\[\LARGE{\color {purple} {PART- A}}\]
\[\color {red} {1.\ Solve:\ \begin{vmatrix}
x & 2 \\
5 & 2 \\
\end{vmatrix}\ = 0}\ \hspace{20cm}\]
\[\color {blue}{Solution:}\ \begin{vmatrix}
x & 2 \\
5 & 2 \\
\end{vmatrix}\ =0\ \hspace{20cm}\]
\[ x(2) – (2)(5) = 0\ \hspace{15cm}\]
\[ 2x – 10 = 0\ \hspace{16cm}\]
\[\implies\ 2x = 10\ \hspace{16cm}\]
\[x = 5\ \hspace{17cm}\]
\[\color {red} {2.\ Find\ ‘x’\ if\ \begin{vmatrix}
x & 3 \\
1 & 2 \\
\end{vmatrix}\ = 0}\ \hspace{20cm}\]
\[\color {blue}{Solution:}\ \begin{vmatrix}
x & 3 \\
1 & 2 \\
\end{vmatrix}\ =0\ \hspace{20cm}\]
\[ x(2) – (3)(1) = 0\ \hspace{15cm}\]
\[2\ x – 3\ =\ 0\ \hspace{16cm}\]
\[2\ x\ =\ 3\ \hspace{16cm}\]
\[x\ =\ \frac{3}{2}\ \hspace{16cm}\]
\[\color {red} {3.\ Find\ ‘x’\ if\ \begin{vmatrix}
x & 4 \\
9 & x \\
\end{vmatrix}\ = 0}\ \hspace{20cm}\]
\[\color {blue}{Solution:}\ \begin{vmatrix}
x & 4 \\
9 & x \\
\end{vmatrix}\ =0\ \hspace{20cm}\]
\[ x(x) – (4)(9) = 0\ \hspace{15cm}\]
\[ x^2 – 36 = 0\ \hspace{16cm}\]
\[x^2 = 36\ \hspace{15cm}\]
\[x = \pm\sqrt{36}\ \hspace{15cm}\]
\[x = \pm 6\ \hspace{15cm}\]
\[\color {red} {4.\ Find\ ‘x’\ if\ \begin{vmatrix}
x & 4 \\
16 & x \\
\end{vmatrix}\ = 0}\ \hspace{20cm}\]
\[\color {blue}{Solution:}\ \begin{vmatrix}
x & 4 \\
16 & x \\
\end{vmatrix}\ =0\ \hspace{20cm}\]
\[ x(x) – (4)(16) = 0\ \hspace{15cm}\]
\[ x^2 – 64\ = 0\ \hspace{16cm}\]
\[x^2\ =\ 64\ \hspace{15cm}\]
\[x = \pm\sqrt{64}\ \hspace{15cm}\]
\[x = \pm 8\ \hspace{15cm}\]
\[\color {red} {5.\ If\ A =\begin{bmatrix}
2 & 3 \\
-1 & 4\\
\end{bmatrix}\ and\ B =\begin{bmatrix}
5 & 0 \\
3 & 6\\
\end{bmatrix}\ then\ find\ A\ +\ B}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ \ A =\begin{bmatrix}
2 & 3 \\
-1 & 4\\
\end{bmatrix}\ and\ B =\begin{bmatrix}
5 & 0 \\
3 & 6\\
\end{bmatrix}\ \hspace{18cm}\]
\[A\ + B = \begin{bmatrix}
2 & 3 \\
-1 & 4\\
\end{bmatrix}\ +\ \begin{bmatrix}
5 & 0 \\
3 & 6\\
\end{bmatrix}\ \hspace{10cm}\]
\[=\begin{bmatrix}
2 + 5 & 3 +0 \\
-1 + 3 & 4 + 6\\
\end{bmatrix}\ \hspace{10cm}\]
\[A\ + B = \begin{bmatrix}
7 & 3 \\
2 & 10\\
\end{bmatrix}\ \hspace{10cm}\]
\[6.\ \color {red}{If\ A =\begin{pmatrix}
5 & – 6 \\
3 & 7\\
\end{pmatrix},\ find\ 3\ A?}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ \ A =\begin{pmatrix}
5 & – 6 \\
3 & 7\\
\end{pmatrix}\ \hspace{15cm}\]
\[ 3A = 3\begin{pmatrix}
5 & – 6 \\
3 & 7\\
\end{pmatrix},\ \hspace{13cm}\]
\[3A = \begin{pmatrix}
15 & – 18 \\
9 & 21\\
\end{pmatrix}\ \hspace{13cm}\]
\[\color {red} {7.\ If\ A =\begin{bmatrix}
1 & 2 \\
3 & 5\\
\end{bmatrix}\ and\ B =\begin{bmatrix}
-5 & 7 \\
0 & 4\\
\end{bmatrix}\ then\ find\ 4A\ +\ B}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ \ A =\begin{bmatrix}
1 & 2 \\
3 & 5\\
\end{bmatrix}\ and\ B =\begin{bmatrix}
-5 & 7 \\
