RD Sharma Class 12 Exercise 7.1 Simultaneous Linear Equation Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 7.1 Simultaneous Linear Equation Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 20, 2022 03:20 PM IST

RD Sharma reading material is one of the best course books for class 12 explicitly for students getting ready for severe tests. Arithmetic is the subject of training. To have great practice, you need the nature of inquiries, and RD Sharma Class 12 meets an excellent prerequisite of questions.

Also Read - RD Sharma Solution for Class 9 to 12 Maths

## Solution of simultaneous linear equations Excercise: 7.1

Solution of Simultaneous Linear Equation exercise 7.1 question 1 subquestion (i)

$x=-1\: \: and\: \: y=4$
Given:
$\left[\begin{array}{ll} 5 & 2 \\ 3 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 3 \\ 5 \end{array}\right]$
Hint:
X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A
Solution:
$\left[\begin{array}{ll} 5 & 2 \\ 3 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 3 \\ 5 \end{array}\right]$
$A X=B\\ |A|=\left[\begin{array}{ll}5 & 2 \\ 3 & 2\end{array}\right]=10-6=4 \neq 0\\ This\; has\; a\; unique\; sol\! ution\; given\; by\; X=A^{-1} B.\\ C_{i j}\; be\; the\; co-\! factor\; o\! f\; the\; elements\; a_{i j}\; in\; A=\left[a_{i j}\right]. \; Then,$
\begin{aligned} &C_{11}=(-1)^{1+1}(2)=2 , \quad C_{12}=(-1)^{1+2}(3)=-3 \\ &C_{21}=(-1)^{2+1}(2)=-2 , \quad C_{22}=(-1)^{2+2}(5)=5 \\ &\operatorname{adj} A=\left[\begin{array}{cc} 2 & -3 \\ -2 & 5 \end{array}\right]^{T} \\ &\quad=\left[\begin{array}{cc} 2 & -2 \\ -3 & 5 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{4}\left[\begin{array}{cc} 2 & -2 \\ -3 & 5 \end{array}\right] \\ X &=A^{-1} B \\ &=\frac{1}{4}\left[\begin{array}{cc} 2 & -2 \\ -3 & 5 \end{array}\right]\left[\begin{array}{l} 3 \\ 5 \end{array}\right] \\ \end{aligned}
\begin{aligned} &{\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{4}\left[\begin{array}{c} 6-10 \\ -9+25 \end{array}\right]=\left[\begin{array}{c} -\frac{4}{4} \\ \frac{16}{4} \end{array}\right]} \\ &\therefore x=-1 \quad, \quad \mathrm{y}=4 \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 1 subquestion (ii)

$x=\frac{9}{2},\; \; y=-\frac{7}{2}$
Given:
$\left[\begin{array}{ll} 5 & 7 \\ 4 & 6 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} -2 \\ -3 \end{array}\right]$
Hint:
X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A .
Solution:
$\left[\begin{array}{ll} 5 & 7 \\ 4 & 6 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} -2 \\ -3 \end{array}\right]$
$A X=B\\ |A|=\left[\begin{array}{ll}5 & 7 \\ 4 & 6\end{array}\right]=30-28=2 \neq 0\\ This\; has\; a\; unique\; solution\; given\; by\; X=A^{-1} B.\\ C_{i i} \; be\; the\; co-\! factor\; o\! f\; the\; elements\; a_{i j}\; in\; A=\left[a_{i j}\right].\; Then,$
\begin{aligned} &C_{11}=(-1)^{1+1}(6)=2 \quad, \quad C_{12}=(-1)^{1+2}(4)=-4 \\ &C_{21}=(-1)^{2+1}(7)=-7 \quad, \quad C_{22}=(-1)^{2+2}(5)=5 \\ &A=\left[\begin{array}{cc} 6 & -4 \\ -7 & 5 \end{array}\right] \end{aligned}
\begin{aligned} \operatorname{adj} A &=\left[\begin{array}{cc} 6 & -4 \\ -7 & 5 \end{array}\right]^{T} \\ &=\left[\begin{array}{cc} 6 & -7 \\ -4 & 5 \end{array}\right] \\ A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{2}\left[\begin{array}{cc} 6 & -7 \\ -4 & 5 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{2}\left[\begin{array}{cc} 6 & -7 \\ -4 & 5 \end{array}\right] \\ X &=A^{-1} B \\ &=\frac{1}{2}\left[\begin{array}{cc} 6 & -7 \\ -4 & 5 \end{array}\right]\left[\begin{array}{l} -2 \\ -3 \end{array}\right] \\ &=\frac{1}{2}\left[\begin{array}{c} -12+21 \\ 8-15 \end{array}\right] \end{aligned}
\begin{aligned} &{\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} \frac{9}{2} \\ -\frac{7}{2} \end{array}\right]} \\ &\therefore x=\frac{9}{2} \quad, \quad \mathrm{y}=-\frac{7}{2} \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 1 subquestion (iii)

$x=-1\; \; ,\; \; y=2$
Given:
$\left[\begin{array}{ll} 3& 4 \\ 1 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 5 \\ -3 \end{array}\right]$
Hint:
X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A .
Solution:
$\left[\begin{array}{ll} 3& 4 \\ 1 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 5 \\ -3 \end{array}\right]$
\begin{aligned} &A X=B \\ &|A|=\left[\begin{array}{cc} 3 & 4 \\ 1 & -1 \end{array}\right]=-3-4=-7 \neq 0 \end{aligned}
$This\; has\; a\; unique\; solution\; given\; by\; X=A^{-1} B.\\ C_{i j}\; be\; the\; co\! -\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
\begin{aligned} &C_{11}=(-1)^{1+1}(-1)=-1 \quad, \quad C_{12}=(-1)^{1+2}(1)=-1 \\ &C_{21}=(-1)^{2+1}(4)=-4 \quad, \quad C_{22}=(-1)^{2+2}(3)=3 \\ &\operatorname{adj} A=\left[\begin{array}{cc} -1 & -1 \\ -4 & 3 \end{array}\right]^{T} \\ &\quad=\left[\begin{array}{cc} -1 & -4 \\ -1 & 3 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{-7}\left[\begin{array}{cc} -1 & -4 \\ -1 & 3 \end{array}\right] \\ X &=A^{-1} B \\ &=\frac{1}{-7}\left[\begin{array}{cc} -1 & -4 \\ -1 & 3 \end{array}\right]\left[\begin{array}{c} 5 \\ -3 \end{array}\right] \\ &=\frac{1}{-7}\left[\begin{array}{c} -5+12 \\ -5-9 \end{array}\right] \end{aligned}
\begin{aligned} &{\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} \frac{7}{-7} \\ \frac{-14}{-7} \end{array}\right]} \\ &\therefore x=-1 \quad, \quad \mathrm{y}=2 \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 1 subquestion (iv)

$x=7\; \; and\; \; y=-2$
Given:
$\left[\begin{array}{ll} 3 & 1 \\ 3 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 19 \\ 23 \end{array}\right]$
Hint:
X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A .
Solution:
$\left[\begin{array}{ll} 3 & 1 \\ 3 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 19 \\ 23 \end{array}\right]$
\begin{aligned} &A X=B \\ &|A|=\left[\begin{array}{cc} 3 & 1 \\ 3 & -1 \end{array}\right]=-3-3=-6 \neq 0 \end{aligned}
$This\; has\; a\; unique\; solution\; given\; by\; X=A^{-1} B.\\ C_{i j} \; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
$\begin{array}{ll} C_{11}=(-1)^{1+1}(-1)=-1 & , \quad C_{12}=(-1)^{1+2}(3)=-3 \\ C_{21}=(-1)^{2+1}(1)=-1 & , \quad C_{22}=(-1)^{2+2}(3)=3 \end{array}$
\begin{aligned} \operatorname{adj} A &=\left[\begin{array}{cc} -1 & -3 \\ -1 & 3 \end{array}\right]^{T} \\ &=\left[\begin{array}{cc} -1 & -1 \\ -3 & 3 \end{array}\right] \\ A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{-6}\left[\begin{array}{cc} -1 & -1 \\ -3 & 3 \end{array}\right] \end{aligned}
\begin{aligned} X &=A^{-1} B \\ &=\frac{1}{-6}\left[\begin{array}{cc} -1 & -1 \\ -3 & 3 \end{array}\right]\left[\begin{array}{l} 19 \\ 23 \end{array}\right] \\ &=\frac{1}{-6}\left[\begin{array}{l} -19-23 \\ -57-69 \end{array}\right] \end{aligned}
\begin{aligned} &{\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{-6}\left[\begin{array}{l} -19-23 \\ -57-69 \end{array}\right]=\left[\begin{array}{c} \frac{-42}{-6} \\ \frac{12}{-6} \end{array}\right]} \\ &\therefore x=7 \quad, \quad \mathrm{y}=-2 \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 1 subquestion (v)

$x=-15\; \; ,\; \; y=7$
Given:
$\left[\begin{array}{ll} 3 & 7 \\ 1 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 4 \\ -1 \end{array}\right]$
Hint:
X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A .
Solution:
$\left[\begin{array}{ll} 3 & 7 \\ 1 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 4 \\ -1 \end{array}\right]$
\begin{aligned} &A X=B \\ &A=\left[\begin{array}{cc} 3 & 7 \\ 1 & 2 \end{array}\right]\\ &|A|=\left[\begin{array}{cc} 3 & 7 \\ 1 & 2 \end{array}\right]=6-7=-1 \neq 0 \end{aligned}
$This\; has\; a\; unique \; solution\; given\; by\; X=A^{-1} B.\\ C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
\begin{aligned} &C_{11}=(-1)^{1+1}(2)=2 \quad, \quad C_{12}=(-1)^{1+2}(1)=-1 \\ &C_{21}=(-1)^{2+1}(7)=-7 \quad, \quad C_{22}=(-1)^{2+2}(3)=3 \\ &\operatorname{adj} A=\left[\begin{array}{cc} 2 & -1 \\ -7 & 3 \end{array}\right]^{T} \\ &\quad=\left[\begin{array}{cc} 2 & -7 \\ -1 & 3 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{-1}\left[\begin{array}{cc} 2 & -7 \\ -1 & 3 \end{array}\right] \end{aligned}
\begin{aligned} X &=A^{-1} B \\ &=\left[\begin{array}{cc} -2 & 7 \\ 1 & -3 \end{array}\right]\left[\begin{array}{c} 4 \\ -1 \end{array}\right] \\ &=\left[\begin{array}{c} -8-7 \\ 4+3 \end{array}\right] \end{aligned}
\begin{aligned} &{\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} -15 \\ 7 \end{array}\right]} \\ &\therefore x=-15 \quad, \quad \mathrm{y}=7 \end{aligned}
$x=\frac{9}{4}\; \; ,\; \; y=\frac{1}{4}$
Given:
$\left[\begin{array}{ll} 3 & 1 \\ 5 & 3 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 7 \\ 12 \end{array}\right]$
Hint:
X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A .
Solution:
$\left[\begin{array}{ll} 3 & 1 \\ 5 & 3 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 7 \\ 12 \end{array}\right]$
\begin{aligned} &A X=B \\ &|A|=\left[\begin{array}{cc} 3 & 1 \\ 5 & 3 \end{array}\right]=9-5=4 \neq 0 \end{aligned}
$This\; has\; a\; unique \; solution\; given\; by\; X=A^{-1} B.\\ C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
\begin{aligned} &C_{11}=(-1)^{1+1}(3)=3 \quad, \quad C_{12}=(-1)^{1+2}(5)=-5 \\ &C_{21}=(-1)^{2+1}(1)=-1 \quad, \quad C_{22}=(-1)^{2+2}(3)=3 \\ &\operatorname{adj} A=\left[\begin{array}{cc} 3 & -5 \\ -1 & 3 \end{array}\right]^{T} \\ &\quad=\left[\begin{array}{cc} 3 & -1 \\ -5 & 3 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{-1}\left[\begin{array}{cc} 3 & -1 \\ -5 & 3 \end{array}\right] \end{aligned}
\begin{aligned} X =A^{-1} B \\ =\frac{1}{4}\left[\begin{array}{cc} 3 & -1 \\ -5 & 3 \end{array}\right]\left[\begin{array}{c} 7 \\ 12 \end{array}\right] \\ =\frac{1}{4}\left[\begin{array}{c} 21-12 \\ -35+36 \end{array}\right] \\ \left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} \frac{9}{4} \\ \frac{1}{4} \end{array}\right] \\ \therefore x =\frac{9}{4} \quad, \quad \mathrm{y}=\frac{1}{4} \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (i)

$x=3\; \; ,\; \; y=1\; \; ,\; \; z=1$
Given:
\begin{aligned} &x+y-z=3 \\ &2 x+3 y+z=10 \\ &3 x-y-7 z=1 \end{aligned}
Hint:
X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A .
Solution:
\begin{aligned} &A=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 3 & -1 & -7 \end{array}\right] \\ &\begin{array}{rc} |A|=\left|\begin{array}{ccc} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 3 & -1 & -7 \end{array}\right| &=1(-2+1)-1(-14-3)-1(-2-9) \\ =-20+17+11 \\ =8 \neq 0 \end{array} \end{aligned}
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].$
\begin{aligned} &C_{11}=(-1)^{1+1}\left|\begin{array}{cc} 3 & 1 \\ -1 & -7 \end{array}\right|=-20 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 2 & 1 \\ 3 & -7 \end{array}\right|=17 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 2 & 3 \\ 3 & -1 \end{array}\right|=-11 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{ll} 1 & -1 \\ -1 & -7 \end{array}\right|=8 \end{aligned}
\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{cc} 1 & -1 \\ 3 & -7 \end{array}\right|=-4 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc} 1 & 1 \\ 3 & -1 \end{array}\right|=4 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{cc} 1 & -1 \\ 3 & 1 \end{array}\right|=4 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc} 1 & -7 \\ 2 & 1 \end{array}\right|=-3 \end{aligned}
\begin{aligned} C_{33} &=(-1)^{3+3}\left|\begin{array}{cc} 1 & 1 \\ 2 & 3 \end{array}\right|=1 \\ \operatorname{adj} A &=\left[\begin{array}{ccc} -20 & 17 & -11 \\ 8 & -4 & 4 \\ 4 & -3 & 1 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} -20 & 8 & 4 \\ 17 & -4 & -3 \\ -11 & 4 & 1 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{8}\left[\begin{array}{ccc} -20 & 8 & 4 \\ 17 & -4 & -3 \\ -11 & 4 & 1 \end{array}\right] \\ X &=A^{-1} B \end{aligned}
\begin{aligned} &=\frac{1}{8}\left[\begin{array}{ccc} -20 & 8 & 4 \\ 17 & -4 & -3 \\ -11 & 4 & 1 \end{array}\right]\left[\begin{array}{c} 3 \\ 10 \\ 1 \end{array}\right] \\ &=\frac{1}{8}\left[\begin{array}{c} -60+80+4 \\ 51-40-3 \\ -33+40+1 \end{array}\right] \\ &=\frac{1}{8}\left[\begin{array}{c} 24 \\ 8 \\ 8 \end{array}\right] \end{aligned}
\begin{aligned} &x=\frac{24}{8} \quad, \quad \mathrm{y}=\frac{8}{8} \quad, \quad z=\frac{8}{8} \\ &\therefore x=3 \quad, \quad \mathrm{y}=1 \quad, \quad z=1 \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (ii)

