Show that each of the following systems of linear equations is consistent and also find their solutions:

Solution:

(i) Here,

$6 x+4 y=2 \quad \cdots(1)$

$9 x+6 y=3 \quad \cdots(2)$

$A X=B$

Here,

$A=\left[\begin{array}{ll}6 & 4 \\ 9 & 6\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{l}2 \\ 3\end{array}\right]$

$\left[\begin{array}{ll}6 & 4 \\ 9 & 6\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}2 \\ 3\end{array}\right]$

$|A|=\left|\begin{array}{ll}6 & 4 \\ 9 & 6\end{array}\right|$

$=36-36$

$=0$

So, $A$ is singular. Thus, the given system of equations is either inconsistent or it is consistent with infinitely many solutions because $(\operatorname{adj} A) B \neq 0$ or $(\operatorname{adj} A)=0$.

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

Then, $C_{11}=6, C_{12}=-9, C_{21}=-4, C_{22}=6$

$\operatorname{adj} A=\left[\begin{array}{cc}6 & -9 \\ -4 & 6\end{array}\right]^{T}$

$=\left[\begin{array}{cc}6 & -4 \\ -9 & 6\end{array}\right]$

$(\operatorname{adj} A) B=\left[\begin{array}{cc}6 & -4 \\ -9 & 6\end{array}\right]\left[\begin{array}{l}2 \\ 3\end{array}\right]$

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

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

If $|A|=0$ and ( $\operatorname{adj} A) B=0$, then the system is consistent and has infinitely many solutions.

Thus, $A X=B$ has infinitely many solutions.

Substituting $y=k$ in the eq. (1), we get

$6 x+4 k=2$

$\Rightarrow 6 x=2-4 k$

$\Rightarrow x=\frac{2-4 k}{6}$

$\Rightarrow x=\frac{1-2 k}{3}$

$\therefore x=\frac{1-2 k}{3}$ and $y=k$

These values of $x$ and $y$ satisfy the third equation.

Thus, $x=\frac{1-2 k}{3}$ and $y=k$ (where $k$ is a real number) satisfy the given system of equations.

(ii) Here,

$2 x+3 y=5 \quad \cdots(1)$

$6 x+9 y=15 \quad \cdots(2)$

or,$A X=B$'

where,'

$A=\left[\begin{array}{ll}2 & 3 \\ 6 & 9\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{c}5 \\ 15\end{array}\right]$

$\Rightarrow\left[\begin{array}{ll}2 & 3 \\ 6 & 9\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}5 \\ 15\end{array}\right]$

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

$=18-18$

$=0$

So, $A$ is singular. Thus, the given system of equations is either inconsistent or it is consistent with infinitely many solutions because $(\operatorname{adj} A) B \neq 0$ or $(\operatorname{adj} A)=0$.

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

$C_{11}=9, C_{12}=-6, C_{21}=-3$ and $C_{22}=2$

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

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

$\Rightarrow(\operatorname{adj} A) B=\left[\begin{array}{cc}9 & -3 \\ -6 & 2\end{array}\right]\left[\begin{array}{c}5 \\ 15\end{array}\right]$

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

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

If $|A|=0$ and (adjA) $\mathrm{B}=0$, then the system is consistent and has infinitely many solutions.

Thus, $\mathrm{AX}=\mathrm{B}$ has infinitely many solutions.

Substituting $y=k$ in eq. (1), we get

$2 x+3 k=5$

$\Rightarrow 2 x=5-3 k$

$\Rightarrow x=\frac{5-3 k}{2}$ and $y=k$

These values of $x$ and $y$ satisfy the third equation.

Thus, $x=\frac{5-3 k}{2}$ and $y=k$ (where $k$ is a real number) satisfy the given system of equations.

(iii) Here,

$5 x+3 y+7 z=4 \quad \ldots(1)$

$3 x+26 y+2 z=9 \quad \ldots(2)$

$7 x+2 y+10 z=5 \quad \ldots(3)$

or,$A X=B$'

where,

$A=\left[\begin{array}{ccc}5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10\end{array}\right], X=\left[\begin{array}{c}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}4 \\ 9 \\ 5\end{array}\right]$

$\left[\begin{array}{ccc}5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}4 \\ 9 \\ 5\end{array}\right]$

$|A|=\left|\begin{array}{ccc}5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10\end{array}\right|$'

$=5(260-4)-3(30-14)+7(6-182)$

$=1280-48-1232$

$=0$

So, $A$ is singular. Thus, the given system of equations is either inconsistent or it is consistent with infinitely many solutions because $(\operatorname{adj} A) B \neq 0$ or $(\operatorname{adj} A) B=0$.

