If $a, b$ and $c$ are non-zero numbers, then the value of the determinant, $D = ∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} c^{2} & b c & b + c c^{2} a^{2} & c a & c + a a^{2} b^{2} & a b & a + b \end{matrix} ∣ ∣ ∣ ∣ ∣$ is

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∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} c^{2} & b c & b + c c^{2} a^{2} & c a & c + a a^{2} b^{2} & a b & a + b \end{matrix} ∣ ∣ ∣ ∣ ∣

Q. If system of equations ax + y + z = a, x + by + z = b and x + y + cz = c is inconsistent, then which of the following is correct?

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A= ⎡⎢⎣b2c2bcb+cc2a2cac+aa2b2aba+b⎤⎥⎦ then |A| =?

The correct option is D 0|A|=∣∣ ∣ ∣∣b2c2bcb+cc2a2acc+aa2b2aba+b∣∣ ∣ ∣∣=b2c2[a2c+abc−abc−a2b]−bc[a3c2+a2bc2−a2b2c−a3b2]+(b+c)[a3bc2−a3b2c]=a2b2c3−a2b3c2−a3bc3−a2b2c3+a2b3c2+a3b3c+a3b2c2+a3bc3−a3b3c−a3b3c−a3b2c2=0

A= $⎡ ⎢ ⎣ \begin{matrix} b^{2} c^{2} & b c & b + c c^{2} a^{2} & c a & c + a a^{2} b^{2} & a b & a + b \end{matrix} ⎤ ⎥ ⎦$ then $| A |$ =?