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  • Please solve RD Sharma solution for Maths Class 12 Chapter 5 Determinants Exercise Very short question Question 13 for maths textbook solution.

Please solve RD Sharma solution for Maths Class 12 Chapter 5 Determinants Exercise Very short question Question 13 for maths textbook solution.

Answers (2)

Answer:6

Hint: Here we use basic concept of diagonal matrix and determinant of matrix

Given:

\begin{aligned} &A \text { is }\left[a_{i j}\right] \text { is } 3 \times 3 \text { diagonal } \mathrm{m} \text { atrix }\\ &a_{11}=1, a_{22}=2 \text { and } a_{33}=3 \end{aligned}

Solution :

\begin{aligned} &\text { Let } A=\left[a_{i j}\right] \text { diagonal matrix }\\ &\text { So }, a_{i j}=0, i \neq j\\ &a_{i j} \neq 0, i=j \end{aligned}

So,

A must be \left[\begin{array}{ccc} a_{11} & 0 & 0 \\ 0 & a_{22} & 0 \\ 0 & 0 & a_{33} \end{array}\right]

\begin{aligned} &\rightarrow a_{11}=1, a_{22}=2 \text { and } a_{33}=3 \\ &{\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right]=A} \end{aligned}

\rightarrow Lets find determinant

\begin{aligned} &|A|=\left|\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right| \\ &=1\left|\begin{array}{ll} 2 & 0 \\ 0 & 3 \end{array}\right|-0\left|\begin{array}{ll} 0 & 0 \\ 0 & 3 \end{array}\right|+0\left|\begin{array}{ll} 0 & 2 \\ 0 & 0 \end{array}\right| \\ &=1(6-0)-0(0-6)+0(0-0) \\ &=1(6)=6 \end{aligned}

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Answer:\frac{1}{2} \sin ^{-1}\left(\frac{x^{2}}{2}\right)+c

Hint Let x^{2}=t

Given: \int \frac{x}{\sqrt{4-x^{4}}} d x

Explanation: \int \frac{x}{\sqrt{4-x^{4}}} d x            ..........(1)

            Let x^{2}=t

           2 x d x=d t                                               (Differentiate w.r.t to t)

           x d x=\frac{d t}{2}

Put in (1)

        \frac{1}{2} \int \frac{d t}{\sqrt{4-t^{2}}}=\frac{1}{2} \sin ^{-1}\left(\frac{t}{2}\right)+c \quad\left[\because \int \frac{d t}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}\right]

                                   =\frac{1}{2} \sin ^{-1}\left(\frac{x^{2}}{2}\right)+c

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