Inverse matrix Method Examples

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6. Inverse matrix Method example ( Enter your problem )

( Enter your problem )

Examples

Other related methods

5. Graphical method
(Previous method)

7. Cramer's Rule method
(Next method)

1. Examples

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)`

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

	`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`

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)`

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

	`-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)`

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

	`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)`

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

	`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`

This material is intended as a summary. Use your textbook for detail explanation.
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5. Graphical method
(Previous method)

7. Cramer's Rule method
(Next method)

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