Let A - a ij -3 - 3- B - b ij - 3 - 3- where b ij -3 i - j a ij and C - c ij -3 - 3- where c ij -4 i - j b ij be any three matrices- If det- A -2 then -det- B -det-C- is equal toA- 3B- 2C- 1D- 4

The correct option is D 4 We have det.B = det.C = det.A = 2, because det.B = a_11 amp; a_12/3 amp; a_13/9 3a_21 amp; a_22 amp; 1/3 a_23 9a_31 amp ...

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Byju's Answer

Standard XII

Mathematics

Factorization Method Form to Remove Indeterminate Form

Let A =[ a ij...

Question

Let $A = [a_{i j}]_{3 \times 3}, B = [b_{i j}]_{3 \times 3},$ where $b_{i j} = 3^{i - j} a_{i j}$ and $C = [c_{i j}]_{3 \times 3},$ where $c_{i j} = 4^{i - j} b_{i j}$ be any three matrices. If det.A = 2 then ‘det.B + det.C’ is equal to

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Solution

The correct option is A
4

We have det.B = det.C = det.A = 2, because det.B = $∣ ∣ ∣ ∣ ∣ \begin{matrix} a_{11} & \frac{a_{12}}{3} & \frac{a_{13}}{9} 3 a_{21} & a_{22} & \frac{1}{3} a_{23} 9 a_{31} & 3 a_{32} & a_{33} \end{matrix} ∣ ∣ ∣ ∣ ∣$

$= \frac{1}{9} \cdot \frac{1}{3} ∣ ∣ ∣ ∣ \begin{matrix} 9 a_{11} & 3 a_{12} & a_{13} 9 a_{21} & 3 a_{22} & a_{23} 9 a_{31} & 3 a_{32} & a_{33} \end{matrix} ∣ ∣ ∣ ∣$

$= \frac{1}{9} \cdot \frac{1}{3} \cdot 3 \cdot 9 ∣ ∣ ∣ ∣ \begin{matrix} a_{11} & a_{12} & a_{13} a_{21} & a_{22} & a_{23} a_{31} & a_{32} & a_{33} \end{matrix} ∣ ∣ ∣ ∣$

$= | A | = 2$

Similarly, det. C = det. B = |A| = 2

Hence, det. B + det. C = 4

Suggest Corrections

Similar questions

Q. If

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| A | = 3, | B | = - 2, | C | = 4,

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Let A=[aij]3×3, B=[bij]3×3, where bij=3i−jaij and C=[cij]3×3, where cij=4i−jbij be any three matrices. If det.A = 2 then ‘det.B + det.C’ is equal to

Let $A = [a_{i j}]_{3 \times 3}, B = [b_{i j}]_{3 \times 3},$ where $b_{i j} = 3^{i - j} a_{i j}$ and $C = [c_{i j}]_{3 \times 3},$ where $c_{i j} = 4^{i - j} b_{i j}$ be any three matrices. If det.A = 2 then ‘det.B + det.C’ is equal to