0 & 4\\
\end{bmatrix}\ \hspace{18cm}\]
\[ 4A = 4\begin{bmatrix}
1 & 2 \\
3 & 5\\
\end{bmatrix}\ \hspace{13cm}\]
\[4A = \begin{bmatrix}
4 & 8 \\
12 & 20\\
\end{bmatrix}\ \hspace{13cm}\]
\[4A\ + B = \begin{bmatrix}
4 & 8 \\
12 & 20\\
\end{bmatrix}\ +\ \begin{bmatrix}
-5 & 7 \\
0 & 4\\
\end{bmatrix}\ \hspace{10cm}\]
\[=\begin{bmatrix}
4 – 5 & 8 +7 \\
12 + 0 & 20 + 4\\
\end{bmatrix}\ \hspace{10cm}\]
\[4A\ + B = \begin{bmatrix}
-1 & 15 \\
12 & 24\\
\end{bmatrix}\ \hspace{10cm}\]
\[8.\ \color {red}{Define\ Non-singular\ matrix}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ a\ square\ matrix\ A\ is\ called\ non – singular\ matrix\ if\ \begin{vmatrix} A \\ \end{vmatrix}\ \neq {0}\ \hspace{5cm}\]
\[\LARGE{\color {purple} {PART- B}}\]
\[9.\ \color {red}{Prove\ that\ the\ matrix \begin{pmatrix}
8 & 16 \\
6 & 12\\
\end{pmatrix},\ is\ singular.}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ Given\ A =\begin{pmatrix}
8 & 16 \\
6 & 12\\
\end{pmatrix},\ \hspace{15cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ = 8(12) – 16(6) \hspace{13cm}\]
\[= 96 – 96 \hspace{12cm}\]
\[= 0 \hspace{13cm}\]
\[A\ is\ a\ singular\ matrix\ \hspace{10cm}\]
\[\color {red} {10.\ Prove\ that\ \begin{pmatrix}
1 & 3 & 5 \\
3 & 5 & 7 \\
17 & 9 & 1 \\
\end{pmatrix}\ is\ a\ Singular\ Matrix.} \hspace{20cm}\]
\[\color {blue}{Solution:}\ Let\ A =\begin{pmatrix}
1 & 3 & 5 \\
3 & 5 & 7 \\
17 & 9 & 1 \\
\end{pmatrix}\ \hspace{18cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =1\begin{vmatrix}
5 & 7 \\
9 & 1 \\
\end{vmatrix}\ -\ 3\begin{vmatrix}
3 & 7 \\
17 & 1 \\
\end{vmatrix}\ +\ 5\begin{vmatrix}
3 & 5 \\
17 & 9 \\
\end{vmatrix}\ \hspace{6cm}\]
\[=1(5\ -\ 63)\ – 3 (3\ -\ 119) + 5(27\ -\ 85)\
\hspace{4cm}\]
\[=1(-58)\ – 3 (-116) + 5(-58)\
\hspace{8cm}\]
\[= -58\ +348 – 290\
\hspace{10cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ = 0\
\hspace{14cm}\]
\[\therefore\ The\ given\ Matrix\ is\ a\ Singular\ Matrix.\ \hspace{15cm}\]
\[\color {red} {11.\ Prove\ that\ \begin{pmatrix}
1 & 2 & -3 \\
3 & 4 & 5 \\
4 & 8 & -12 \\
\end{pmatrix}\ is\ a\ Singular\ Matrix.} \hspace{20cm}\]
\[\color {blue}{Solution:}\ Let\ A =\begin{pmatrix}
1 & 2 & -3 \\
3 & 4 & 5 \\
4 & 8 & -12 \\
\end{pmatrix}\ \hspace{18cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =1\begin{vmatrix}
4 & 5 \\
8 & -12 \\
\end{vmatrix}\ -\ 2\begin{vmatrix}
3 & 5 \\
4 & -12 \\
\end{vmatrix}\ -\ 3\begin{vmatrix}
3 & 4 \\
4 & 8 \\
\end{vmatrix}\ \hspace{6cm}\]
\[=1(-48\ -\ 40)\ – 2 (-36\ -\ 20) -\ 3(24\ -\ 16)\
\hspace{4cm}\]
\[=1(-88)\ – 3 (-56) -\ 3(8)\
\hspace{8cm}\]
\[= -88\ +\ 112\ -\ 24\
\hspace{10cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ = 0\
\hspace{14cm}\]
\[\therefore\ The\ given\ Matrix\ is\ a\ Singular\ Matrix.\ \hspace{15cm}\]
