# Show that each one of the following systems of linear equation is inconsistent:

Question:

Show that each one of the following systems of linear equation is inconsistent:

(i) $2 x+5 y=7$

$6 x+15 y=13$

(ii) $2 x+3 y=5$

$6 x+9 y=10$

(iii) $4 x-2 y=3$

$6 x-3 y=5$

(iv) $4 x-5 y-2 z=2$

$5 x-4 y+2 z=-2$

$2 x+2 y+8 z=-1$

(v) $3 x-y-2 z=2$

$2 y-z=-1$

$3 x-5 y=3$

(vi) $x+y-2 z=5$

$x-2 y+z=-2$

$-2 x+y+z=4$

Solution:

(i) The given system of equations can be expressed as follows:

$A X=B$

Here,

$A=\left[\begin{array}{cc}2 & 5 \\ 6 & 15\end{array}\right], X=\left[\begin{array}{c}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{c}7 \\ 13\end{array}\right]$

Now,

$|A|=\left|\begin{array}{cc}2 & 5 \\ 6 & 15\end{array}\right|$

$=(30-30)$

$=0$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}=\left[a_{i j}\right]$. Then,

$C_{11}=-1^{1+1}(15)=15, C_{12}=-1^{1+2}(6)=-6$

$C_{21}=-1^{2+1}(5)=-5, C_{22}=-1^{2+2}(2)=2$

$\operatorname{adj} A=\left[\begin{array}{cc}15 & -6 \\ -5 & 2\end{array}\right]^{\mathrm{T}}$

$=\left[\begin{array}{cc}15 & -5 \\ -6 & 2\end{array}\right]$

$(\operatorname{adj} A) B=\left[\begin{array}{cc}15 & -5 \\ -6 & 2\end{array}\right]\left[\begin{array}{c}7 \\ 13\end{array}\right]$

$=\left[\begin{array}{l}105-65 \\ -42+26\end{array}\right]$

$=\left[\begin{array}{c}40 \\ -16\end{array}\right] \neq 0$

Hence, the given system of equations is inconsistent.

(ii) The given system of equations can be expressed as follows:

$A X=B$

Here,

$A=\left[\begin{array}{ll}2 & 3 \\ 6 & 9\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{c}5 \\ 10\end{array}\right]$

Now,

Here,

$A=\left[\begin{array}{ll}2 & 3 \\ 6 & 9\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{c}5 \\ 10\end{array}\right]$

Now,

$|A|=\left|\begin{array}{ll}2 & 3 \\ 6 & 9\end{array}\right|$

$=(18-18)$

$=0$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}=\left[a_{i j}\right] .$ Then,

$C_{11}=(-1)^{1+1}(9)=9, C_{12}=(-1)^{1+2}(6)=-6$

$C_{21}=(-1)^{2+1}(3)=-3, C_{22}=(-1)^{2+2}(2)=2$

$\operatorname{adj} A=\left[\begin{array}{cc}9 & -6 \\ -3 & 2\end{array}\right]^{\mathrm{T}}$

$=\left[\begin{array}{cc}9 & -3 \\ -6 & 2\end{array}\right]$

$(\operatorname{adj} A) B=\left[\begin{array}{cc}9 & -3 \\ -6 & 2\end{array}\right]\left[\begin{array}{c}5 \\ 10\end{array}\right]$

$=\left[\begin{array}{c}45-30 \\ -30+20\end{array}\right]$

$=\left[\begin{array}{c}15 \\ -10\end{array}\right] \neq 0$

Hence, the given system of equations is inconsistent.

(iii) The given system of equations can be expressed as follows:

$A X=B$

Here,

$A=\left[\begin{array}{rr}4 & -2 \\ 6 & -3\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{l}3 \\ 5\end{array}\right]$

$|A|=\left|\begin{array}{ll}4 & -2 \\ 6 & -3\end{array}\right|$

$=(-12+12)$

$=0$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}=\left[a_{i j}\right]$. Then,

$\begin{array}{ll}C_{11}=-(1)^{1+1}(-3)=-3, & C_{12}=-(1)^{1+2}(6)=-6 \\ C_{21}=-(1)^{2+1}(-2)=2, & C_{22}=-(1)^{2+2}(4)=4\end{array}$

adj $A=\left[\begin{array}{cc}-3 & -6 \\ 2 & 4\end{array}\right]^{\mathrm{T}}$

$=\left[\begin{array}{ll}-3 & 2 \\ -6 & 4\end{array}\right]$

$(a d j A) B=\left[\begin{array}{ll}-3 & 2 \\ -6 & 4\end{array}\right]\left[\begin{array}{l}3 \\ 5\end{array}\right]$

$=\left[\begin{array}{c}-9+10 \\ -18+20\end{array}\right]$

$=\left[\begin{array}{l}1 \\ 2\end{array}\right] \neq 0$

Hence, the given system of equations is inconsistent.