$x=-\frac{8}{7}\; \; ,\; \; y=\frac{10}{7}\; \; ,\; \; z=\frac{19}{7}$
Given:
\begin{aligned} &x+y+z=3 \\ &2 x-y+z=-1 \\ &2 x+y-3 z=-9 \end{aligned}
Hint:
X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A .
Solution:
\begin{aligned} &{\left[\begin{array}{ccc} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 3 & -1 & -7 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 3 \\ -1 \\ -9 \end{array}\right]} \\ &A X=B \end{aligned}
\begin{aligned} |A|=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 2 & 1 & -3 \end{array}\right| &=1(3-1)-1(6-2)+1(2+2) \\ &=2+8+4 \\ &=14 \neq 0 \end{aligned}
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].$
\begin{aligned} &C_{11}=(-1)^{1+1}\left|\begin{array}{cc} -1 & 1 \\ 1 & -3 \end{array}\right|=2 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 2 & 1 \\ 2 & -3 \end{array}\right|=8 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 2 & -1 \\ 2 & 1 \end{array}\right|=4 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{cc} 1 & 1 \\ 1 & -3 \end{array}\right|=4 \end{aligned}
\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{cc} 1 & 1 \\ 2 & -3 \end{array}\right|=-5 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll} 1 & 1 \\ 2 & 1 \end{array}\right|=-1 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array}\right|=2 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll} 1 & 1 \\ 2 & 1 \end{array}\right|=1 \end{aligned}
\begin{aligned} &C_{33}=(-1)^{3+3}\left|\begin{array}{cc} 1 & 1 \\ 2 & -1 \end{array}\right|=-3 \\ &\operatorname{adj} A=\left[\begin{array}{ccc} 2 & 4 & 2 \\ 8 & -5 & 1 \\ 4 & 1 & -3 \end{array}\right]^{T} \\ &A^{-1}=\frac{1}{|A|} a d j A \end{aligned}
\begin{aligned} &=\frac{1}{14}\left[\begin{array}{ccc} 2 & 4 & 2 \\ 8 & -5 & 1 \\ 4 & 1 & -3 \end{array}\right] \\ X=& A^{-1} B \end{aligned}
\begin{aligned} &=\frac{1}{14}\left[\begin{array}{ccc} 2 & 4 & 2 \\ 8 & -5 & 1 \\ 4 & 1 & -3 \end{array}\right]\left[\begin{array}{c} 3 \\ -1 \\ 9 \end{array}\right] \\ &=\frac{1}{14}\left[\begin{array}{c} 6-4-18 \\ 24+5-9 \\ 12-1+27 \end{array}\right] \\ &=\frac{1}{14}\left[\begin{array}{c} -16 \\ 20 \\ 38 \end{array}\right] \end{aligned}
$x=-\frac{16}{14}\; \; ,\; \; y=\frac{20}{14}\; \; ,\; \; z=\frac{38}{14} \\ x=-\frac{8}{7}\; \; ,\; \; y=\frac{10}{7}\; \; ,\; \; z=\frac{19}{7}$

Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (iii)

$x=\frac{1}{2}\; \; ,\; \; y=\frac{1}{3}\; \; ,\; \; z=\frac{1}{5}$
Given:
\begin{aligned} &6x-12y+25z=4 \\ &4 x+15 y-20z=3 \\ &2 x+18y+15 z=10 \end{aligned}
Hint:
X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A .
Solution:
$A=\left[\begin{array}{ccc} 6 & -12 & 25 \\ -12 & 15 & -20 \\ 2 & 18 & 15 \end{array}\right]$
$\begin{gathered} |A|=\left|\begin{array}{ccc} 6 & -12 & 25 \\ -12 & 15 & -20 \\ 2 & 18 & 15 \end{array}\right|=6(225+360)+12(60+40)+25(72-30) \\ = 3510+1200+1050 \\ =5760 \end{gathered}$
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
\begin{aligned} &C_{11}=(-1)^{1+1}\left|\begin{array}{cc} 15 & -20 \\ 18 & 15 \end{array}\right|=585 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 4 & -20 \\ 2 & 15 \end{array}\right|=-100 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{ll} 4 & 15 \\ 2 & 18 \end{array}\right|=42 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{cc} 12 & 25 \\ 18 & 15 \end{array}\right|=630 \end{aligned}
\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{ll} 6 & 25 \\ 2 & 15 \end{array}\right|=40 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc} 6 & -12 \\ 2 & 8 \end{array}\right|=-132 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{cc} -12 & 25 \\ 15 & -20 \end{array}\right|=-135 \quad , \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc} 6 & 25 \\ 4 & -20 \end{array}\right|=220 \end{aligned}
\begin{aligned} C_{33} &=(-1)^{3+3}\left|\begin{array}{cc} 6 & -12 \\ 4 & 15 \end{array}\right|=138 \\ \operatorname{adjA} &=\left[\begin{array}{ccc} 585 & -100 & 42 \\ 630 & 40 & -132 \\ -135 & 220 & 138 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 585 & 630 & -135 \\ -100 & 40 & 220 \\ 42 & -132 & 138 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{5760}\left[\begin{array}{ccc} 585 & 630 & -135 \\ -100 & 40 & 220 \\ 42 & -132 & 138 \end{array}\right] \end{aligned}
\begin{aligned} X &=A^{-1} B \\ &=\frac{1}{5760}\left[\begin{array}{ccc} 585 & 630 & -135 \\ -100 & 40 & 220 \\ 42 & -132 & 138 \end{array}\right]\left[\begin{array}{c} 3 \\ -1 \\ 9 \end{array}\right] \\ &=\frac{1}{5760}\left[\begin{array}{c} 6-4-18 \\ 24+5-9 \\ 12-1+27 \end{array}\right] \\ &=\frac{1}{5760}\left[\begin{array}{c} -16 \\ 20 \\ 38 \end{array}\right] \end{aligned}
\begin{aligned} &x=\frac{2880}{5760} \quad, \quad y=\frac{1920}{5760} \quad, \quad z=\frac{1152}{5760} \\ &x=\frac{1}{2} \quad, \quad y=\frac{1}{3} \quad, \quad z=\frac{1}{5} \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (iv)

\begin{aligned} &x=1 \quad, \quad y=1 \quad, \quad z=1 \end{aligned}
Given:
\begin{aligned} &3 x+2 y+7 z=14 \\ &2 x-y+3 z=4 \\ &x+2 y-3 z=0 \end{aligned}
Hint:
X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.
Solution:
\begin{aligned} &A=\left[\begin{array}{ccc} 3 & 2 & 7 \\ 2 & -1 & 3 \\ 1 & 2 & -3 \end{array}\right] \\ &{\left[\begin{array}{ccc} 3 & 2 & 7 \\ 2 & -1 & 3 \\ 1 & 2 & -3 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 14 \\ 4 \\ 0 \end{array}\right]} \end{aligned}
A\; X\; =B\\ \begin{aligned} |A|=\left|\begin{array}{ccc} 3 & 2 & 7 \\ 2 & -1 & 3 \\ 1 & 2 & -3 \end{array}\right| &=3(3-6)-4(-6-3)+7(4+1) \\ &=-9+36+35 \\ &=62 \neq 0 \end{aligned}
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].$
$\begin{array}{ll} C_{11}=(-1)^{1+1}(3-6)=-3 & , \quad C_{12}=(-1)^{1+2}(-6-3)=9 \\ C_{13}=(-1)^{1+3}(4+1)=5 & , \quad C_{21}=(-1)^{2+1}(-12-14)=26 \end{array}$
$\begin{array}{ll} C_{22}=(-1)^{2+2}(-3-7)=-10 \quad, & C_{23}=(-1)^{2+3}(6-4)=-2 \\ C_{31}=(-1)^{3+1}(12+7)=19 & , \quad C_{32}=(-1)^{3+2}(9-14)=5 \end{array}$
\begin{aligned} C_{33}=(-1)^{3+3}(-3-8)=-11 \\ \operatorname{adjA} =\left[\begin{array}{ccc} -3 & 9 & 5 \\ 26 & -5 & -2 \\ 19 & 5 & -11 \end{array}\right]^{T} \\ =\left[\begin{array}{ccc} -3 & 26 & 19 \\ 9 & -16 & 5 \\ 5 & -2 & -11 \end{array}\right] \end{aligned}
\begin{aligned} X &=A^{-1} B \\ &=\frac{1}{62}\left[\begin{array}{ccc} -3 & 26 & 19 \\ 9 & -16 & 5 \\ 5 & -2 & -11 \end{array}\right]\left[\begin{array}{c} 14 \\ 4 \\ 0 \end{array}\right] \\ &=\frac{1}{62}\left[\begin{array}{c} -42+104+0 \\ 126-64+0 \\ 70-8+0 \end{array}\right] \end{aligned}
\begin{aligned} &=\frac{1}{62}\left[\begin{array}{l} 62 \\ 62 \\ 62 \end{array}\right] \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]} \\ &x=1 \quad, \quad y=1 \quad, \quad z=1 \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (v)

\begin{aligned} &x=\frac{1}{2} \quad, \quad y=\frac{1}{3} \quad, \quad z=\frac{1}{5} \end{aligned}
Given:
\begin{aligned} &\frac{2}{x}-\frac{3}{y}+\frac{3}{z}=10 \\ &\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=10 \\ &\frac{3}{x}-\frac{1}{y}+\frac{3}{z}=13 \end{aligned}
Hint:
X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.
Solution:
$\text { Let } \frac{1}{x} \text { be } a, \frac{1}{y} \text { be } b, \frac{1}{z} \text { be } c$
\begin{aligned} &A=\left[\begin{array}{ccc} 2 & -3 & 3 \\ 1 & 1 & 1 \\ 3 & -1 & 2 \end{array}\right] \\ &{\left[\begin{array}{ccc} 2 & -3 & 3 \\ 1 & 1 & 1 \\ 3 & -1 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 14 \\ 4 \\ 0 \end{array}\right]} \end{aligned}
A\; X=B\\ \begin{aligned} |A|=\left|\begin{array}{ccc} 2 & -3 & 3 \\ 1 & 1 & 1 \\ 3 & -1 & 2 \end{array}\right| &=2(2+1)+3(2-3)+3(-1-3) \\ &=6-3-12 \\ &=-9 \end{aligned}
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].$
\begin{aligned} &C_{11}=(-1)^{1+1}\left|\begin{array}{cc} 1 & 1 \\ -1 & 2 \end{array}\right|=3 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{ll} 1 & 1 \\ 3 & 2 \end{array}\right|=1 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 1 & 1 \\ 3 & -1 \end{array}\right|=-4 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{ll} -3 & 3 \\ -1 & 2 \end{array}\right|=3 \end{aligned}
\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{ll} 2 & 3 \\ 3 & 2 \end{array}\right|=-5 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc} 2 & -3 \\ 3 & -1 \end{array}\right|=-7 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{cc} -3 & 3 \\ 1 & 1 \end{array}\right|=-6 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc} 2 & 3 \\ 1 & 1 \end{array}\right|=1 \end{aligned}
\begin{aligned} C_{33} &=(-1)^{3+3}\left|\begin{array}{cc} 2 & -3 \\ 1 & 1 \end{array}\right|=5 \\ \operatorname{adjA} &=\left[\begin{array}{ccc} 3 & 1 & -4 \\ 3 & -5 & -7 \\ -6 & 1 & 5 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 3 & 3 & -6 \\ 1 & -5 & 1 \\ -4 & -7 & 5 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{-9}\left[\begin{array}{ccc} 3 & 3 & -6 \\ 1 & -5 & 1 \\ -4 & -7 & 5 \end{array}\right] \\ X=& A^{-1} B \end{aligned}
\begin{aligned} &=-\frac{1}{9}\left[\begin{array}{ccc} 3 & 3 & -6 \\ 1 & -5 & 1 \\ -4 & -7 & 5 \end{array}\right]\left[\begin{array}{c} 10 \\ 10 \\ 13 \end{array}\right] \\ &=-\frac{1}{9}\left[\begin{array}{c} 30+30-78 \\ 10-50+13 \\ -40-70+65 \end{array}\right] \\ &=-\frac{1}{9}\left[\begin{array}{r} -18 \\ -27 \\ -45 \end{array}\right] \end{aligned}
\begin{aligned} &\frac{1}{x}=a=\frac{-9}{-18}, \frac{1}{y}=b=\frac{-9}{-27}, \frac{1}{z}=c=\frac{-9}{-45} \\ &x=\frac{1}{2} \quad, \quad y=\frac{1}{3} \quad, \quad z=\frac{1}{5} \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (vi)

\begin{aligned} &x=1 \quad, \quad y=2 \quad, \quad z=5 \end{aligned}
Given:
\begin{aligned} &5 x+3 y+z=16 \\ &2 x+y+3 z=19 \\ &x+2 y+4 z=25 \end{aligned}
Hint:
X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.
Solution:
\begin{aligned} &A=\left[\begin{array}{lll} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array}\right] \\ &\begin{aligned} |A|=\left|\begin{array}{lll} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array}\right|=5(4-6)-3(8-3)+1(4-1) \\ \end{aligned} \end{aligned}
$=-10-15+3 \\ =-22$
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].$
\begin{aligned} &C_{11}=(-1)^{1+1}\left|\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right|=-2 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{ll} 2 & 3 \\ 1 & 4 \end{array}\right|=-5 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right|=3 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{ll} 3 & 1 \\ 2 & 4 \end{array}\right|=-10 \end{aligned}
\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{ll} 5 & 1 \\ 1 & 4 \end{array}\right|=19 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll} 5 & 3 \\ 1 & 2 \end{array}\right|=-7 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{ll} 3 & 1 \\ 1 & 3 \end{array}\right|=8 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll} 5 & 1 \\ 2 & 3 \end{array}\right|=-13 \end{aligned}
\begin{aligned} C_{33} &=(-1)^{3+3}\left|\begin{array}{cc} 5 & 3 \\ 2 & 1 \end{array}\right|=-1 \\ \operatorname{adj} A &=\left[\begin{array}{ccc} -2 & -5 & 3 \\ -10 & 19 & -7 \\ 8 & -13 & -1 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} \text { adjA } \\ &=\frac{1}{-22}\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right] \end{aligned}
\begin{aligned} X &=A^{-1} B \\ &=-\frac{1}{22}\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right]\left[\begin{array}{l} 16 \\ 19 \\ 25 \end{array}\right] \\ &=-\frac{1}{22}\left[\begin{array}{c} -32-190+200 \\ -80+361-325 \\ 48-133-25 \end{array}\right] \end{aligned}
\begin{aligned} &=-\frac{1}{22}\left[\begin{array}{c} -22 \\ -44 \\ -110 \end{array}\right] \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 5 \end{array}\right]} \\ &x=1, \quad y=2 \quad, \quad z=5 \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (vii)

\begin{aligned} &\ x=-2 \quad, \quad \mathrm{y}=3 \quad, \quad z=1 \end{aligned}
Given:
\begin{aligned} &3 x+4 y+2 z=8 \\ &2 y-3 z=3 \\ &x-2 y+6 z=-2 \end{aligned}
Hint:
X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.
Solution:
\begin{aligned} & A=\left[\begin{array}{ccc} 3 & 4 & 2 \\ 0 & 2 & -3 \\ 1 & -2 & 6 \end{array}\right] \\ &{\left[\begin{array}{ccc} 3 & 4 & 2 \\ 0 & 2 & -3 \\ 1 & -2 & 6 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 8 \\ 3 \\ -2 \end{array}\right]} \end{aligned}\\ A\: \: X=B
\begin{aligned} |A|=\left|\begin{array}{ccc} 3 & 4 & 2 \\ 0 & 2 & -3 \\ 1 & -2 & 6 \end{array}\right| &=3(12-6)-4(0+3)+2(0-2) \\ &=18-12-4 \\ &=2 \end{aligned}
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].$
\begin{aligned} &C_{11}=(-1)^{1+1}\left|\begin{array}{cc} 2 & -3 \\ -2 & 6 \end{array}\right|=6 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 0 & -3 \\ 1 & 6 \end{array}\right|=-3 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 0 & 2 \\ 1 & -2 \end{array}\right|=-2 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{cc} 4 & 2 \\ -2 & 6 \end{array}\right|=-28 \end{aligned}
\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{ll} 3 & 2 \\ 1 & 6 \end{array}\right|=16 \quad, \quad \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc} 3 & 4 \\ 1 & -2 \end{array}\right|=10 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{cc} 4 & 2 \\ 2 & -3 \end{array}\right|=-16 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc} 3 & 2 \\ 0 & -3 \end{array}\right|=9 \\ &C_{33}=(-1)^{3+3}\left|\begin{array}{ll} 3 & 4 \\ 0 & 2 \end{array}\right|=6 \end{aligned}
\begin{aligned} \operatorname{adjA} &=\left[\begin{array}{ccc} 6 & -3 & -2 \\ -28 & 16 & 10 \\ -16 & 9 & 6 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 6 & -28 & -16 \\ -3 & 16 & 9 \\ -2 & 10 & 6 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} \operatorname{adjA} \\ &=\frac{1}{2}\left[\begin{array}{ccc} 6 & -28 & -16 \\ -3 & 16 & 9 \\ -2 & 10 & 6 \end{array}\right] \end{aligned}
\begin{aligned} X &=A^{-1} B \\ &=\frac{1}{2}\left[\begin{array}{ccc} 6 & -28 & -16 \\ -3 & 16 & 9 \\ -2 & 10 & 6 \end{array}\right]\left[\begin{array}{c} 8 \\ 3 \\ -2 \end{array}\right] \\ &=\frac{1}{2}\left[\begin{array}{c} 48-84+32 \\ -24+48-18 \\ -16+30-12 \end{array}\right] \end{aligned}
\begin{aligned} &=\frac{1}{2}\left[\begin{array}{c} -4 \\ 6 \\ 2 \end{array}\right] \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} -2 \\ 3 \\ 1 \end{array}\right]} \\ &x=-2, \quad y=3 \quad, \quad z=1 \end{aligned}

### Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (viii)

\begin{aligned} & x=1 \quad, \quad \mathrm{y}=1 \quad, \quad z=-1 \end{aligned}
Given:
\begin{aligned} &2 x+y+z=2 \\ &x+3 y-z=5 \\ &3 x+y-2 z=6 \end{aligned}
Hint:
X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.
Solution:
\begin{aligned} &{\left[\begin{array}{ccc} 2 & 1 & 1 \\ 1 & 3 & -1 \\ 3 & 1 & -2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 2 \\ 5 \\ 6 \end{array}\right]} \\ &A X=B \end{aligned}
\begin{aligned} |A|=\left|\begin{array}{ccc} 2 & 1 & 1 \\ 1 & 3 & -1 \\ 3 & 1 & -2 \end{array}\right|=& 2(-6+1)-1(-2+3)+1(1-9) \\ &=10-1-8 \\ &=-19 \end{aligned}
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].$
\begin{aligned} &C_{11}=(-1)^{1+1}\left|\begin{array}{ll} 3 & -1 \\ 1 & -2 \end{array}\right|=-5 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 1 & -1 \\ 3 & -2 \end{array}\right|=-1 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{ll} 1 & 3 \\ 3 & 1 \end{array}\right|=-8 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{cc} 1 & 1 \\ 1 & -2 \end{array}\right|=3 \end{aligned}
\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{cc} 2 & -1 \\ 3 & -2 \end{array}\right|=-7 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc} 2 & 1 \\ 3 & 1 \end{array}\right|=1 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{cc} 1 & 1 \\ 3 & -1 \end{array}\right|=-4 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc} 2 & 1 \\ 1 & -1 \end{array}\right|=3 \\ &C_{33}=(-1)^{3+3}\left|\begin{array}{ll} 2 & 1 \\ 1 & 3 \end{array}\right|=5 \end{aligned}
\begin{aligned} \operatorname{adj} A &=\left[\begin{array}{ccc} -5 & -1 & -8 \\ 3 & -7 & 1 \\ -4 & 3 & 5 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} -5 & 3 & -4 \\ -1 & -7 & 3 \\ -8 & 1 & 5 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} \operatorname{adjA} \\ &=\frac{1}{-19}\left[\begin{array}{ccc} -5 & 3 & -4 \\ -1 & -7 & 3 \\ -8 & 1 & 5 \end{array}\right] \end{aligned}
\begin{aligned} X &=A^{-1} B \\ &=\frac{1}{-19}\left[\begin{array}{ccc} -5 & 3 & -4 \\ -1 & -7 & 3 \\ -8 & 1 & 5 \end{array}\right]\left[\begin{array}{l} 2 \\ 5 \\ 6 \end{array}\right] \\ &=\frac{1}{-19}\left[\begin{array}{c} -10+15-24 \\ -2-35+18 \\ -16+5+30 \end{array}\right] \end{aligned}
\begin{aligned} &=\frac{1}{-19}\left[\begin{array}{c} -19 \\ -19 \\ 19 \end{array}\right] \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 1 \\ 1 \\ -1 \end{array}\right]} \\ &x=1 \quad, \quad y=1 \quad, \quad z=-1 \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (ix)

\begin{aligned} & x=-2 \quad, \quad \mathrm{y}=1 \quad, \quad z=2 \end{aligned}
Given:
\begin{aligned} &2 x+6 y=2 \\ &3 x-z=-8 \\ &2 x-y+z=-3 \end{aligned}
Hint:
X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.
Solution:
\begin{aligned} &{\left[\begin{array}{ccc} 2 & 6 & 0 \\ 3 & 0 & -1 \\ 2 & -1 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 2 \\ -8 \\ -3 \end{array}\right]} \\ &A X=B \end{aligned}
\begin{aligned} |A|=\left|\begin{array}{ccc} 2 & 6 & 0 \\ 3 & 0 & -1 \\ 2 & -1 & 1 \end{array}\right| &=2(0-1)-6(3+2)+0(-3+0) \\ &=-2-30 \\ &=-32 \end{aligned}
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].$
\begin{aligned} &C_{11}=(-1)^{1+1}\left|\begin{array}{cc} 0 & -1 \\ -1 & 1 \end{array}\right|=-1 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 3 & -1 \\ 2 & 1 \end{array}\right|=-5 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 3 & 0 \\ 2 & -1 \end{array}\right|=-3 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{cc} 6 & 0 \\ -1 & 1 \end{array}\right|=-6 \end{aligned}
\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{ll} 2 & 0 \\ 2 & 1 \end{array}\right|=2 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc} 2 & 6 \\ 2 & -1 \end{array}\right|=14 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{cc} 6 & 0 \\ 0 & -1 \end{array}\right|=-6 \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc} 2 & 0 \\ 3 & -1 \end{array}\right|=2 \\ &C_{33}=(-1)^{3+3}\left|\begin{array}{ll} 2 & 6 \\ 3 & 0 \end{array}\right|=-18 \end{aligned}
\begin{aligned} \operatorname{adjA} &=\left[\begin{array}{ccc} -1 & -5 & -3 \\ -1 & 2 & 14 \\ -6 & 2 & -18 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} -1 & -6 & -6 \\ -5 & 2 & 2 \\ -3 & 14 & -18 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=-\frac{1}{32}\left[\begin{array}{ccc} -1 & -6 & -6 \\ -5 & 2 & 2 \\ -3 & 14 & -18 \end{array}\right] \end{aligned}
\begin{aligned} X &=A^{-1} B \\ &=-\frac{1}{32}\left[\begin{array}{ccc} -1 & -6 & -6 \\ -5 & 2 & 2 \\ -3 & 14 & -18 \end{array}\right]\left[\begin{array}{c} 2 \\ -8 \\ -3 \end{array}\right] \\ &=-\frac{1}{32}\left[\begin{array}{c} -2+48+18 \\ -10-16-6 \\ -6-112+54 \end{array}\right] \end{aligned}
\begin{aligned} &=-\frac{1}{32}\left[\begin{array}{c} 64 \\ -32 \\ -64 \end{array}\right] \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} -2 \\ 1 \\ 2 \end{array}\right]} \\ &x=-2, \quad y=1 \quad, \quad z=2 \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (x)

\begin{aligned} & x=1 \quad, \quad \mathrm{y}=2 \quad, \quad z=3 \end{aligned}
Given:
\begin{aligned} &x-y+z=2 \\ &2 x-y=0 \\ &2 y-z=1 \end{aligned}
Hint:
X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.
Solution:
$\begin{array}{r} {\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 0 & 2 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 2 \\ 0 \\ 1 \end{array}\right]} \end{array}\\ \ A\; X=B$
\begin{aligned} |A|=\left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 0 & 2 & -1 \end{array}\right| &=1(1-0)+1(-2-0)+1(4-0) \\ &=1-2+4 \\ &=3 \end{aligned}
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right]$
\begin{aligned} &C_{11}=(-1)^{1+1}\left|\begin{array}{cc} -1 & 0 \\ 2 & -1 \end{array}\right|=1 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 2 & 0 \\ 0 & -1 \end{array}\right|=2 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 2 & -1 \\ 0 & 2 \end{array}\right|=4 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{cc} -1 & 1 \\ 2 & -1 \end{array}\right|=1 \end{aligned}
\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{cc} 1 & 1 \\ 0 & -1 \end{array}\right|=-1 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc} 1 & -1 \\ 0 & 2 \end{array}\right|=-2 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{ll} -1 & 1 \\ -1 & 0 \end{array}\right|=1 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll} 1 & 1 \\ 2 & 0 \end{array}\right|=2 \\ &C_{33}=(-1)^{3+3}\left|\begin{array}{ll} 1 & -1 \\ 2 & -1 \end{array}\right|=1 \end{aligned}
\begin{aligned} \operatorname{adjA} &=\left[\begin{array}{ccc} 1 & 2 & 4 \\ 1 & -1 & -2 \\ 1 & 2 & 1 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 1 & 1 & 1 \\ 2 & -1 & 2 \\ 4 & -2 & 1 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{3}\left[\begin{array}{ccc} 1 & 1 & 1 \\ 2 & -1 & 2 \\ 4 & -2 & 1 \end{array}\right] \end{aligned}
\begin{aligned} X &=A^{-1} B \\ &=\frac{1}{3}\left[\begin{array}{ccc} 1 & 1 & 1 \\ 2 & -1 & 2 \\ 4 & -2 & 1 \end{array}\right]\left[\begin{array}{l} 2 \\ 0 \\ 1 \end{array}\right] \\ &=\frac{1}{3}\left[\begin{array}{l} 2+1 \\ 4+2 \\ 8+1 \end{array}\right] \end{aligned}
\begin{aligned} =\frac{1}{3}\left[\begin{array}{l} 3 \\ 6 \\ 9 \end{array}\right] \end{aligned}
\begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]} \\ &x=1 \quad, \quad y=2 \quad, \quad z=3 \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (xi)