\[12.\ \color {red}{If\ A =\begin{bmatrix}
2 & 3 \\
-1 & 2 \\
\end{bmatrix}\ ,\ B =\begin{bmatrix}
-1 & 2 \\
1 & -5 \\
\end{bmatrix}\ ,\ Find\ the\
value\ of\ AB\ and\ BA}\ \hspace{12cm}\]
\[\color {blue}{Solution:}\ \hspace{18cm}\]
\[AB =\begin{bmatrix}
2 & 3 \\
-1 & 2 \\
\end{bmatrix}\ \begin{bmatrix}
-1 & 2 \\
1 & -5 \\
\end{bmatrix}\ \hspace{12cm}\]
\[ = \begin{bmatrix}
2 × -1\ +\ 3 × 1\ & 2\ ×\ 2\ +\ 3\ ×\ -5 \\
-1\ ×\ -1 +\ 2 × 1\ & -1\ ×\ 2\ +\ 2\ ×\ -5\
\end{bmatrix}\ \hspace{9cm}\]
\[ = \begin{bmatrix}
-2\ +\ 3 & 4\ -\ 15 \\
1\ +\ 2 & -2\ -\ 10\\
\end{bmatrix}\ \hspace{10cm}\]
\[ AB = \begin{bmatrix}
1 & -11 \\
3 & -12\\
\end{bmatrix}\ \hspace{12cm}\]
\[BA =\begin{bmatrix}
-1 & 2 \\
1 & -5 \\
\end{bmatrix}\ \begin{bmatrix}
2 & 3 \\
-1 & 2 \\
\end{bmatrix}\ \hspace{12cm}\]
\[ = \begin{bmatrix}
-1 ×\ 2\ +\ 2 ×\ -1\ & -1\ ×\ 3\ +\ 2\ ×\ 2 \\
1\ ×\ 2 +\ -5 ×\ -1\ & 1\ ×\ 3\ +\ -5\ ×\ 2\
\end{bmatrix}\ \hspace{9cm}\]
\[ = \begin{bmatrix}
-2\ -\ 2 & -3\ +\ 4 \\
2\ +\ 5 & 3\ -\ 10\\
\end{bmatrix}\ \hspace{10cm}\]
\[BA\ = \begin{bmatrix}
-4 & 1 \\
7 & -7\\
\end{bmatrix}\ \hspace{12cm}\]
\[13.\ \color {red}{If\ A =\begin{pmatrix}
3 & 6 \\
1 & -2 \\
\end{pmatrix}\ and\ \ B =\begin{pmatrix}
5 & 0 \\
2 & 3 \\
\end{pmatrix}\ , Show\ that\ (AB)^T\ =\ B^T\ A^T}\ \hspace{12cm}\]
\[\color {blue}{Solution:}\ \hspace{18cm}\]
\[Let\ A =\begin{pmatrix}
3 & 6 \\
1 & -2 \\
\end{pmatrix}\ and\ \ B =\begin{pmatrix}
5 & 0 \\
2 & 3 \\
\end{pmatrix}\ \hspace{12cm}\]
\[AB =\begin{pmatrix}
3 & 6 \\
1 & -2 \\
\end{pmatrix}\ \begin{pmatrix}
5 & 0 \\
2 & 3 \\
\end{pmatrix}\ \hspace{12cm}\]
\[ = \begin{pmatrix}
3\ ×\ 5\ +\ 6 ×\ 2\ & 3\ ×\ 0\ +\ 6\ ×\ 3 \\
1\ ×\ 5\ +\ -2 ×\ 2\ & 1\ ×\ 0\ +\ -2\ ×\ 3\
\end{pmatrix}\ \hspace{9cm}\]
\[ = \begin{pmatrix}
15\ +\ 12 & 0\ +\ 18 \\
5\ -\ 4 & 0\ -\ 6\\
\end{pmatrix}\ \hspace{10cm}\]
\[ AB = \begin{pmatrix}
27 & 18 \\
1 & -6\\
\end{pmatrix}\ \hspace{12cm}\]
\[(AB)^T\ = \begin{pmatrix}
27 & 1 \\
18 & -6\\
\end{pmatrix}\ ————– (1)\ \hspace{10cm}\]
\[Let\ B^T =\begin{pmatrix}
5 & 2 \\
0 & 3 \\
\end{pmatrix}\ and\ \ A^T\ =\begin{pmatrix}
3 & 1 \\
6 & -2 \\
\end{pmatrix}\ \hspace{12cm}\]
\[B^T\ A^T\ =\begin{pmatrix}
5 & 2 \\
0 & 3 \\
\end{pmatrix}\ \begin{pmatrix}
3 & 1 \\
6 & -2 \\
\end{pmatrix}\ \hspace{12cm}\]
\[ = \begin{pmatrix}
5\ ×\ 3\ +\ 2 ×\ 6\ & 5\ ×\ 1\ +\ 2\ ×\ -2 \\
0\ ×\ 3\ +\ 3 ×\ 6\ & 0\ ×\ 1\ +\ 3\ ×\ -2\
\end{pmatrix}\ \hspace{9cm}\]
\[ = \begin{pmatrix}
15\ +\ 12 & 5\ -\ 4 \\
0\ +\ 18 & 0\ -\ 6\\
\end{pmatrix}\ \hspace{10cm}\]
\[B^T\ A^T\ = \begin{pmatrix}
27 & 1 \\
18 & -6\\
\end{pmatrix}\ ————– (2)\ \hspace{10cm}\]
\[From\ (1)\ and\ (2)\ (AB)^T\ =\ B^T\ A^T\ \hspace{10cm}\]
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