MyQuestionIcon

201

You visited us 201 times! Enjoying our articles? Unlock Full Access!

Adjoint of a Matrix

Vbegin vmatri...

Question

$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ =$

A

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

(1 + a^{2} + b^{2})^{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

D

(1 + a^{2} + b^{2})^{2} (1 - a^{2} - b^{2})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $(1 + a^{2} + b^{2})^{3}$
$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$R_{1} \to R_{1} + b R_{3}; R_{2} \to R_{2} - a R_{3}$
$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} + b^{2} & 0 & - b (1 + a^{2} + b^{2}) 0 & 1 + a^{2} + b^{2} & a (1 + a^{2} + b^{2}) 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & - b 0 & 1 & a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ (1 + a^{2} + b^{2})^{2} R_{3} \to R_{3} - 2 b R - 1 + 2 a R - 2$
$= (1 + a^{2} + b^{2})^{2} ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & - b 0 & 1 & a 0 & 0 & 1 + a^{2} + b^{2} \end{matrix} ∣ ∣ ∣ ∣ = (1 + a^{2} + b^{2})^{3}$

Suggest Corrections

thumbs-up

0

Similar questions

Q. The value of determinant

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

(1 + i) (1 + 2 i) (1 + 3 i) . . . (1 + n i) = a + i b, then 2 \times 5 \times 10 \times . . . . \times (1 + n^{2})

is equal to
(a)

\sqrt{a^{2} + b^{2}}

\sqrt{a^{2} - b^{2}}

a^{2} + b^{2}

a^{2} - b^{2}

a + b

If (1+i)(1+2i)(1+3i)......(1+ni) = a+ib, then

2.5.10.....(1+ $n^{2}$ ) is equal to

$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (1 + a^{2} + b^{2})^{3}$

If $(1 + i) (1 + 2 i) (1 + 3 i) . . . . (1_{n} i) = a + i b, then 2 \times 5 \times 10 \times . . . . \times (1 + n^{2}) is equal to$

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Adjoint and Inverse of a Matrix

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Adjoint of a Matrix

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon