Q.Solve the following equations by the methods of inversion. $x+y+z=1,\ 2x+3y+2z=2$ and $ax+ay+2az=4$

Compute $A^{-1}$ via cofactors/adjoint, then $X=A^{-1}B$.

$A=\begin{bmatrix}1 & -1 & 1\\ 2 & 1 & -3\\ 1 & 1 & 1\end{bmatrix}$, $B=\begin{bmatrix}4\\0\\2\end{bmatrix}$

$|A| = 1(1+3)-(-1)(2+3)+1(2-1) = 4+5+1=10$

Cofactors:

$C_{11}=4,\ C_{12}=-5,\ C_{13}=1$

$C_{21}=2,\ C_{22}=0,\ C_{23}=-2$

$C_{31}=2,\ C_{32}=5,\ C_{33}=3$

$\text{adj}(A) = \begin{bmatrix}4 & 2 & 2\\ -5 & 0 & 5\\ 1 & -2 & 3\end{bmatrix}\quad(\text{transpose of the cofactor matrix})$

$A^{-1} = \dfrac1{10}\begin{bmatrix}4 & 2 & 2\\ -5 & 0 & 5\\ 1 & -2 & 3\end{bmatrix}$

$X = A^{-1}B = \dfrac1{10}\begin{bmatrix}4(4)+2(0)+2(2)\\ -5(4)+0(0)+5(2)\\ 1(4)+(-2)(0)+3(2)\end{bmatrix} = \dfrac1{10}\begin{bmatrix}20\\-10\\10\end{bmatrix} = \begin{bmatrix}2\\-1\\1\end{bmatrix}$

Check: $x-y+z=2+1+1=4$ ✓; $2x+y-3z=4-1-3=0$ ✓; $x+y+z=2-1+1=2$ ✓

Translate the word problem into 3 linear equations, write as $AX=B$, and solve by elimination (equivalent to matrix reduction).

Let the numbers be $x, y, z$.

"If three numbers are added, their sum is 2": $\quad x+y+z=2$ ... (i)

"If two times the second number is subtracted from the sum of first and third we get 8": $\quad (x+z)-2y=8 \implies x-2y+z=8$ ... (ii)

"If three times the first number is added to the sum of second and third we get 4": $\quad 3x+(y+z)=4 \implies 3x+y+z=4$ ... (iii)

In matrix form $AX=B$: $A=\begin{bmatrix}1&1&1\\1&-2&1\\3&1&1\end{bmatrix}$, $X=\begin{bmatrix}x\\y\\z\end{bmatrix}$, $B=\begin{bmatrix}2\\8\\4\end{bmatrix}$

Reducing: (ii) $-$ (i): $-3y = 6 \implies y=-2$

(iii) $-$ (i): $2x = 2 \implies x=1$

From (i): $1+(-2)+z=2 \implies z=3$

Check: (ii): $1-2(-2)+3=1+4+3=8$ ✓   (iii): $3(1)+(-2)+3=4$ ✓

So $x=1,\ y=-2,\ z=3$.

Set up $AX=B$ from the three cost equations and solve using Cramer's rule (determinants).

Let $x,y,z$ = cost per dozen of pencils, pens, erasers respectively.

$4x+3y+2z=60$

$2x+4y+6z=90$

$6x+2y+3z=70$

In matrix form $AX=B$ with $A=\begin{bmatrix}4&3&2\\2&4&6\\6&2&3\end{bmatrix}$, $X=\begin{bmatrix}x\\y\\z\end{bmatrix}$, $B=\begin{bmatrix}60\\90\\70\end{bmatrix}$.

$\det(A)$:

$\det A=4(4\cdot3-6\cdot2)-3(2\cdot3-6\cdot6)+2(2\cdot2-4\cdot6)$

$=4(0)-3(-30)+2(-20)=0+90-40=50$

$\det(A_x)$ (replace column 1 with $B$):

$A_x=\begin{bmatrix}60&3&2\\90&4&6\\70&2&3\end{bmatrix}, \quad \det A_x=60(0)-3(270-420)+2(180-280)=450-200=250$

$x=\frac{250}{50}=5$

$\det(A_y)$ (replace column 2 with $B$):

$A_y=\begin{bmatrix}4&60&2\\2&90&6\\6&70&3\end{bmatrix}, \quad \det A_y=4(270-420)-60(6-36)+2(140-540)=-600+1800-800=400$

$y=\frac{400}{50}=8$

$\det(A_z)$ (replace column 3 with $B$):

$A_z=\begin{bmatrix}4&3&60\\2&4&90\\6&2&70\end{bmatrix}, \quad \det A_z=4(280-180)-3(140-540)+60(4-24)=400+1200-1200=400$

$z=\frac{400}{50}=8$

Verification: $4(5)+3(8)+2(8)=20+24+16=60$ ✓; $6(5)+2(8)+3(8)=30+16+24=70$ ✓.