\begin{aligned} &x=1 \quad, \quad y=1 \quad, \quad z=2 \end{aligned}
Given:
\begin{aligned} &8 x+4 y+3 z=18 \\ &2 x+y+z=5 \\ &x+2 y+z=5 \end{aligned}
Hint:
X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.
Solution:
\begin{aligned} &{\left[\begin{array}{lll} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 18 \\ 5 \\ 5 \end{array}\right]} \\ &A X=B \end{aligned}
\begin{aligned} |A|=\left|\begin{array}{lll} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{array}\right|=& 8(1-2)-4(2-1)+3(4-1) \\ &=-8-4+9 \\ &=-3 \end{aligned}
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].$
\begin{aligned} &C_{11}=(-1)^{1+1}\left|\begin{array}{ll} 1 & 1 \\ 2 & 1 \end{array}\right|=-1 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 2 & 1 \\ 1 & 1 \end{array}\right|=-1 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right|=3 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{ll} 4 & 3 \\ 2 & 1 \end{array}\right|=2 \end{aligned}
\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{ll} 8 & 3 \\ 1 & 1 \end{array}\right|=5 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll} 8 & 4 \\ 1 & 2 \end{array}\right|=-12 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{ll} 4 & 3 \\ 1 & 1 \end{array}\right|=1 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll} 8 & 3 \\ 2 & 1 \end{array}\right|=-2 \\ &C_{33}=(-1)^{3+3}\left|\begin{array}{ll} 8 & 4 \\ 2 & 1 \end{array}\right|=0 \end{aligned}
\begin{aligned} \text { adjA } &=\left[\begin{array}{ccc} -1 & -1 & 3 \\ 2 & 5 & -12 \\ -1 & -2 & 0 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} -1 & 2 & 1 \\ -1 & 5 & -2 \\ 3 & -12 & 0 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=-\frac{1}{3}\left[\begin{array}{ccc} -1 & 2 & 1 \\ -1 & 5 & -2 \\ 3 & -12 & 0 \end{array}\right] \end{aligned}
\begin{aligned} X &=A^{-1} B \\ &=-\frac{1}{3}\left[\begin{array}{ccc} -1 & 2 & 1 \\ -1 & 5 & -2 \\ 3 & -12 & 0 \end{array}\right]\left[\begin{array}{c} 18 \\ 5 \\ 5 \end{array}\right] \\ &=-\frac{1}{3}\left[\begin{array}{c} -18+10+5 \\ -18+25-10 \\ 54-60 \end{array}\right] \end{aligned}
\begin{aligned} &=-\frac{1}{3}\left[\begin{array}{r} -3 \\ -3 \\ -6 \end{array}\right] \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right]} \\ &x=1 \quad, \quad y=1 \quad, \quad z=2 \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (xii)

\begin{aligned} &x=3 \quad, \quad y=1 \quad, \quad z=2 \end{aligned}
Given:
\begin{aligned} &x+y+z=6 \\ &x+2 z=7 \\ &3 x+y+z=12 \end{aligned}
Hint:
X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.
Solution:
\begin{aligned} &{\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 6 \\ 7 \\ 12 \end{array}\right]} \\ &A X=B \end{aligned}
\begin{aligned} |A|=\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \end{array}\right|=& 1(0-2)-1(1-6)+1(1-0) \\ &=-2+5+1 \\ &=4 \end{aligned}
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].$
\begin{aligned} &C_{11}=(-1)^{1+1}\left|\begin{array}{ll} 0 & 2 \\ 1 & 1 \end{array}\right|=-2 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{ll} 1 & 2 \\ 3 & 1 \end{array}\right|=5 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{ll} 1 & 0 \\ 3 & 1 \end{array}\right|=1 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right|=0 \end{aligned}
\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{ll} 1 & 1 \\ 3 & 1 \end{array}\right|=-2 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc} 1 & 1 \\ 3 & 1 \end{array}\right|=2 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{ll} 1 & 1 \\ 0 & 2 \end{array}\right|=2 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc} 1 & 1 \\ 0 & 2 \end{array}\right|=-1 \\ &C_{33}=(-1)^{3+3}\left|\begin{array}{ll} 1 & 1 \\ 1 & 0 \end{array}\right|=-1 \end{aligned}
\begin{aligned} \operatorname{adj} A &=\left[\begin{array}{ccc} -2 & 5 & 1 \\ 0 & -2 & 2 \\ 2 & -1 & -1 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} -2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{4}\left[\begin{array}{ccc} -2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1 \end{array}\right] \end{aligned}
\begin{aligned} X &=A^{-1} B \\ &=\frac{1}{4}\left[\begin{array}{ccc} -2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1 \end{array}\right]\left[\begin{array}{c} 6 \\ 7 \\ 12 \end{array}\right] \\ &=\frac{1}{4}\left[\begin{array}{c} -12+0+24 \\ 30-14-12 \\ 6+14-12 \end{array}\right] \end{aligned}
\begin{aligned} &=\frac{1}{4}\left[\begin{array}{c} 12 \\ 4 \\ 8 \end{array}\right] \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 3 \\ 1 \\ 2 \end{array}\right]} \\ &x=3, \quad y=1 \quad, \quad z=2 \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (xiii)

\begin{aligned} &x=2 \quad, \quad y=3 \quad, \quad z=5 \end{aligned}
Given:
\begin{aligned} &\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4 \\ &\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1 \\ &\frac{6}{x}+\frac{9}{y}-\frac{-20}{z}=2 \end{aligned}
Hint:
X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.
Solution:
\begin{aligned} &\text { Let } \frac{1}{x}=a, \frac{1}{y}=b, \frac{1}{z}=c\\ &\left[\begin{array}{ccc} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 4 \\ 1 \\ 2 \end{array}\right]\\ &\mathrm{A} \mathrm{X}=\mathrm{B} \end{aligned}
\begin{aligned} |A|=\left | \begin{array}{ccc} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{array} \right | &=2(120-45)-3(-80-30)+10(36+36) \\ &=150+330+720 \\ &=1200 \end{aligned}
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].$
\begin{aligned} &C_{11}=(-1)^{1+1}\left|\begin{array}{cc} -6 & 5 \\ 9 & -20 \end{array}\right|=75 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 4 & 5 \\ 6 & -20 \end{array}\right|=110 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 4 & -6 \\ 6 & 9 \end{array}\right|=72 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{cc} 3 & 10 \\ 9 & -20 \end{array}\right|=150 \end{aligned}
\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{cc} 2 & 10 \\ 6 & -20 \end{array}\right|=-100 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll} 2 & 3 \\ 6 & 9 \end{array}\right|=0 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{cc} 3 & 10 \\ -6 & 5 \end{array}\right|=75 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc} 2 & 10 \\ 4 & 5 \end{array}\right|=30 \end{aligned}
\begin{aligned} C_{33}=(-1)^{3+3}\left|\begin{array}{cc} 2 & 3 \\ 4 & -6 \end{array}\right|=-24 \\ \operatorname{adjA} =\left[\begin{array}{ccc} 75 & 110 & 72 \\ 150 & -100 & 0 \\ 75 & 30 & -24 \end{array}\right]^{T} \\ =\left[\begin{array}{ccc} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{1200}\left[\begin{array}{ccc} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{array}\right] \end{aligned}
\begin{aligned} X &=A^{-1} B \\ &=\frac{1}{1200}\left[\begin{array}{ccc} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{array}\right]\left[\begin{array}{l} 4 \\ 1 \\ 2 \end{array}\right] \\ &=\frac{1}{1200}\left[\begin{array}{c} 300+150+150 \\ 440-100+60 \\ 288-48 \end{array}\right] \end{aligned}
\begin{aligned} &=\frac{1}{1200}\left[\begin{array}{l} 600 \\ 400 \\ 240 \end{array}\right] \\ &\frac{1}{x}=a=\frac{1200}{600}, \frac{1}{y}=b=\frac{1200}{400}, \frac{1}{z}=c=\frac{1200}{240} \\ &x=2, \quad y=3 \quad, \quad z=5 \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 2 subquestion (xiv)

\begin{aligned} &x=2 \quad, \quad y=1 \quad, \quad z=3 \end{aligned}
Given:
\begin{aligned} &x-y+2 z=7 \\ &3 x+4 y-5 z=-5 \\ &2 x-y+3 z=12 \end{aligned}
Hint:
X=A-1B is used to solve this problem. First we find the determinant of matrix A i.e |A| then will find the co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.
Solution:
\begin{aligned} &A=\left[\begin{array}{ccc} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{array}\right] \\ \end{aligned}
\begin{aligned} |A|=\left|\begin{array}{ccc} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{array}\right| &=1(12-5)+1(9+10)+2(-3-8) \\ &=7+19-22 \\ &=4 \end{aligned}
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].$
\begin{aligned} &C_{11}=(-1)^{1+1}\left|\begin{array}{cc} 4 & -5 \\ -1 & 3 \end{array}\right|=7 & \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 3 & -5 \\ 2 & 3 \end{array}\right|=-19 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 3 & 4 \\ 2 & -1 \end{array}\right|=-11 & , \quad C_{21}=(-1)^{2+1}\left|\begin{array}{rr} -1 & 2 \\ -1 & 3 \end{array}\right|=1 \end{aligned}
\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right|=-1 \quad \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc} 1 & -1 \\ 2 & -1 \end{array}\right|=-1 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{cc} -1 & 2 \\ 4 & -5 \end{array}\right|=-3 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc} 1 & 2 \\ 3 & -5 \end{array}\right|=11 \\ &C_{33}=(-1)^{3+3}\left|\begin{array}{cc} 1 & -1 \\ 3 & 4 \end{array}\right|=7 \end{aligned}
\begin{aligned} \operatorname{adj} A &=\left[\begin{array}{ccc} 7 & -19 & -11 \\ 1 & -1 & -1 \\ -3 & 11 & 7 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} \text { adjA } \\ &=\frac{1}{4}\left[\begin{array}{ccc} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{array}\right] \end{aligned}
\begin{aligned} X &=A^{-1} B \\ &=\frac{1}{4}\left[\begin{array}{ccc} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{array}\right]\left[\begin{array}{c} 7 \\ -5 \\ 12 \end{array}\right] \\ &=\frac{1}{4}\left[\begin{array}{c} 49-5-36 \\ -133+5+132 \\ -77+5+84 \end{array}\right] \end{aligned}
\begin{aligned} &=\frac{1}{4}\left[\begin{array}{c} 8 \\ 4 \\ 12 \end{array}\right] \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 2 \\ 1 \\ 3 \end{array}\right]} \\ &x=2, \quad y=1 \quad, \quad z=3 \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 3 subquestion (i)

\begin{aligned} x=\frac{1-2 k}{3}, y=k \\ \end{aligned}
Given:
\begin{aligned} 6 x+4 y=2,\: \: 9 x+6 y=3 \end{aligned}
Hint:
A system of two linear equations can have one solution, an infinite number of solution, if a system has no solution it’s called inconsistent
Solution:
\begin{aligned} &6 x+4 y=2 \; \; \; \; \; \; \;....(i) \\ &9 x+6 y=3\; \; \; \; \; \; \;....(ii) \end{aligned}
\begin{aligned} &A X=B\\ &\text { Here, }\\ &A=\left[\begin{array}{ll} 6 & 4 \\ 9 & 6 \end{array}\right], \quad X=\left[\begin{array}{l} x \\ y \end{array}\right] \text { and } B=\left[\begin{array}{l} 2 \\ 3 \end{array}\right]\\ &\left[\begin{array}{ll} 6 & 4 \\ 9 & 6 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 2 \\ 3 \end{array}\right] \end{aligned}
\begin{aligned} |A| &=\left|\begin{array}{ll} 6 & 4 \\ 9 & 6 \end{array}\right| \\ &=36-36 \\ |A| &=0 \end{aligned}
So, A is singular. Thus the given system of equation is either inconsistent or it is consistent with indefinitely many solutions because
$(\operatorname{adj} A) B \neq 0 \text { or }(a d j A)=0$
$Let\; C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
\begin{aligned} &C_{11}=6, C_{12}=-9, C_{21}=-4, C_{22}=6 \\ &\operatorname{adj} A=\left[\begin{array}{cc} 6 & -9 \\ -4 & 6 \end{array}\right]^{T} \\ &\quad=\left[\begin{array}{cc} 6 & -4 \\ -9 & 6 \end{array}\right] \end{aligned}
\begin{aligned} (\text { adjA }) B &=\left[\begin{array}{cc} 6 & -4 \\ -9 & 6 \end{array}\right]\left[\begin{array}{l} 2 \\ 3 \end{array}\right] \\ &=\left[\begin{array}{c} 12-12 \\ -18+18 \end{array}\right] \\ &=\left[\begin{array}{l} 0 \\ 0 \end{array}\right] \end{aligned}
If |A| = 0 and (adjA)B = 0 then the system is consistent and has infinitely many solutions.
Thus, AX = B has infinitely many solutions
Substituting y = k in eqn (i), We get
\begin{aligned} &6 x+4 k=2 \\ &6 x=2-4 k \\ &x=\frac{2-4 k}{6} \\ &x=\frac{1-2 k}{3} \\ &x=\frac{1-2 k}{3} \text { and } y=k \end{aligned}
The values of x and y satisfy the third equation.
\begin{aligned} Thus\; x=\frac{1-2 k}{3} \text { and } y=k \end{aligned}
Where ‘ $k$ ’ is a real number satisfy the given system of equations.

