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Byju's Answer

Standard VII

Mathematics

Applying Algebra in Calendars

In a square m...

Question

In a square matrix $A$ of order $3$ the elements, $a_{i i} s^{^{'}}$ are the sum of the roots of the equation $x^{2} - (a + b) x + a b = 0; a_{1, i + 1} s^{^{'}}$ are the product of the roots, $a_{1, i - i} s^{^{'}}$ are all unity and the rest of the elements all zero. The value of the det. $(A)$ is equal to

0

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(a + b)^{3}

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a^{3} - b^{3}

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(a^{2} + b^{2})

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Solution

The correct option is B $(a + b)^{3}$
Given, square matrix $A$ of order $3$ with $a_{i i} = (a + b)$ (sum of roots)

$a_{1 i + 1} = a_{12} = a b$ (product of roots)

$a_{1 i - 1} = a_{10} = 1$

$| A | = ∣ ∣ ∣ ∣ ∣ \begin{matrix} (a + b) & 0 & 0 1 & (a + b) & a b 0 & 0 & (a + b) \end{matrix} ∣ ∣ ∣ ∣ ∣ = (a + b) [(a + b) (a + b) - 0]$

$= (a + b)^{3}$

$∴$ Option $B$ is correct.

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In a square matrix A of order 3 the elements, ai is′ are the sum of the roots of the equation x2−(a+b)x+ab=0; a1, i+1s′ are the product of the roots, a1, i−is′ are all unity and the rest of the elements all zero. The value of the det. (A) is equal to

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