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

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)

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

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