Mathematics Test - 13

Area of asquare = side × side

Perimeter of a square = 4 × side

Given, wire when bent in the form of a square encloses an area of 484 sq.cm

Side = √484 = 22 cm

Perimeter = 88 cm

Same wire is used to make a circle.

Perimeter of a circle of radius r = 2πr

⇒ 2πr = 88

⇒$\frac{44r}{7}$ = 88

⇒ r = 14 cm

Area of a circle of radius r = πr²

Area enclosed by the circle=$\frac{22}{7}\times {14}^2$

⇒ Area enclosed by the circle = 616 sq. cm.

$\frac{\left(1+i\right)\left(2+i\right)}{3-i}=\frac{1+3i}{3-i}$

$=\frac{1+3i}{3-i}\times \frac{3+i}{3+i}=\frac{10i}{10}=ior0+i$

Argument,$\theta$tan^-1$\left(\frac{1}{0}\right)$= tan^-1$\left(\tan \frac{\mathrm{\pi }}{2}\right)=\frac{\mathrm{\pi }}{2}$

$A=\left[\begin{array}{ccc}0& 1& 2\\ -1& 0& -3\\ -2& 3& 0\end{array}\right]$

$A^T=\left[\begin{array}{ccc}0& -1& -2\\ 1& 0& 3\\ 2& -3& 0\end{array}\right]=-\left[\begin{array}{ccc}0& 1& 2\\ -1& 0& -3\\ -2& 3& 0\end{array}\right]=-A$

Hence, A is skew symmetric matrix

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

Therefore A is non-invertible.

I. \(A B=\left|\begin{array}{cc}1 & 2 \\ 2 & 1 \\ 1 & 1\end{array}\right|_{3 \times 2}\left|\begin{array}{ccc}1 & 2 & -4 \\ 2 & 1 & -4\end{array}\right|_{2 \times 3}\)

\(=\left|\begin{array}{ccc}5 & 4 & -12 \\ 4 & 5 & -12 \\ 3 & 3 & -8\end{array}\right|\)

II. \(\mathrm{BA}=\left|\begin{array}{ccc}1 & 2 & -4 \\ 2 & 1 & -4\end{array}\right|_{2 \times 3}\left|\begin{array}{cc}1 & 2 \\ 2 & 1 \\ 1 & 1\end{array}\right|_{3 \times 2}\)

\(=\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|_{2 \times 2}\)

Here, \(\mathrm{B}\) is not the right inverse of \(\mathrm{A}\) but \(\mathrm{B}\) is the left inverse of \(\mathrm{A}\).

\(\left|\begin{array}{ccc}x+a & b & c \\ a & x+b & c \\ a & b & x+c\end{array}\right|=0\)

Applying, \(C_{1} \rightarrow C_{1}+C_{2}+C_{3}\)

\(\left|\begin{array}{ccc}(a+b+c+x) & b & c \\ (a+b+c+x) & x+b & c \\ (a+b+c+x) & b & c+x\end{array}\right|=0\)

\((a+b+c+x)\left|\begin{array}{ccc}1 & b & c \\ 1 & x+b & c \\ 1 & b & c+x\end{array}\right|=0\)

\(C_{2} \rightarrow C_{2}-C_{1}, C_{3} \rightarrow C_{3}-C_{1}\)

\((a+b+c+x)\left|\begin{array}{ccc}1 & b & c \\ 0 & x & 0 \\ 0 & 0 & x\end{array}\right|=0\)

\((a+b+c+x) 1 \cdot x^{2}=0\)

\(x=0,-(a+b+c)(\because x \neq 0)\)

A A is defined only when A is a matrix of order \(\mathrm{m} \times \mathrm{n}\) where \(\mathrm{m}=\mathrm{n}\) \(A \times A=(m \times n)(m \times n)=(m \times n)(n \times n)\) if \(m=n\)

\(=m \times n=n \times n\) or \(m \times m\)

\(=A\) is a square matrix.

We know that, elements of principal diagonals of a skew-symmetric matrix are all zero. \(\mathrm{A}=\left|\begin{array}{ccc}0 & a & b \\ -a & 0 & -c \\ -b & c & 0\end{array}\right|_{3 \times 3} \Rightarrow|A|\left|\begin{array}{ccc}0 & a & b \\ -a & 0 & -c \\ -b & c & 0\end{array}\right|\)

\(=a b c-a b c=0\)

If any two adjacent rows or columns of a determinant are interchanged in position, the value of the determinant changes its sign.