(iv) The given system of equations can be written as follows:

$A X=B$

Here,

$A=\left[\begin{array}{ccc}4 & -5 & -2 \\ 5 & -4 & 2 \\ 2 & 2 & 8\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}2 \\ -2 \\ -1\end{array}\right]$

$|A|=\left|\begin{array}{ccc}4 & -5 & -2 \\ 5 & -4 & 2 \\ 2 & 2 & 8\end{array}\right|$

$=4(-32-4)+5(40-4)-2(10+8)$

$=-144+180-36$

$=0$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}-4 & 2 \\ 2 & 8\end{array}\right|=28, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{ll}5 & 2 \\ 2 & 8\end{array}\right|=-36, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{cc}5 & -4 \\ 2 & 2\end{array}\right|=18$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}-5 & -2 \\ 2 & 8\end{array}\right|=36, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{cc}4 & -2 \\ 2 & 8\end{array}\right|=36, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc}4 & -5 \\ 2 & 2\end{array}\right|=-18$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}-5 & -2 \\ -4 & 2\end{array}\right|=-18, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc}4 & -2 \\ 5 & 2\end{array}\right|=-18, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{cc}4 & -5 \\ 5 & -4\end{array}\right|=9$

$\operatorname{adj} A=\left[\begin{array}{ccc}28 & -36 & 18 \\ 36 & 36 & -18 \\ -18 & -18 & 9\end{array}\right]^{\mathrm{T}}$

$=\left[\begin{array}{ccc}28 & 36 & -18 \\ -36 & 36 & -18 \\ 18 & -18 & 9\end{array}\right]$

$(\operatorname{adj} A) B=\left[\begin{array}{ccc}28 & 36 & -18 \\ -36 & 36 & -18 \\ 18 & -18 & 9\end{array}\right]\left[\begin{array}{c}2 \\ -2 \\ -1\end{array}\right]$

$=\left[\begin{array}{c}56-72+18 \\ -72-72+18 \\ 36+36-9\end{array}\right]$

$=\left[\begin{array}{c}2 \\ -126 \\ 63\end{array}\right] \neq 0$

Hence, the given system of equations is consistent.

(v) The given system of equations can be written as follows:

$A X=B$

Here,

$A=\left[\begin{array}{ccc}3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}2 \\ -1 \\ 3\end{array}\right]$

$|A|=\left|\begin{array}{ccc}3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right|$

$=3(0-5)+1(0+3)-2(0-6)$

$=-15+3+12$

$=0$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}2 & -1 \\ -5 & 0\end{array}\right|=-5, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc}0 & -1 \\ 3 & 0\end{array}\right|=-3, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{cc}0 & 2 \\ 3 & -5\end{array}\right|=-6$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}-1 & -2 \\ -5 & 0\end{array}\right|=10, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{cc}3 & -2 \\ 3 & 0\end{array}\right|=6, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll}3 & -1 \\ 3 & -5\end{array}\right|=12$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}-1 & -2 \\ 2 & -1\end{array}\right|=5, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll}3 & -2 \\ 0 & -1\end{array}\right|=3, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{cc}3 & -1 \\ 0 & 2\end{array}\right|=6$

$\operatorname{adj} A=\left[\begin{array}{ccc}-5 & -3 & -6 \\ 10 & 6 & 12 \\ 5 & 3 & 6\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}-5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6\end{array}\right]$

$(\operatorname{adj} A) B=\left[\begin{array}{ccc}-5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6\end{array}\right]\left[\begin{array}{c}2 \\ -1 \\ 3\end{array}\right]$

$=\left[\begin{array}{c}-10-10+15 \\ -6-6+9 \\ -12-12+18\end{array}\right]$

$=\left[\begin{array}{l}-5 \\ -3 \\ -6\end{array}\right] \neq 0$

Hence, the given system of equations is consistent.

(vi) The given system of equations can be written as follows:

$A X=B$

Here,

$A=\left[\begin{array}{ccc}1 & 1 & -2 \\ 1 & -2 & 1 \\ -2 & 1 & 1\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}5 \\ -2 \\ 4\end{array}\right]$

$|A|=\left|\begin{array}{ccc}1 & 1 & -2 \\ 1 & -2 & 1 \\ -2 & 1 & 1\end{array}\right|$

$=1(-2-1)-1(1+2)-2(1-4)$

$=-3-3+6$

$=0$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}-2 & 1 \\ 1 & 1\end{array}\right|=-3, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc}1 & 1 \\ -2 & 1\end{array}\right|=-3, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{cc}1 & -2 \\ -2 & 1\end{array}\right|=-3$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}1 & -2 \\ 1 & 1\end{array}\right|=-3, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{cc}1 & -2 \\ -2 & 1\end{array}\right|=-3, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc}1 & 1 \\ -2 & 1\end{array}\right|=-3$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}1 & -2 \\ -2 & 1\end{array}\right|=-3, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc}1 & -2 \\ 1 & 1\end{array}\right|=-3, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{cc}1 & 1 \\ 1 & -2\end{array}\right|=-3$

$\operatorname{adj} A=\left[\begin{array}{rrr}-3 & -3 & -3 \\ -3 & -3 & -3 \\ -3 & -3 & -3\end{array}\right]^{T}$

$=\left[\begin{array}{lll}-3 & -3 & -3 \\ -3 & -3 & -3 \\ -3 & -3 & -3\end{array}\right]$

$(\operatorname{adj} A) B=\left[\begin{array}{rrr}-3 & -3 & -3 \\ -3 & -3 & -3 \\ -3 & -3 & -3\end{array}\right]\left[\begin{array}{c}5 \\ -2 \\ 4\end{array}\right]$

$=\left[\begin{array}{l}-15+6-12 \\ -15+6-12 \\ -15+6-12\end{array}\right]$

$=\left[\begin{array}{l}-21 \\ -21 \\ -21\end{array}\right] \neq 0$

Hence, the given system of equations is consistent.