# 1.1 MATRIX AND DETERMINANTS ( Exercise Problems)

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