Solution of Simultaneous Linear Equation exercise 7.1 question 3 subquestion (ii)

\begin{aligned} x=\frac{5-3 k}{2} \text { and } y=k\\ \end{aligned}
Given:
\begin{aligned} &2 x+3 y=5\\ &6 x+9 y=15 \end{aligned}
Hint:
Consistent equation means two or more equations that are possible to solve based on using set of values for the variables.
Solution:
Here,
\begin{aligned} &2 x+3 y=5\; \; \; \; \; \; ......(i)\\ &6 x+9 y=15\; \; \; \; \; ......(ii) \end{aligned} \\ AX=B \\ Where
\begin{aligned} &A=\left[\begin{array}{ll} 2 & 3 \\ 6 & 9 \end{array}\right], X=\left[\begin{array}{l} x \\ y \end{array}\right] \text { and } B=\left[\begin{array}{c} 5 \\ 15 \end{array}\right] \\ &{\left[\begin{array}{ll} 2 & 3 \\ 6 & 9 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 5 \\ 15 \end{array}\right]} \\ &\begin{aligned} |A| &=\left|\begin{array}{ll} 2 & 3 \\ 6 & 9 \end{array}\right| \\ &=18-18 \\ |A| &=0 \end{aligned} \end{aligned}
So, A is singular. Thus the given system of equation is either inconsistent or it is consistent with indefinitely many solutions because
$(\operatorname{adj} A) B \neq 0 \text { or }(a d j A)=0$
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
\begin{aligned} &C_{11}=9, C_{12}=-6, C_{21}=-3, C_{22}=2 \\ &\operatorname{adj} A=\left[\begin{array}{cc} 9 & -6 \\ -3 & 2 \end{array}\right]^{T} \\ &\quad=\left[\begin{array}{cc} 9 & -3 \\ -6 & 2 \end{array}\right] \end{aligned}
\begin{aligned} (\operatorname{adjA}) B &=\left[\begin{array}{cc} 9 & -3 \\ -6 & 2 \end{array}\right]\left[\begin{array}{c} 5 \\ 15 \end{array}\right] \\ &=\left[\begin{array}{c} 45-45 \\ -30+30 \end{array}\right] \\ &=\left[\begin{array}{l} 0 \\ 0 \end{array}\right] \end{aligned}
If |A| = 0 and (adjA)B = 0 then the system is consistent and has infinitely many solutions.
Thus, AX = B has infinitely many solutions
Substituting y = k in eqn (i), We get
\begin{aligned} &2 x+3 k=5 \\ &2 x=5-3 k \\ &x=\frac{5-3 k}{2} \\ &\text { And } y=k \end{aligned}
The values of x and y satisfy the third equation.
Thus\; \; \begin{aligned} x=\frac{5-3 k}{2} \text { and } y=k\\ \end{aligned}
Where ‘ $k$ ’ is a real number satisfy the given system of equations.

Solution of Simultaneous Linear Equation exercise 7.1 question 3 subquestion (iii)

$x=\frac{7-16 k}{11}, y=\frac{3+k}{11} \text { and } z=k$
Given:
\begin{aligned} &5 x+3 y+7 z=4 \\ &3 x+26 y+2 z=9 \\ &7 x+2 y+10 z=5 \end{aligned}
Hint:
Consistent equation means two or more equations that are possible to solve based on using set of values for the variables.
Solution: Here,
\begin{aligned} 5 x+3 y+7 z=4 \; \; \; \; \; ......(i)\\ 3 x+26 y+2 z=9 \; \; \; \; \; ......(ii)\\ 7 x+2 y+10 z=5 \; \; \; \; \; ......(iii) \end{aligned} \\ \\ AX=B \\ \\ Where
\begin{aligned} &A=\left[\begin{array}{ccc} 5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10 \end{array}\right], X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{l} 4 \\ 9 \\ 5 \end{array}\right] \\ &{\left[\begin{array}{ccc} 5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 4 \\ 9 \\ 5 \end{array}\right]} \end{aligned}
\begin{aligned} |A| &=\left|\begin{array}{ccc} 5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10 \end{array}\right| \\ &=5(260-4)-3(30-14)+7(6-182) \\ &=1280-48-1232=0 \\ |A| &=0 \end{aligned}
So, A is singular. Thus the given system of equation is either inconsistent or it is consistent with indefinitely many solutions because
$(\operatorname{adj} A) B \neq 0 \text { or }(a d j A)=0$
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
\begin{aligned} &C_{11}=(-1)^{1+1}\left|\begin{array}{cc} 26 & 2 \\ 2 & 10 \end{array}\right|=256 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 3 & 2 \\ 7 & 10 \end{array}\right|=-16 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 3 & 26 \\ 7 & 2 \end{array}\right|=-176 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{cc} 3 & 7 \\ 2 & 10 \end{array}\right|=-16 \end{aligned}
\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{cc} 5 & 7 \\ 7 & 10 \end{array}\right|=1 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll} 5 & 3 \\ 7 & 2 \end{array}\right|=11 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{cc} 3 & 7 \\ 26 & 2 \end{array}\right|=-176 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll} 5 & 7 \\ 3 & 2 \end{array}\right|=11 \\ &C_{33}=(-1)^{3+3}\left|\begin{array}{cc} 5 & 3 \\ 3 & 26 \end{array}\right|=121 \end{aligned}
\begin{aligned} \operatorname{adj} A &=\left[\begin{array}{ccc} 256 & -16 & -176 \\ -16 & 1 & 11 \\ -176 & 11 & 121 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 256 & -16 & -176 \\ -16 & 1 & 11 \\ -176 & 11 & 121 \end{array}\right] \end{aligned}
\begin{aligned} (\operatorname{adj} A) B &=\left[\begin{array}{ccc} 256 & -16 & -176 \\ -16 & 1 & 11 \\ -176 & 11 & 121 \end{array}\right]\left[\begin{array}{l} 4 \\ 9 \\ 5 \end{array}\right] \\ &=\left[\begin{array}{c} 1024-144-880 \\ -64+9+55 \\ -704+9+605 \end{array}\right] \\ &=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] \end{aligned}
If |A| = 0 and (adjA)B = 0 then the system is consistent and has infinitely many solutions.
Thus, AX = B has infinitely many solutions
Substituting z = k in eqn (i) and eqn (ii), We get
\begin{aligned} &5 x+3 y=4-7 k \text { and } 3 x+26 y=9-2 k \\ &{\left[\begin{array}{cc} 5 & 3 \\ 3 & 26 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 4-7 k \\ 9-2 k \end{array}\right]} \end{aligned}
\begin{aligned} |A| &=\left|\begin{array}{cc} 5 & 3 \\ 3 & 26 \end{array}\right| \\ &=130-9 \\ &=121 \neq 0 \\ \operatorname{adj} A &=\left|\begin{array}{cc} 26 & -3 \\ -3 & 5 \end{array}\right| \\ A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{121}\left[\begin{array}{cc} 26 & -3 \\ -3 & 5 \end{array}\right] \end{aligned}
\begin{aligned} &X=A^{-1} B \\ &{\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{121}\left[\begin{array}{cc} 26 & -3 \\ -3 & 5 \end{array}\right]\left[\begin{array}{l} 4-7 k \\ 9-2 k \end{array}\right]} \\ &{\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{121}\left[\begin{array}{l} 104-182 k-27+6 k \\ -12+21 k+45-10 k \end{array}\right]} \end{aligned}
\begin{aligned} &{\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} \frac{77-176 k}{121} \\ \frac{33+11 k}{121} \end{array}\right]} \\ &x=\frac{11(7-16 k)}{121}, y=\frac{11(3+k)}{121} \text { and } z=k \\ &x=\frac{7-16 k}{11}, y=\frac{3+k}{11} \text { and } z=k \end{aligned}
The values of x and y and z satisfy the third equation.
$Thus\; x=\frac{7-16 k}{11}, y=\frac{3+k}{11} \text { and } z=k$
Where ‘ $k$ ’ is a real number satisfy the given system of equations.

Solution of Simultaneous Linear Equation exercise 7.1 question 3 subquestion (iv)

\begin{aligned} x=\frac{5}{3}, y=\frac{3 k-4}{3} \text { and } z=k\\ \end{aligned}
Given:
\begin{aligned} &x-y+z=3\\ &2 x+y-z=2\\ &-x-2 y+2 z=1 \end{aligned}
Hint:
Consistent equation means two or more equations that are possible to solve based on using set of values for the variables.
Solution: Here,
\begin{aligned} &x-y+z=3\; \; \; \; \; ....(i)\\ &2 x+y-z=2\; \; \; \; \; ....(ii)\\ &-x-2 y+2 z=1\; \; \; \; \; ....(iii) \end{aligned}
\begin{aligned} &A X=B \\ &\qquad A=\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ -1 & -2 & 2 \end{array}\right], X=\left[\begin{array}{c} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{l} 3 \\ 2 \\ 1 \end{array}\right] \\ &{\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ -1 & -2 & 2 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \\ 1 \end{array}\right]} \\ &\begin{aligned} |A| &=\left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ -1 & -2 & 2 \end{array}\right| \\ &=1(2-2)+1(4-1)+1(-4+1) \\ &=0+3-3=0 \\ |A| &=0 \end{aligned} \end{aligned}
So, A is singular. Thus the given system of equation is either inconsistent or it is consistent with indefinitely many solutions because
$(\operatorname{adj} A) B \neq 0 \text { or }(a d j A)=0 \\ C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
\begin{aligned} &C_{11}=(-1)^{1+1}\left|\begin{array}{cc} 1 & -1 \\ -2 & 2 \end{array}\right|=0 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 2 & -1 \\ -1 & 2 \end{array}\right|=-3 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{cc} 2 & 1 \\ -1 & -2 \end{array}\right|=-3 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{cc} -1 & 1 \\ 2 & 2 \end{array}\right|=0 \end{aligned}
\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{cc} 1 & 1 \\ -1 & 2 \end{array}\right|=3 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc} 1 & -1 \\ -1 & -2 \end{array}\right|=3 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{cc} -1 & 1 \\ 1 & -1 \end{array}\right|=0 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc} 1 & 1 \\ 2 & -1 \end{array}\right|=3 \\ &C_{33}=(-1)^{3+3}\left|\begin{array}{cc} 1 & -1 \\ 2 & 1 \end{array}\right|=3 \end{aligned}
\begin{aligned} \operatorname{adj} A &=\left[\begin{array}{ccc} 0 & -3 & -3 \\ 0 & 3 & 3 \\ 0 & 3 & 3 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 0 & 0 & 0 \\ -3 & 3 & 3 \\ -3 & 3 & 3 \end{array}\right] \end{aligned}
\begin{aligned} (\text { adj } A) B &=\left[\begin{array}{ccc} 0 & 0 & 0 \\ -3 & 3 & 3 \\ -3 & 3 & 3 \end{array}\right]\left[\begin{array}{l} 3 \\ 2 \\ 1 \end{array}\right] \\ &=\left[\begin{array}{c} 0 \\ -9+6+3 \\ -9+6+3 \end{array}\right] \\ &=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] \end{aligned}

If |A| = 0 and (adjA)B = 0 then the system is consistent and has infinitely many solutions.
Thus, AX = B has infinitely many solutions
Substituting z = k in eqn (i) and eqn (ii), We get
\begin{aligned} &x-y=3-k \text { and } 2 x+y=2+k\\ &\left[\begin{array}{cc} 1 & -1 \\ 2 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 3-k \\ 2+k \end{array}\right]\\ &\text { Now, }\\ &|A|=\left|\begin{array}{cc} 1 & -1 \\ 2 & 1 \end{array}\right|\\ &=1+2\\ &=3 \neq 0 \end{aligned}
\begin{aligned} \operatorname{adj} A &=\left|\begin{array}{cc} 1 & 2 \\ -1 & 1 \end{array}\right| \\ A^{-1} &=\frac{1}{|A|} \text { adjA } \\ &=\frac{1}{3}\left[\begin{array}{cc} 1 & 2 \\ -1 & 1 \end{array}\right] \\ X=& A^{-1} B \end{aligned}
\begin{aligned} &{\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{3}\left[\begin{array}{cc} 1 & 2 \\ -1 & 1 \end{array}\right]\left[\begin{array}{l} 3-k \\ 2+k \end{array}\right]} \\ &{\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{3}\left[\begin{array}{c} 3-k+2+k \\ -6+2 k+2+k \end{array}\right]} \end{aligned}
\begin{aligned} &{\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} \frac{5}{3} \\ \frac{3 k-4}{3} \end{array}\right]} \\ &x=\frac{5}{3}, y=\frac{3 k-4}{3} \text { and } z=k \end{aligned}
The values of x and y and z satisfy the third equation.
\begin{aligned} Thus,\; x=\frac{5}{3}, y=\frac{3 k-4}{3} \text { and } z=k \end{aligned}
Where ‘ $k$ ’ is a real number satisfy the given system of equations.