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

$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}26 & 2 \\ 2 & 10\end{array}\right|=256, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc}3 & 2 \\ 7 & 10\end{array}\right|=-16, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{cc}3 & 26 \\ 7 & 2\end{array}\right|=-176$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}3 & 7 \\ 2 & 10\end{array}\right|=-16, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{cc}5 & 7 \\ 7 & 10\end{array}\right|=1, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll}5 & 3 \\ 7 & 2\end{array}\right|=11$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}3 & 7 \\ 26 & 2\end{array}\right|=-176, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc}5 & 7 \\ 3 & 2\end{array}\right|=11, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{cc}5 & 3 \\ 3 & 26\end{array}\right|=121$

$\operatorname{adj} A=\left[\begin{array}{ccc}256 & -16 & -176 \\ -16 & 1 & 11 \\ -176 & 11 & 121\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}256 & -16 & -176 \\ -16 & 1 & 11 \\ -176 & 11 & 121\end{array}\right]$

$(\operatorname{adj} A) B=\left[\begin{array}{ccc}256 & -16 & -176 \\ -16 & 1 & 11 \\ -176 & 11 & 121\end{array}\right]\left[\begin{array}{l}4 \\ 9 \\ 5\end{array}\right]$

$=\left[\begin{array}{c}1024-144-880 \\ -64+9+55 \\ -704+99+605\end{array}\right]$

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

if $|A|=0$ and (adjA)B $=0$, then the system is consistent and has infinitely many solutions.

Thus, $A X=B$ has infinitely many solutions.

Substituting $z=k$ in eq. (1) and eq. (2), we get

$5 x+3 y=4-7 k$ and $3 x+26 y=9-2 k$

$\left[\begin{array}{cc}5 & 3 \\ 3 & 26\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}4-7 k \\ 9-2 k\end{array}\right]$

Now,

$|A|=\left|\begin{array}{cc}5 & 3 \\ 3 & 26\end{array}\right|$

$=130-9$

$=121 \neq 0$

$\operatorname{adj} A=\left|\begin{array}{cc}26 & -3 \\ -3 & 5\end{array}\right|$

$\Rightarrow A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$=\frac{1}{121}\left[\begin{array}{cc}26 & -3 \\ -3 & 5\end{array}\right]$

$\therefore X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{121}\left[\begin{array}{cc}26 & -3 \\ -3 & 5\end{array}\right]\left[\begin{array}{c}4-7 k \\ 9-2 k\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{121}\left[\begin{array}{c}104-182 k-27+6 k \\ -12+21 k+45-10 k\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}\frac{77-176 k}{121} \\ \frac{33+11 k}{121}\end{array}\right]$

$\Rightarrow x=\frac{11(7-16 k)}{121}, y=\frac{11(3+k)}{121}$ and $z=k$

$\therefore x=\frac{7-16 k}{11}, y=\frac{3+k}{11}$ and $z=k$

These values of $x, y$ and $z$ also satisfy the third equation.

Thus, $x=\frac{7-16 k}{11}, y=\frac{3+k}{11}$ and $z=k$ (where $k$ is a real number) satisfy the given system of equations.

(iv) Here,

$x-y+z=3 \quad \ldots(1)$

$2 x+y-z=2$      ....(2)

$-x-2 y+2 z=1$     ....(3)

or, $A X=B$

where,

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

$\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -1 \\ -1 & -2 & 2\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}3 \\ 2 \\ 1\end{array}\right]$

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

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

$=0+3-3$

$=0$

So, $A$ is singular. Thus, the given system of equations is either inconsistent or it is consistent with infinitely many solutions because $(\operatorname{adj} A) B \neq 0$ or $(\operatorname{adj} A) B=0$.

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

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

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

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

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

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

$(\operatorname{adj} A) B=\left[\begin{array}{ccc}0 & 0 & 0 \\ -3 & 3 & 3 \\ -3 & 3 & 3\end{array}\right]\left[\begin{array}{l}3 \\ 2 \\ 1\end{array}\right]$

$=\left[\begin{array}{c}0 \\ -9+6+3 \\ -9+6+3\end{array}\right]$

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

If $|A|=0$ and $(\operatorname{adj} A) B=0$, then the system is consistent and has infinitely many solutions.

