1. Solve linear equations x+y=2 and 2x+3y=4 using Inverse matrix Method

Solution:
Here `x+y=2`
`2x+3y=4`

Now converting given equations into matrix form
`[[1,1],[2,3]] [[x],[y]]=[[2],[4]]`

Now, A = `[[1,1],[2,3]]`, X = `[[x],[y]]` and B = `[[2],[4]]`

`:.AX = B`

`:.X = A^-1 B`

`|A|`

=

`1`   `1`   `2`   `3`

`=1 × 3 - 1 × 2`

`=3 -2`

`=1`


`"Here, " |A| = 1 != 0`

`:. A^(-1) " is possible."`

`Adj(A)`

=

Adj `1` `1` `2` `3`

=	`+(3)` `-(2)` `-(1)` `+(1)` T

=	`3` `-2` `-1` `1` T

=	`3` `-1` `-2` `1`

`"Now, "A^(-1)=1/|A| × Adj(A)`

`"Here, "X = A^(-1) × B`

`:. X = 1/|A| × Adj(A) × B`

=

`1/(1)` ×

`3` `-1` `-2` `1`

×

`2` `4`

=	`3×2-1×4` `-2×2+1×4`

=

`2` `0`

`:.[[x],[y]]=[[2],[0]]`

`:.x=2,y=0`

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2. Solve linear equations 2x+7y-11=0 and 3x-y-5=0 using Inverse matrix Method

Solution:
Here `2x+7y-11=0`
`3x-y-5=0`

Now converting given equations into matrix form
`[[2,7],[3,-1]] [[x],[y]]=[[11],[5]]`

Now, A = `[[2,7],[3,-1]]`, X = `[[x],[y]]` and B = `[[11],[5]]`

`:.AX = B`

`:.X = A^-1 B`

`|A|`

=

`2`   `7`   `3`   `-1`

`=2 × (-1) - 7 × 3`

`=-2 -21`

`=-23`


`"Here, " |A| = -23 != 0`

`:. A^(-1) " is possible."`

`Adj(A)`

=

Adj `2` `7` `3` `-1`

=	`+(-1)` `-(3)` `-(7)` `+(2)` T

=	`-1` `-3` `-7` `2` T

=	`-1` `-7` `-3` `2`

`"Now, "A^(-1)=1/|A| × Adj(A)`

`"Here, "X = A^(-1) × B`

`:. X = 1/|A| × Adj(A) × B`

=

`1/(-23)` ×

`-1` `-7` `-3` `2`

×

`11` `5`

=

`-1/23` ×

`-1×11-7×5` `-3×11+2×5`

=

`-1/23` ×

`-46` `-23`

=

`2` `1`

`:.[[x],[y]]=[[2],[1]]`

`:.x=2,y=1`

---

3. Solve linear equations 3x-y=3 and 7x+2y=20 using Inverse matrix Method

Solution:
Here `3x-y=3`
`7x+2y=20`

Now converting given equations into matrix form
`[[3,-1],[7,2]] [[x],[y]]=[[3],[20]]`

Now, A = `[[3,-1],[7,2]]`, X = `[[x],[y]]` and B = `[[3],[20]]`

`:.AX = B`

`:.X = A^-1 B`

`|A|`

=

`3`   `-1`   `7`   `2`

`=3 × 2 - (-1) × 7`

`=6 +7`

`=13`


`"Here, " |A| = 13 != 0`

`:. A^(-1) " is possible."`

`Adj(A)`

=

Adj `3` `-1` `7` `2`

=	`+(2)` `-(7)` `-(-1)` `+(3)` T

=	`2` `-7` `1` `3` T

=	`2` `1` `-7` `3`

`"Now, "A^(-1)=1/|A| × Adj(A)`

`"Here, "X = A^(-1) × B`

`:. X = 1/|A| × Adj(A) × B`

=

`1/(13)` ×

`2` `1` `-7` `3`

×

`3` `20`

=

`1/13` ×

`2×3+1×20` `-7×3+3×20`

=

`1/13` ×

`26` `39`

=

`2` `3`

`:.[[x],[y]]=[[2],[3]]`

`:.x=2,y=3`

---

4. Solve linear equations 2x-y=11 and 5x+4y=1 using Inverse matrix Method

Solution:
Here `2x-y=11`
`5x+4y=1`

Now converting given equations into matrix form
`[[2,-1],[5,4]] [[x],[y]]=[[11],[1]]`

Now, A = `[[2,-1],[5,4]]`, X = `[[x],[y]]` and B = `[[11],[1]]`

`:.AX = B`

`:.X = A^-1 B`

`|A|`

=

`2`   `-1`   `5`   `4`

`=2 × 4 - (-1) × 5`

`=8 +5`

`=13`


`"Here, " |A| = 13 != 0`

`:. A^(-1) " is possible."`

`Adj(A)`

=

Adj `2` `-1` `5` `4`

=	`+(4)` `-(5)` `-(-1)` `+(2)` T

=	`4` `-5` `1` `2` T

=	`4` `1` `-5` `2`

`"Now, "A^(-1)=1/|A| × Adj(A)`

`"Here, "X = A^(-1) × B`

`:. X = 1/|A| × Adj(A) × B`

=

`1/(13)` ×

`4` `1` `-5` `2`

×

`11` `1`

=

`1/13` ×

`4×11+1×1` `-5×11+2×1`

=

`1/13` ×

`45` `-53`

=	`45/13` `-53/13`

`:.[[x],[y]]=[[45/13],[-53/13]]`

`:.x=45/13,y=-53/13`

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This material is intended as a summary. Use your textbook for detail explanation.
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