Solution of Simultaneous Linear Equation exercise 7.1 question 3 subquestion (v)

$x=k-2,\; y=8-2k\; and\; z=k$
Given:
\begin{aligned} &x+y+z=6 \\ &x+2 y+3 z=14 \\ &x+4 y+7 z=30 \end{aligned}
Hint:
Consistent equation means two or more equations that are possible to solve based on using set of values for the variables.
Solution: Here,
\begin{aligned} &x+y+z=6 \; \; \; \; \; .....(i)\\ &x+2 y+3 z=14 \; \; \; \; \; .....(ii) \\ &x+4 y+7 z=30 \; \; \; \; \; .....(iii) \end{aligned}
\begin{aligned} &A X=B\\ &\text { Where }\\ &A=\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \end{array}\right], \quad X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{c} 6 \\ 14 \\ 30 \end{array}\right] \end{aligned}
\begin{aligned} {\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 6 \\ 14 \\ 30 \end{array}\right]} \\ |A|=\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \end{array}\right| \end{aligned}\\ =1(14-12)-1(7-3)+1(4-2) \\ =2-4+2=0 \\ \left | A \right |=0
So, A is singular. Thus the given system of equation is either inconsistent or it is consistent with indefinitely many solutions because
$(\operatorname{adj} A) B \neq 0 \text { or }(a d j A)=0 \\ C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
\begin{aligned} &C_{11}=(-1)^{1+1}\left|\begin{array}{ll} 2 & 3 \\ 4 & 7 \end{array}\right|=2 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{ll} 1 & 3 \\ 1 & 7 \end{array}\right|=-4 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{ll} 1 & 2 \\ 1 & 4 \end{array}\right|=2 \quad , \quad C_{21}=(-1)^{2+1}\left|\begin{array}{ll} 1 & 1 \\ 2 & 7 \end{array}\right|=-3 \end{aligned}
\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{ll} 1 & 1 \\ 1 & 7 \end{array}\right|=6 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll} 1 & 1 \\ 1 & 4 \end{array}\right|=-3 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{ll} 1 & 1 \\ 2 & 3 \end{array}\right|=1 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll} 1 & 1 \\ 1 & 3 \end{array}\right|=-2 \\ &C_{33}=(-1)^{3+3}\left|\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right|=1 \end{aligned}
\begin{aligned} \operatorname{adj} A &=\left[\begin{array}{ccc} 2 & -4 & 2 \\ -3 & 6 & -3 \\ 1 & -2 & 1 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 2 & -3 & 1 \\ -4 & 6 & -2 \\ 2 & -3 & 1 \end{array}\right] \end{aligned}
\begin{aligned} (\operatorname{adj} A) B &=\left[\begin{array}{ccc} 2 & -3 & 1 \\ -4 & 6 & -2 \\ 2 & -3 & 1 \end{array}\right]\left[\begin{array}{c} 6 \\ 14 \\ 30 \end{array}\right] \\ &=\left[\begin{array}{c} 12-42+30 \\ -24+84-60 \\ 12-42+30 \end{array}\right] \\ &=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] \end{aligned}
If |A| = 0 and (adjA)B = 0 then the system is consistent and has infinitely many solutions.
Thus, AX = B has infinitely many solutions
Substituting z = k in eqn (i) and eqn (ii), We get
\begin{aligned} &x+y=6-k \text { and } x+2 y=14-3 k\\ &\left[\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 6-k \\ 14-3 k \end{array}\right]\\ &\text { Now, }\\ &|A|=\left|\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right|\\ &=2-1\\ &=1 \neq 0 \end{aligned}
\begin{aligned} \operatorname{adj} A &=\left|\begin{array}{cc} 2 & -1 \\ -1 & 1 \end{array}\right| \\ A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{1}\left[\begin{array}{cc} 2 & -1 \\ -1 & 1 \end{array}\right] \\ X=& A^{-1} B \end{aligned}
\begin{aligned} &{\left[\begin{array}{c} x \\ y \end{array}\right]=\frac{1}{1}\left[\begin{array}{cc} 2 & -1 \\ -1 & 1 \end{array}\right]\left[\begin{array}{c} 6-k \\ 14-3 k \end{array}\right]} \\ &{\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{1}\left[\begin{array}{c} 12-2 k-14+3 k \\ -6+k+14-3 k \end{array}\right]} \\ &{\left[\begin{array}{c} x \\ y \end{array}\right]=\left[\begin{array}{c} \frac{k-2}{1} \\ \frac{8-2 k}{1} \end{array}\right]} \end{aligned}
$x=k-2,\; y=8-2k\; and\; z=k$

The values of x and y and z satisfy the third equation.
$Thus\; x=k-2,\; y=8-2k\; and\; z=k$
Where ‘ $k$ ’ is a real number satisfy the given system of equations.

Solution of Simultaneous Linear Equation exercise 7.1 question 3 subquestion (vi)

\begin{aligned} x=\frac{1-2 k}{2}, y=k \text { and } z=0 \end{aligned}
Given:
\begin{aligned} &2 x+2 y-2 z=1\\ &4 x+4 y-z=2\\ &6 x+6 y+2 z=3 \end{aligned}
Hint:
Consistent equation means two or more equations that are possible to solve based on using set of values for the variables.
Solution: Here,
\begin{aligned} &2 x+2 y-2 z=1\; \; \; \; \; .....(i)\\ &4 x+4 y-z=2\; \; \; \; \; .....(ii)\\ &6 x+6 y+2 z=3\; \; \; \; \; .....(iii) \end{aligned}
\begin{aligned} &A X=B\\ &\text { Where }\\ &A=\left[\begin{array}{ccc} 2 & 2 & -2 \\ 4 & 4 & -1 \\ 6 & 6 & 2 \end{array}\right], \quad X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right] \end{aligned}
\begin{aligned} &{\left[\begin{array}{ccc} 2 & 2 & -2 \\ 4 & 4 & -1 \\ 6 & 6 & 2 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]} \\ &|A|=\left|\begin{array}{ccc} 2 & 2 & -2 \\ 4 & 4 & -1 \\ 6 & 6 & 2 \end{array}\right| \\ &= 2(8+6)-2(8+6)-2(24-24) \\ & =28-28=0 \end{aligned}
So, A is singular. Thus the given system of equation is either inconsistent or it is consistent with indefinitely many solutions because
$(\operatorname{adj} A) B \neq 0 \text { or }(a d j A)=0$
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
\begin{aligned} &C_{11}=(-1)^{1+1}\left|\begin{array}{cc} 4 & -1 \\ 6 & 2 \end{array}\right|=14 \quad, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc} 4 & -1 \\ 6 & 2 \end{array}\right|=-14 \\ &C_{13}=(-1)^{1+3}\left|\begin{array}{ll} 4 & 4 \\ 6 & 6 \end{array}\right|=0 \quad, \quad C_{21}=(-1)^{2+1}\left|\begin{array}{cc} 2 & -2 \\ 6 & 2 \end{array}\right|=-16 \end{aligned}
\begin{aligned} &C_{22}=(-1)^{2+2}\left|\begin{array}{cc} 2 & -2 \\ 6 & 2 \end{array}\right|=16 \quad, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll} 2 & 2 \\ 6 & 6 \end{array}\right|=0 \\ &C_{31}=(-1)^{3+1}\left|\begin{array}{ll} 2 & -2 \\ 4 & -1 \end{array}\right|=6 \quad, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll} 2 & -2 \\ 4 & -1 \end{array}\right|=-6 \\ &C_{33}=(-1)^{3+3}\left|\begin{array}{ll} 2 & 2 \\ 4 & 4 \end{array}\right|=0 \end{aligned}
\begin{aligned} \operatorname{adj} A &=\left[\begin{array}{ccc} -14 & -14 & 0 \\ -16 & 16 & 0 \\ 6 & -6 & 0 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 14 & -16 & 6 \\ -14 & 16 & -6 \\ 0 & 0 & 0 \end{array}\right] \end{aligned}
\begin{aligned} (\text { adjA }) B &=\left[\begin{array}{ccc} 14 & -16 & 6 \\ -14 & 16 & -6 \\ 0 & 0 & 0 \end{array}\right]\left[\begin{array}{c} 1 \\ 2 \\ 3 \end{array}\right] \\ &=\left[\begin{array}{c} 14-32+18 \\ -14+32-18 \\ 0 \end{array}\right] \\ &=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] \end{aligned}

If |A| = 0 and (adjA)B = 0 then the system is consistent and has infinitely many solutions.
Thus, AX = B has infinitely many solutions
Substituting y = k in eqn (i) and eqn (ii), We get
\begin{aligned} &2 x+2 z=1-2 k \text { and } 4 x+z=2-4 k \\ &{\left[\begin{array}{ll} 2 & -2 \\ 4 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 1-2 k \\ 2-4 k \end{array}\right]} \\ &\begin{aligned} |A| &=\left|\begin{array}{ll} 2 & -2 \\ 4 & -1 \end{array}\right| \\ &=-2+8 \\ &=6 \neq 0 \end{aligned} \end{aligned}
\begin{aligned} \operatorname{adj} A &=\left|\begin{array}{rr} -1 & 2 \\ -4 & 2 \end{array}\right| \\ A^{-1} &=\frac{1}{|A|} \text { adjA } \\ &=\frac{1}{6}\left[\begin{array}{rr} -1 & 2 \\ -4 & 2 \end{array}\right] \\ X=& A^{-1} B \end{aligned}
\begin{aligned} &{\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{6}\left[\begin{array}{ll} -1 & 2 \\ -4 & 2 \end{array}\right]\left[\begin{array}{l} 1-2 k \\ 2-4 k \end{array}\right]} \\ &{\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{6}\left[\begin{array}{l} -1+2 k+4-8 k \\ -4+8 k+4-8 k \end{array}\right]} \\ &{\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} \frac{3-6 k}{6} \\ 0 \end{array}\right]} \end{aligned}
$x=\frac{1}{2} -k ,\; y=k\; and\; z=0$
The values of x and y and z satisfy the third equation.
$x=\frac{1}{2} -k ,\; y=k\; and\; z=0$
Where ‘ $k$ ’ is a real number satisfy the given system of equations.

Solution of Simultaneous Linear Equation exercise 7.1 question 4 subquestion (i)

Inconsistent
Given:
$2x+5y=7 \: \: ,\: \: 6x+15y=13$
Hint:
Inconsistent means two or more equations that are impossible to solve based on using one set of values for variables.
Solution:
The given system of equations can be expressed as follows
$A X=B \\ Here, \\ A=\left[\begin{array}{cc}2 & 5 \\ 6 & 15\end{array}\right] \quad, \quad X=\left[\begin{array}{l}x \\ y\end{array}\right] \quad and \quad B=\left[\begin{array}{c}7 \\ 13\end{array}\right] \\ Now, \\ |A|=\left|\begin{array}{cc}2 & 5 \\ 6 & 15\end{array}\right|=|30-30|=0$
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
\begin{aligned} &C_{11}=-1^{1+1}(15)=15 \quad, \quad C_{12}=-1^{1+2}(6)=-6 \\ &C_{21}=-1^{2+1}(5)=-5 \quad, \quad C_{22}=-1^{2+2}(2)=2 \\ &(a d j A) B=\left[\begin{array}{cc} 15 & -5 \\ -6 & 2 \end{array}\right]\left[\begin{array}{c} 7 \\ 13 \end{array}\right]=\left[\begin{array}{c} 105-65 \\ -42+26 \end{array}\right]=\left[\begin{array}{c} 40 \\ -16 \end{array}\right] \neq 0 \end{aligned}
Hence, the given system of equation is inconsistent.

Solution of Simultaneous Linear Equation exercise 7.1 question 4 subquestion (ii)

Inconsistent
Given:
$2x+3y=5 \quad , \quad 6x+9y=10$
Hint:
Inconsistent means two or more equations that are impossible to solve based on using one set of values for variables.
Solution:
The given system of equations can be expressed as follows
$A X=B \\ Here, \\ A=\left[\begin{array}{cc}2 & 3 \\ 6 & 9\end{array}\right] \quad, \quad X=\left[\begin{array}{l}x \\ y\end{array}\right] \quad and \quad B=\left[\begin{array}{c}5 \\ 10\end{array}\right] \\ Now, \\ |A|=\left|\begin{array}{cc}2 & 3 \\ 6 & 9\end{array}\right|=|18-18|=0$
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
\begin{aligned} &C_{11}=-1^{1+1}(9)=9 \quad, \quad C_{12}=-1^{1+2}(6)=-6 \\ &C_{21}=-1^{2+1}(3)=-3 \quad, \quad C_{22}=-1^{2+2}(2)=2 \\ &(a d j A) B=\left[\begin{array}{cc} 9 & -3 \\ -6 & 2 \end{array}\right]\left[\begin{array}{c} 5 \\ 10 \end{array}\right]=\left[\begin{array}{c} 45-30 \\ -30+20 \end{array}\right]=\left[\begin{array}{c} 15 \\ -10 \end{array}\right] \neq 0 \end{aligned}
Hence, the given system of equation is inconsistent.

Solution of Simultaneous Linear Equation exercise 7.1 question 4 subquestion (iii)

Inconsistent
Given:
$4x-2y=3 \quad , \quad 6x-3y=5$
Hint:
Inconsistent means two or more equations that are impossible to solve based on using one set of values for variables.
Solution:
The given system of equations can be expressed as follows
$A X=B \\ Here, \\ A=\left[\begin{array}{ll}4 & -2 \\ 6 & -3\end{array}\right] \quad, \quad X=\left[\begin{array}{c}x \\ y\end{array}\right] \quad and \quad B=\left[\begin{array}{l}3 \\ 5\end{array}\right] \\ Now, \\ |A|=\left|\begin{array}{ll}4 & -2 \\ 6 & -3\end{array}\right|=|-12+12|=0$
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
\begin{aligned} &C_{11}=-1^{1+1}(-3)=-3 \quad, \quad C_{12}=-1^{1+2}(6)=-6 \\ &C_{21}=-1^{2+1}(-2)=2 \quad, \quad C_{22}=-1^{2+2}(4)=4 \\ &(\text { adjA }) B=\left[\begin{array}{ll} -3 & 2 \\ -6 & 4 \end{array}\right]\left[\begin{array}{l} 3 \\ 5 \end{array}\right]=\left[\begin{array}{c} -9+10 \\ -18+20 \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \end{array}\right] \neq 0 \end{aligned}
Hence, the given system of equation is inconsistent.