Thus, $A X=B$ has infinitely many solutions.

Substituting $z=k$ in eq. (1) and eq. (2), we get

$x-y=3-k$ and $2 x+y=2+k$

$\left[\begin{array}{cc}1 & -1 \\ 2 & 1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}3-k \\ 2+k\end{array}\right]$

Now,

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

$=1+2=3 \neq 0$

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

$\Rightarrow A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$=\frac{1}{3}\left[\begin{array}{cc}1 & 1 \\ -2 & 1\end{array}\right]$

$\therefore X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{3}\left[\begin{array}{cc}1 & 1 \\ -2 & 1\end{array}\right]\left[\begin{array}{l}3-k \\ 2+k\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}3-k+2+k \\ -6+2 k+2+k\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}\frac{5}{3} \\ \frac{3 k-4}{3}\end{array}\right]$

$\therefore x=\frac{5}{3}, y=\frac{3 k-4}{3}$ and $z=k$

These values of $x, y$ and $z$ also satisfy the third equation.

Thus, $x=\frac{5}{3}, y=\frac{3 k-4}{3}$ and $z=k$ (where $k$ is a real number) satisfy the given system of equations.

(v) Here,

$x+y+z=6 \quad \ldots(1)$

$x+2 y+3 z=14 \quad \ldots(2)$

$x+4 y+7 z=30$         ...(3)
or, $A X=B$
where,

$A=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}6 \\ 14 \\ 30\end{array}\right]$

$\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}6 \\ 14 \\ 30\end{array}\right]$

$|A|=\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7\end{array}\right|$

$=1(14-12)-1(7-3)+1(4-2)$

$=2-4+2$

$=0$

So, $A$ is singular. Thus, the given system of equations is either inconsistent or it is consistent with infinitely many solutions because $(\operatorname{adj} A) B \neq 0$ or $(\operatorname{adj} A)=0$.

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

$C_{11}=(-1)^{1+1}\left|\begin{array}{ll}2 & 3 \\ 4 & 7\end{array}\right|=2, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{ll}1 & 3 \\ 1 & 7\end{array}\right|=-4, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{ll}1 & 2 \\ 1 & 4\end{array}\right|=2$

$C_{21}=(-1)^{2+1}\left|\begin{array}{ll}1 & 1 \\ 4 & 7\end{array}\right|=-3, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{ll}1 & 1 \\ 1 & 7\end{array}\right|=6, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll}1 & 1 \\ 1 & 4\end{array}\right|=-3$

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

$\operatorname{adj} A=\left[\begin{array}{ccc}2 & -4 & 2 \\ -3 & 6 & -3 \\ 1 & -2 & 1\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}2 & -3 & 1 \\ -4 & 6 & -2 \\ 2 & -3 & 1\end{array}\right]$

$(\operatorname{adj} A) B=\left[\begin{array}{ccc}2 & -3 & 1 \\ -4 & 6 & -2 \\ 2 & -3 & 1\end{array}\right]\left[\begin{array}{c}6 \\ 14 \\ 30\end{array}\right]$

$=\left[\begin{array}{c}12-42+30 \\ -24+84-60 \\ 12-42+30\end{array}\right]$

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

If $|A|=0$ and $(\operatorname{ad} j A) B=0$, then the system is consistent and has infinitely many solutions.

Thus, $A X=B$ has infinitely many solutions.

Subs tituting $z=k$ in eq. (1) and eq. (2), we get

$x+y=6-k$ and $x+2 y=14-3 k$

$\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}6-k \\ 14+3 k\end{array}\right]$

Now,

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

$=2-1=1 \neq 0$

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

$\Rightarrow A^{-1}=\frac{1}{|A|}$ adj $A$

$=\frac{1}{1}\left[\begin{array}{cc}2 & -1 \\ -1 & 1\end{array}\right]$

$\therefore X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{1}\left[\begin{array}{cc}2 & -1 \\ -1 & 1\end{array}\right]\left[\begin{array}{c}6-k \\ 14-3 k\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{1}\left[\begin{array}{c}12-2 k-14+3 k \\ -6+k+14-3 k\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}\frac{k-2}{1} \\ \frac{8-2 k}{1}\end{array}\right]$

$\therefore x=k-2, y=8-2 k$ and $z=k$

These values of $x, y$ and $z$ also satisfy the third equation.

Thus, $x=k-2, y=8-2 k$ and $z=k$ (where $k$ is a real number) satisfy the given system of equations.

(vi) Here,

$2 x+2 y-2 z=1 \quad \ldots(1)$

$4 x+4 y-z=2 \quad \ldots(2)$

$6 x+6 y+2 z=3 \quad \ldots(3)$

or, $A X=B$

where, $A=\left[\begin{array}{ccc}2 & 2 & -2 \\ 4 & 4 & -1 \\ 6 & 6 & 2\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$

$\left[\begin{array}{ccc}2 & 2 & -2 \\ 4 & 4 & -1 \\ 6 & 6 & 2\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$

$|A|=\left|\begin{array}{ccc}2 & 2 & -2 \\ 4 & 4 & -1 \\ 6 & 6 & 2\end{array}\right|$

$=2(8+6)-2(8+6)-2(24-24)$

$=28-28-0$

$=0$

So, $A$ is singular. Thus, the system of equations is either inconsistent or it is consistent with infinitely many solutions because (adj $A$ ) $B \neq 0$ or $($ adj $A) B=0$.

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

$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}4 & -1 \\ 6 & 2\end{array}\right|=14, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc}4 & -1 \\ 6 & 2\end{array}\right|=-14, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{ll}4 & 4 \\ 6 & 6\end{array}\right|=0$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}2 & -2 \\ 6 & 2\end{array}\right|=-16, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{cc}2 & -2 \\ 6 & 2\end{array}\right|=16, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll}2 & 2 \\ 6 & 6\end{array}\right|=0$

$C_{31}=(-1)^{3+1}\left|\begin{array}{ll}2 & -2 \\ 4 & -1\end{array}\right|=6, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{ll}2 & -2 \\ 4 & -1\end{array}\right|=-6, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{ll}2 & 2 \\ 4 & 4\end{array}\right|=0$

adj $A=\left[\begin{array}{ccc}14 & -14 & 0 \\ -16 & 16 & 0 \\ 6 & -6 & 0\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}14 & -16 & 6 \\ -14 & 16 & -6 \\ 0 & 0 & 0\end{array}\right]$

$(\operatorname{adj} A) B=\left[\begin{array}{ccc}14 & -16 & 6 \\ -14 & 16 & -6 \\ 0 & 0 & 0\end{array}\right]\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$

$=\left[\begin{array}{c}14-32+18 \\ -14+32-18 \\ 0\end{array}\right]$

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

If $|A|=0$ and (adj $A) B=0$, then the system is consistent and has infinitely many solutions.

Thus, $A X=B$ has infinitely many solutions.

Substituting $y=k$ in eq. (1) and eq. (2), we get

$2 x-2 z=1-2 k$ and $4 x-z=2-4 k$

$\left[\begin{array}{ll}2 & -2 \\ 4 & -1\end{array}\right]\left[\begin{array}{l}x \\ z\end{array}\right]=\left[\begin{array}{l}1-2 k \\ 2-4 k\end{array}\right]$

Now,

$|A|=\left|\begin{array}{ll}2 & -2 \\ 4 & -1\end{array}\right|$

$=-2+8=6 \neq 0$

$\operatorname{adj} A=\left|\begin{array}{ll}-1 & 2 \\ -4 & 2\end{array}\right|$

$\Rightarrow A^{-1}=\frac{1}{|A|}$ adj $A$

$=\frac{1}{6}\left[\begin{array}{ll}-1 & 2 \\ -4 & 2\end{array}\right]$

$\therefore X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ z\end{array}\right]=\frac{1}{6}\left[\begin{array}{ll}-1 & 2 \\ -4 & 2\end{array}\right]\left[\begin{array}{l}1-2 k \\ 2-4 k\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ z\end{array}\right]=\frac{1}{6}\left[\begin{array}{l}-1+2 k+4-8 k \\ -4+8 k+4-8 k\end{array}\right]$

$\Rightarrow\left[\begin{array}{c}x \\ z\end{array}\right]=\left[\begin{array}{c}\frac{3-6 k}{6} \\ 0\end{array}\right]$

$\therefore x=\frac{1-2 k}{2}, y=k$ and $z=0$

These values of $x, y$ and $z$ satisfy the third equation.

Thus, $x=\frac{1-2 k}{2}, y=k$ and $z=0$ (where $k$ is real number) satisfy the given system of equations.