Solution of Simultaneous Linear Equation exercise 7.1 question 4 subquestion (iv)

Inconsistent
Given:
$4 x-5 y-2 z=2 \quad, \quad 5 x-4 y+2 z=-2 \quad, \quad 2 x+2 y+8 z=-1$
Hint:
Inconsistent means two or more equations that are impossible to solve based on using one set of values for variables.
Solution:
The given system of equations can be expressed as follows
$A X=B \\ Here, \\ A=\left[\begin{array}{ccc}4 & -5 & -2 \\ 5 & -4 & 2 \\ 2 & 2 & 8\end{array}\right] \quad, \quad X=\left[\begin{array}{c}x \\ y \\ z\end{array}\right] \quad and \quad B=\left[\begin{array}{c}2 \\ -2 \\ -1\end{array}\right] \\ Now,$
\begin{aligned} |A|=\left|\begin{array}{ccc} 4 & -5 & -2 \\ 5 & -4 & 2 \\ 2 & 2 & 8 \end{array}\right| &=4(-32-4)+5(40-4)-2(10+8) \\ &=-144+180-36 \\ &=0 \end{aligned}
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
\begin{aligned} &C_{11}=-1^{1+1}\left|\begin{array}{cc} -4 & 2 \\ 2 & 8 \end{array}\right|=28 \quad, \quad C_{21}=-1^{2+1}\left|\begin{array}{cc} -5 & -2 \\ 2 & 8 \end{array}\right|=36 \\ &C_{12}=-1^{1+2}\left|\begin{array}{ll} 5 & 2 \\ 2 & 8 \end{array}\right|=-36 \quad, \quad C_{22}=-1^{2+2}\left|\begin{array}{cc} 4 & -2 \\ 2 & 8 \end{array}\right|=36 \end{aligned}
\begin{aligned} &C_{13}=-1^{1+3}\left|\begin{array}{cc} 5 & -4 \\ 2 & 2 \end{array}\right|=18 \quad, \quad C_{23}=-1^{2+3}\left|\begin{array}{cc} 4 & -5 \\ 2 & 2 \end{array}\right|=-18 \\ &C_{31}=-1^{3+1}\left|\begin{array}{cc} -5 & -2 \\ -4 & 2 \end{array}\right|=-18 \quad, \quad C_{32}=-1^{3+2}\left|\begin{array}{cc} 4 & -2 \\ 5 & 2 \end{array}\right|=-18 \\ &C_{33}=-1^{3+3}\left|\begin{array}{cc} 4 & -5 \\ 5 & -4 \end{array}\right|=9 \end{aligned}
\begin{aligned} (\text { adjA }) B &=\left[\begin{array}{ccc} 28 & 36 & -18 \\ -36 & 36 & -18 \\ 18 & -18 & 9 \end{array}\right]\left[\begin{array}{c} 2 \\ -2 \\ -1 \end{array}\right] \\ &=\left[\begin{array}{c} 56-72+18 \\ -72-72+18 \\ 36+36-9 \end{array}\right] \\ &=\left[\begin{array}{c} 2 \\ -126 \\ 63 \end{array}\right] \neq 0 \end{aligned}
Hence, the given system of equation is inconsistent.

Solution of Simultaneous Linear Equation exercise 7.1 question 4 subquestion (v)

Inconsistent
Given:
$3 x-y-2 z=2 \quad, \quad 2 y-z=-1 \quad, \quad 3 x-5 y=3$
Hint:
Inconsistent means two or more equations that are impossible to solve based on using one set of values for variables.
Solution:
The given system of equations can be expressed as follows
\begin{aligned} &A X=B\\ &\text { Here, }\\ &A=\left[\begin{array}{ccc} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array}\right] \quad, \quad X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right] \end{aligned}
\begin{aligned} &\text { Now, }\\ &\begin{aligned} |A|=\left|\begin{array}{ccc} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array}\right| &=3(0-5)+1(0+3)-2(0-6) \\ &=-15+3+12 \\ &=0 \end{aligned} \end{aligned}
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
\begin{aligned} &C_{11}=-1^{1+1}\left|\begin{array}{cc} 2 & -1 \\ -5 & 0 \end{array}\right|=-5 \quad, \quad C_{21}=-1^{2+1}\left|\begin{array}{cc} 0 & -1 \\ 3 & 0 \end{array}\right|=-3 \\ &C_{12}=-1^{1+2}\left|\begin{array}{cc} 0 & 2 \\ 3 & -5 \end{array}\right|=-6 \quad, \quad C_{22}=-1^{2+2}\left|\begin{array}{cc} -1 & -2 \\ -5 & 0 \end{array}\right|=10 \end{aligned}
\begin{aligned} &C_{13}=-1^{1+3}\left|\begin{array}{cc} 3 & -2 \\ 3 & 0 \end{array}\right|=6 \quad, \quad C_{23}=-1^{2+3}\left|\begin{array}{ll} 3 & -1 \\ 3 & -5 \end{array}\right|=12 \\ &C_{31}=-1^{3+1}\left|\begin{array}{ll} -1 & -2 \\ 2 & -1 \end{array}\right|=5 \quad, \quad C_{32}=-1^{3+2}\left|\begin{array}{ll} 3 & -2 \\ 0 & -1 \end{array}\right|=3 \\ &C_{33}=-1^{3+3}\left|\begin{array}{cc} 3 & -1 \\ 0 & 2 \end{array}\right|=6 \end{aligned}
$\begin{gathered} \operatorname{adjA}=\left[\begin{array}{ccc} -5 & -3 & -6 \\ 10 & 6 & 12 \\ 5 & 3 & 6 \end{array}\right]^{T} \\ (\text { adjA }) B=\left[\begin{array}{ccc} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{array}\right]\left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right] \\ =\left[\begin{array}{c} -10-10+15 \\ -6-6+9 \\ -12-12+18 \end{array}\right] \\ =\left[\begin{array}{c} -5 \\ -3 \\ -6 \end{array}\right] \neq 0 \end{gathered}$
Hence, the given system of equation is inconsistent.

Solution of Simultaneous Linear Equation exercise 7.1 question 4 subquestion (vi)

Inconsistent
Given:
$x+y-2 z=5 \quad, \quad x-2 y+z=-2 \quad, \quad-2 x+y+z=4$
Hint:
Inconsistent means two or more equations that are impossible to solve based on using one set of values for variables.
Solution:
The given system of equations can be expressed as follows
\begin{aligned} &A X=B\\ &\text { Here, }\\ &A=\left[\begin{array}{ccc} 1 & 1 & -2 \\ 1 & -2 & 1 \\ -2 & 1 & 1 \end{array}\right] \quad, \quad X=\left[\begin{array}{c} x \\ y \\ z \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{c} 5 \\ -2 \\ 4 \end{array}\right] \end{aligned}
\begin{aligned} &\text { Now, }\\ &\begin{aligned} |A|=\left|\begin{array}{ccc} 1 & 1 & -2 \\ 1 & -2 & 1 \\ -2 & 1 & 1 \end{array}\right| &=1(-2-1)-1(1+2)-2(1-4) \\ &=-3-3+6 \\ &=0 \end{aligned} \end{aligned}
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
\begin{aligned} &C_{11}=-1^{1+1}\left|\begin{array}{cc} -2 & 1 \\ 1 & 1 \end{array}\right|=-3 \quad, \quad C_{21}=-1^{2+1}\left|\begin{array}{cc} 1 & -2 \\ 1 & 1 \end{array}\right|=-3 \\ &C_{12}=-1^{1+2}\left|\begin{array}{cc} 1 & 1 \\ -2 & 1 \end{array}\right|=-3 \quad, \quad C_{22}=-1^{2+2}\left|\begin{array}{cc} 1 & -2 \\ -2 & 1 \end{array}\right|=-3 \end{aligned}
\begin{aligned} &C_{13}=-1^{1+3}\left|\begin{array}{cc} 1 & -2 \\ -2 & 1 \end{array}\right|=-3 \quad, \quad C_{23}=-1^{2+3}\left|\begin{array}{cc} 1 & 1 \\ -2 & 1 \end{array}\right|=-3 \\ &C_{31}=-1^{3+1}\left|\begin{array}{cc} 1 & -2 \\ -2 & 1 \end{array}\right|=-3 \quad, \quad C_{32}=-1^{3+2}\left|\begin{array}{cc} 1 & -2 \\ 1 & 1 \end{array}\right|=-3 \\ &C_{33}=-1^{3+3}\left|\begin{array}{cc} 1 & 1 \\ 1 & -2 \end{array}\right|=-3 \end{aligned}
\begin{aligned} &\operatorname{adjA}=\left[\begin{array}{rrr} -3 & -3 & -3 \\ -3 & -3 & -3 \\ -3 & -3 & -3 \end{array}\right]^{T} \\ &(\operatorname{adj} A) B=\left[\begin{array}{rrr} -3 & -3 & -3 \\ -3 & -3 & -3 \\ -3 & -3 & -3 \end{array}\right]\left[\begin{array}{c} 5 \\ -2 \\ 4 \end{array}\right] \\ &=\left[\begin{array}{r} -15+6-12 \\ -15+6-12 \\ -15+6-12 \end{array}\right] \\ &=\left[\begin{array}{r} -21 \\ -21 \\ -21 \end{array}\right] \neq 0 \end{aligned}
Hence, the given system of equation is inconsistent.

Solution of Simultaneous Linear Equation exercise 7.1 question 5

$x=2 \quad , \quad y=-1 \quad , \quad z=4$
Given:
\begin{aligned} &A=\left[\begin{array}{ccc} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right] \quad, \quad B=\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right] \backslash \\ &x-y=3, \quad 2 x+3 y+4 z=17 \quad, \quad y+2 z=7 \end{aligned}
Hint:
For matrix multiply matrix A with matrix B, Then X=A-1B is formula for which is used to solve this problem.
Solution:
\begin{aligned} A B &=\left[\begin{array}{ccc} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right]\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right] \\ &=\left[\begin{array}{lll} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}\right] \\ A B &=6\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=6 I_{3} \end{aligned}
\begin{aligned} &\frac{1}{6} A B=I_{3} \\ &\left(\frac{1}{6} B\right) A=I_{3} \quad(\therefore A B=B A) \\ &A^{-1}=\frac{1}{6} B \\ &A^{-1}=\frac{1}{6}\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right] \end{aligned}
\begin{aligned} X &=A^{-1} B \\ &=\frac{1}{6}\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right]\left[\begin{array}{c} 3 \\ 17 \\ 7 \end{array}\right] \\ &=\frac{1}{6}\left[\begin{array}{c} 6+34-28 \\ -12+34-28 \\ 6-17+35 \end{array}\right] \end{aligned}
\begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{6}\left[\begin{array}{c} 12 \\ -6 \\ 24 \end{array}\right]} \\ &x=2, y=-1, z=4 \end{aligned}
\begin{aligned} &x=1, y=2, z=3 \end{aligned}
Given:
\begin{aligned} &2 x-3 y+5 z=11 \\ &3 x+2 y-4 z=-5 \\ &x+y+2 z=-z \\ &A=\left[\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right] \end{aligned}
Hint:
X=A-1B is used to solve this problem. First we find the determinant and co-factor of matrix A, take it’s transpose that will be Adj A using Adj A calculate A-1.
Solution:
\begin{aligned} &A=\left[\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right] \\ &|A|=\left|\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right| \end{aligned}
\begin{aligned} |A| &=2(-4+4)+3(-6+4)+5(3-2) \\ &=0-6+5=-1 \end{aligned}
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
\begin{aligned} &C_{11}=-1^{1+1}\left|\begin{array}{ll} 2 & -4 \\ 1 & -2 \end{array}\right|=0 \quad, \quad C_{21}=-1^{2+1}\left|\begin{array}{cc} -3 & 5 \\ 1 & -2 \end{array}\right|=-1 \\ &C_{12}=-1^{1+2}\left|\begin{array}{ll} 3 & -4 \\ 1 & -2 \end{array}\right|=2 \quad, \quad C_{22}=-1^{2+2}\left|\begin{array}{cc} 2 & 5 \\ 1 & -2 \end{array}\right|=-9 \end{aligned}
\begin{aligned} &C_{13}=-1^{1+3}\left|\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right|=1 \quad, \quad C_{23}=-1^{2+3}\left|\begin{array}{cc} 2 & -3 \\ 1 & 1 \end{array}\right|=-5 \\ &C_{31}=-1^{3+1}\left|\begin{array}{cc} -3 & 5 \\ 2 & -4 \end{array}\right|=2 \quad, \quad C_{32}=-1^{3+2}\left|\begin{array}{cc} 2 & 5 \\ 3 & -4 \end{array}\right|=23 \\ &C_{33}=-1^{3+3}\left|\begin{array}{cc} 2 & -3 \\ 3 & 2 \end{array}\right|=13 \end{aligned}
\begin{aligned} &\operatorname{adj} A=\left[\begin{array}{ccc} 0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 23 & 13 \end{array}\right]^{T}=\left[\begin{array}{ccc} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} a d j A \end{aligned}
$\begin{gathered} =\frac{1}{-1}\left[\begin{array}{ccc} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{array}\right] \\ {\left[\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 11 \\ -5 \\ -3 \end{array}\right]} \end{gathered}$
\begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 0+5-6 \\ 22+45-69 \\ 11+25-39 \end{array}\right]} \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]} \\ &x=1, \mathrm{y}=2, z=3 \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 7

\begin{aligned} &x=-1, \mathrm{y}=-2, z=3 \end{aligned}
Given:
\begin{aligned} &A=\left[\begin{array}{ccc} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{array}\right] \\ &x+2 y+5 z=10 \\ &x-y-z=-2 \\ &2 x+3 y-z=-11 \end{aligned}
Hint:
X=A-1B is used to solve this problem. First we find the determinant and co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.
Solution:
\begin{aligned} A &=\left[\begin{array}{ccc} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{array}\right] \\ |A| &=\left|\begin{array}{ccc} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{array}\right| \\ |A| &=1(1+3)-2(-1+2)+5(3+2) \\ &=4-2+25=27 \end{aligned}
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
$\begin{array}{ll} C_{11}=-1^{1+1}\left|\begin{array}{cc} -1 & -1 \\ 3 & -1 \end{array}\right|=4 \quad, \quad C_{21}=-1^{2+1}\left|\begin{array}{cc} 2 & 5 \\ 3 & -1 \end{array}\right|=-1 \\ C_{12}=-1^{1+2}\left|\begin{array}{ll} 1 & -1 \\ 2 & -1 \end{array}\right|=-1 \quad, \quad C_{22}=-1^{2+2}\left|\begin{array}{cc} 1 & 5 \\ 2 & -1 \end{array}\right|=17 \end{array}$
\begin{aligned} &C_{13}=-1^{1+3}\left|\begin{array}{cc} 1 & -1 \\ -2 & 3 \end{array}\right|=5 \quad, \quad C_{23}=-1^{2+3}\left|\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right|=1 \\ &C_{31}=-1^{3+1}\left|\begin{array}{cc} 2 & 5 \\ -1 & -1 \end{array}\right|=3 \quad, \quad C_{32}=-1^{3+2}\left|\begin{array}{cc} 1 & 5 \\ 1 & -1 \end{array}\right|=6 \end{aligned}
\begin{aligned} &C_{33}=-1^{3+3}\left|\begin{array}{cc} 1 & 2 \\ 1 & -1 \end{array}\right|=-3 \\ &\operatorname{adjA}=\left[\begin{array}{ccc} 4 & -1 & 5 \\ 17 & -11 & 1 \\ 3 & 6 & -3 \end{array}\right]^{T}=\left[\begin{array}{ccc} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{array}\right] \\ &A^{-1}=\frac{1}{|A|} a d j A \end{aligned}
$\begin{gathered} =\frac{1}{27}\left[\begin{array}{ccc} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{array}\right] \\ {\left[\begin{array}{ccc} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 10 \\ -2 \\ -11 \end{array}\right]} \\ A X=B \end{gathered}$
\begin{aligned} &X=A^{-1} B \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{27}\left[\begin{array}{ccc} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{array}\right]\left[\begin{array}{c} 10 \\ -2 \\ -11 \end{array}\right]} \end{aligned}
\begin{aligned} &\frac{1}{27}\left[\begin{array}{c} 40-34-33 \\ -10+22-66 \\ 50-2+33 \end{array}\right] \\ &=\frac{1}{27}\left[\begin{array}{c} -27 \\ -54 \\ 81 \end{array}\right] \\ &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} \frac{-27}{27} \\ \frac{-54}{27} \\ \frac{81}{27} \end{array}\right]=\left[\begin{array}{c} -1 \\ -2 \\ 3 \end{array}\right]} \end{aligned}
\begin{aligned} &x=-1, \mathrm{y}=-2, z=3 \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 8 subquestion (i)

$x=4,\; \; y=-3,\: \: z=1$
Given:
\begin{aligned} &A=\left[\begin{array}{ccc} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{array}\right] \\ &x-2 y=10 \\ &2 x+y+3 y=8 \\ &-2 y+z=7 \end{aligned}
Hint:
X=A-1B is used to solve this problem. First we find the determinant and co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.
Solution:
\begin{aligned} A &=\left[\begin{array}{ccc} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{array}\right] \\ |A| &=1(1+6)+2(2-0)+0(-4-0) \\ &=7+4+0 \\ &=11 \end{aligned}
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
$\begin{array}{ll} C_{11}=-1^{1+1}\left|\begin{array}{cc} 1 & 3 \\ -2 & 1 \end{array}\right|=7 \quad , \quad C_{21}=-1^{2+1}\left|\begin{array}{ll} -2 & 0 \\ -2 & 1 \end{array}\right|=2 \\ C_{12}=-1^{1+2}\left|\begin{array}{ll} 2 & 3 \\ 0 & 1 \end{array}\right|=-2 \quad, \quad C_{22}=-1^{2+2}\left|\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right|=1 \end{array}$
\begin{aligned} &C_{13}=-1^{1+3}\left|\begin{array}{cc} 2 & 1 \\ 0 & -2 \end{array}\right|=-4 \quad, \quad C_{23}=-1^{2+3}\left|\begin{array}{cc} 1 & -2 \\ 0 & -2 \end{array}\right|=2 \\ &C_{31}=-1^{3+1}\left|\begin{array}{cc} -2 & 0 \\ 1 & 3 \end{array}\right|=-6 \quad, \quad C_{32}=-1^{3+2}\left|\begin{array}{cc} 1 & 0 \\ 2 & 3 \end{array}\right|=3 \\ &C_{33}=-1^{3+3}\left|\begin{array}{cc} 1 & -2 \\ 2 & 1 \end{array}\right|=5 \end{aligned}
\begin{aligned} \operatorname{adj} A &=\left[\begin{array}{ccc} 7 & -2 & -4 \\ 2 & 1 & 2 \\ -6 & -3 & 5 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} 7 & 2 & 6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{array}\right] \end{aligned}
\begin{aligned} A^{-1} &=\frac{1}{|A|} a d j A \\ &=\frac{1}{11}\left[\begin{array}{ccc} 7 & 2 & 6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{array}\right] \end{aligned}
\begin{aligned} X &=A^{-1} B \\ &=\frac{1}{11}\left[\begin{array}{ccc} 7 & 2 & 6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{array}\right]\left[\begin{array}{c} 10 \\ 8 \\ -1 \end{array}\right] \\ &=\frac{1}{11}\left[\begin{array}{c} 70+16-42 \\ -20+8-21 \\ -40+16+35 \end{array}\right]=\frac{1}{11}\left[\begin{array}{c} 44 \\ -33 \\ 11 \end{array}\right] \end{aligned}
\begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 4 \\ -3 \\ 1 \end{array}\right]} \\ &x=4, y=-3, z=1 \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 8 subquestion (ii)

\begin{aligned} &x=3, y=2, z=-1 \end{aligned}
Given:
$A=\left[\begin{array}{ccc} 3 & -4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1 \end{array}\right]$
Hint:
X=A-1B is used to solve this problem. First we find the determinant and co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.
Solution:
\begin{aligned} A &=\left[\begin{array}{ccc} 3 & -4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1 \end{array}\right] \\ |A| &=3(3-0)+4(2-5)+2(0-3) \\ &=9-12-6 \\ &=-9 \end{aligned}
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
\begin{aligned} &C_{11}=-1^{1+1}\left|\begin{array}{ll} 3 & 5 \\ 0 & 1 \end{array}\right|=3 \quad, \quad C_{21}=-1^{2+1}\left|\begin{array}{cc} -4 & 2 \\ 0 & 1 \end{array}\right|=4 \\ &C_{12}=-1^{1+2}\left|\begin{array}{ll} 2 & 5 \\ 1 & 1 \end{array}\right|=3 \quad, \quad C_{22}=-1^{2+2}\left|\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right|=1 \end{aligned}
\begin{aligned} &C_{13}=-1^{1+3}\left|\begin{array}{ll} 2 & 3 \\ 1 & 0 \end{array}\right|=-3 \quad, \quad C_{23}=-1^{2+3}\left|\begin{array}{cc} 3 & -4 \\ 1 & 0 \end{array}\right|=-4 \\ &C_{31}=-1^{3+1}\left|\begin{array}{cc} -4 & 2 \\ 3 & 5 \end{array}\right|=-26 \quad, \quad C_{32}=-1^{3+2}\left|\begin{array}{ll} 3 & 2 \\ 2 & 5 \end{array}\right|=-11 \\ &C_{33}=-1^{3+3}\left|\begin{array}{cc} 3 & -4 \\ 2 & 3 \end{array}\right|=17 \end{aligned}
\begin{aligned} \operatorname{adj} A &=\left[\begin{array}{ccc} 3 & 4 & -26 \\ 3 & 1 & -11 \\ -3 & -4 & 17 \end{array}\right] \\ A^{-1} &=\frac{1}{|A|} \operatorname{adjA} \\ &=\frac{1}{-9}\left[\begin{array}{ccc} 3 & 4 & -26 \\ 3 & 1 & -11 \\ -3 & -4 & 17 \end{array}\right] \end{aligned}
\begin{aligned} X &=A^{-1} B \\ &=\frac{1}{-9}\left[\begin{array}{ccc} 3 & 4 & -26 \\ 3 & 1 & -11 \\ -3 & -4 & 17 \end{array}\right]\left[\begin{array}{c} -1 \\ 7 \\ 2 \end{array}\right] \\ &=\frac{1}{-9}\left[\begin{array}{c} -3+28-52 \\ -3+7-22 \\ 3-28+34 \end{array}\right]=\frac{1}{-9}\left[\begin{array}{c} -27 \\ -18 \\ 9 \end{array}\right] \end{aligned}
\begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 3 \\ 2 \\ -1 \end{array}\right]} \\ &x=3, y=2, z=-1 \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 8 subquestion (iii)

\begin{aligned} &x=4, y=-3, z=1 \end{aligned}
Given:
\begin{aligned} &A=\left[\begin{array}{ccc} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{array}\right] \quad, \quad B=\left[\begin{array}{ccc} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 1 & 5 \end{array}\right] \\ &x-2 y=10 \\ &2 x+y+3 y=8 \\ &-2 y+z=7 \end{aligned}
Hint:
X=A-1B is used to solve this problem. First we find the determinant and co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.
Solution:
\begin{aligned} A B &=\left[\begin{array}{ccc} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{array}\right]\left[\begin{array}{ccc} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 1 & 5 \end{array}\right] \\ &=\left[\begin{array}{ccc} 7+4+0 & 2-2+0 & -6+6+0 \\ 14-2-12 & 4+1+6 & -12-3+15 \\ 0+4-4 & 0-1+2 & 0+6+5 \end{array}\right] \\ &=\left[\begin{array}{ccc} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{array}\right] \\ A B &=11\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned}
\begin{aligned} &A B=11 I_{3} \\ &{\left[\frac{1}{11}\right] A B=I_{3}} \\ &{\left[\frac{1}{11} B\right] A=I_{3}} \\ &A^{-1}=\frac{1}{11} B \\ &\quad=\frac{1}{11}\left[\begin{array}{ccc} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 1 & 5 \end{array}\right] \end{aligned}
\begin{aligned} X &=A^{-1} B \\ &=\frac{1}{11}\left[\begin{array}{ccc} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 1 & 5 \end{array}\right]\left[\begin{array}{c} 10 \\ 8 \\ 7 \end{array}\right] \\ &=\frac{1}{11}\left[\begin{array}{c} 70+16-42 \\ -20+8-21 \\ -40+16+35 \end{array}\right]=\frac{1}{11}\left[\begin{array}{c} 44 \\ -33 \\ 11 \end{array}\right] \end{aligned}
\begin{aligned} &x=4, y=-3, z=1 \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 8 subquestion (iv)

$x=0,\: \: y=-5,\: \: z=-3$
Given:
\begin{aligned} &A=\left[\begin{array}{ccc} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{array}\right] \\ &x-2 y=10 \\ &2 x-y-z=8 \\ &-2 y+z=7 \end{aligned}
Hint:
X=A-1B is used to solve this problem. First we find the determinant and co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.
Solution:
\begin{aligned} A &=\left[\begin{array}{ccc} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{array}\right] \\ |A| &=1(-1-2)+2(2) \\ &=-3+4 \\ &=1 \end{aligned}
$C_{i j}\; be\; the\; co-\! f\! actor\; o\! f\; the\; elements\; a_{i j}\; in \; A=\left[a_{i j}\right].\; Then,$
\begin{aligned} &C_{11}=-1^{1+1}\left|\begin{array}{cc} -1 & -2 \\ -1 & 1 \end{array}\right|=-3 \quad, \quad C_{21}=-1^{2+1}\left|\begin{array}{cc} 2 & 0 \\ -1 & 1 \end{array}\right|=-2 \\ &C_{12}=-1^{1+2}\left|\begin{array}{cc} -2 & -2 \\ 0 & 1 \end{array}\right|=2 \quad, \quad C_{22}=-1^{2+2}\left|\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right|=1 \end{aligned}
\begin{aligned} &C_{13}=-1^{1+3}\left|\begin{array}{cc} -2 & -1 \\ 0 & -1 \end{array}\right|=2 \quad, \quad C_{23}=-1^{2+3}\left|\begin{array}{cc} 1 & 2 \\ 0 & -1 \end{array}\right|=1 \\ &C_{31}=-1^{3+1}\left|\begin{array}{cc} 2 & 0 \\ -1 & -2 \end{array}\right|=-4 \quad, \quad C_{32}=-1^{3+2}\left|\begin{array}{cc} 1 & 0 \\ -2 & -2 \end{array}\right|=2 \\ &C_{33}=-1^{3+3}\left|\begin{array}{cc} 1 & 2 \\ -2 & -1 \end{array}\right|=3 \end{aligned}
\begin{aligned} \operatorname{adj} A &=\left[\begin{array}{rcc} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{array}\right]^{T} \\ &=\left[\begin{array}{ccc} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{array}\right] \end{aligned}
\begin{aligned} &\left(A^{T}\right)^{-1}=\left(A^{-1}\right)^{T} \\ &\text { i.e } C=\left[\begin{array}{ccc} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{array}\right] \\ &C^{-1}=\left[\begin{array}{rrr} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{array}\right] \end{aligned}
$\begin{gathered} {\left[\begin{array}{ccc} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 10 \\ 8 \\ 7 \end{array}\right]} \\ C X=B \\ X=C^{-1} B \end{gathered}$
\begin{aligned} =\left[\begin{array}{ccc} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{array}\right]\left[\begin{array}{c} 10 \\ 8 \\ 7 \end{array}\right] \\ =\left[\begin{array}{c} -30+16+14 \\ -20+8+7 \\ -40+16+21 \end{array}\right] \\ x=0, y=-5, z=-3 \end{aligned}

Solution of Simultaneous Linear Equation exercise 7.1 question 8 subquestion (v)

\begin{aligned} x=2, y=-1, z=4 \end{aligned}
Given:
\begin{aligned} &A=\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right] \quad, \quad B=\left[\begin{array}{ccc} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right] \\ &y+2 z=7 \\ &x-y=3 \\ &2 x+3 y+11 z=17 \end{aligned}
Hint:
X=A-1B is used to solve this problem. First we find the determinant and co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.
Solution:
\begin{aligned} A &=\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right], B=\left[\begin{array}{ccc} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right] \\ B A &=\left[\begin{array}{ccc} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right]\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right] \\ &=\left[\begin{array}{ccc} 2+4+0 & 2-2+0 & -4+4+0 \\ 4-12+8 & 4+6-4 & -8-12+20 \\ 0-4+4 & 0+2-2 & 0-4+10 \end{array}\right] \end{aligned}
\begin{aligned} &=\left[\begin{array}{lll} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}\right] \\ &B A=6\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ &B A=6 I_{3} \\ &B\left[\frac{1}{6} A\right]=I_{3} \end{aligned}
\begin{aligned} &{\left[\frac{1}{11} B\right] A=I_{3}} \\ &B^{-1}=\frac{1}{6} A \\ &\quad=\frac{1}{6}\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right] \\ &B X=C \\ &X=B^{-1} C \end{aligned}
$\begin{gathered} =\frac{1}{6}\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right]\left[\begin{array}{c} 3 \\ 17 \\ 7 \end{array}\right] \\ =\frac{1}{6}\left[\begin{array}{c} 6+34-28 \\ -12+34-28 \\ 6-17+35 \end{array}\right] \end{gathered}$
\begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{6}\left[\begin{array}{l} 12 \\ -6 \\ 24 \end{array}\right]} \\ &x=2, y=-1, z=4 \end{aligned}

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