# Given `A= ((4, 3, 2),(5,6,3),(3,5,2));B = ((a, b, c),(d,e,f),(g,h,i))`Find B so that AB = I = BA as follows: First equate entries on the two sides of the equation AB = I . Then solve the...

Given `A= ((4, 3, 2),(5,6,3),(3,5,2));B = ((a, b, c),(d,e,f),(g,h,i))`

Find B so that AB = I = BA as follows:

First equate entries on the two sides of the equation AB = I . Then solve the resulting nine equations for a, b, c, d, e, f, g, h and i.

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### 1 Answer

You need to multiply the matrices A and B such that:

`((4,3,2),(5,6,3),(3,5,2))*((a,b,c),(d,e,f),(g,h,i)) = ((1,0,0),(0,1,0),(0,0,1))`

`((4a+3d+2g, 4b+3e+2h, 4c+3f+2i),(5a+6d+3g, 5b+6e+3h, 5c+6f+3i),(3a+5d+2g, 3b+5+2h, 3c+5f+2i)) = ((1,0,0),(0,1,0),(0,0,1))`

Equating corresponding terms yields:

`4a+3d+2g = 1 `

`4b+3e+2h = 0`

`4c+3f+2i = 0 `

`5a+6d+3g = 0 `

`5b+6e+3h = 1 `

`5c+6f+3i = 0 `

`3a+5d+2g = 0 `

`3b+5e+2h = 0 `

`3c+5f+2i = 1`

You need to consider the equations that contain a,d,g such that:

`4a+3d+2g = 1 `

`5a+6d+3g = 0 `

`3a+5d+2g = 0`

You need to subtract the third equation from the first such that:

`a - 2d = 1`

You need to multiply the first equation by 3 and the second equation by -2 and then you should add the new equations such that:

`12a + 9d + 6g = 3`

` -10a - 12d - 6g = 0`

`2a - 3d = 3`

You need to multiply the equation a - 2d = 1 by -2 such that:

`-2a + 4d = -2`

Adding this equation to 2a - 3d = 3 yields:

`2a - 3d - 2a + 4d= 3 - 2`

`d = 1 =gt a - 2 = 1 =gt a = 3`

`15+6+3g = 0 =gt 3g = -21 =gt g = -7`

You need to consider the equations that contain b,e,h such that:

`4b+3e+2h = 0 `

`5b+6e+3h = 1 `

`3b+5e+2h = 0`

Subtracting the third equation from the first yields:

`b - 2e = 0 =gt b = 2e`

Substituting 2e for b in the first and second equations yields:

`11e + 2h = 0 `

`16e + 3h = 1`

You need to multiply by 3 the equation 11e + 2h = 0 and by -2 the equation 16e + 3h = 1 such that:

`33e + 6h = 0 `

`-32e - 6h = -2`

`33e + 6h -32e - 6h = 0-2`

`e = -2 =gt b = -4`

`-16-6+2h = 0 =gt 2h = 22 =gt h = 11`

You need to consider the equations that contain c,f,i such that:

`4c+3f+2i = 0 `

`5c+6f+3i = 0 `

`3c+5f+2i = 1`

`4c+3f+2i - 3c-5f-2i = -1`

` c - 2f = -1 =gt c = 2f - 1`

Substituting 2f - 1 for c in the first and second equations yields:

`4(2f-1)+3f+2i = 0=gt 11f + 2i = 4 `

`5(2f-1)+6f+3i = 0 =gt 16f + 3i = 5`

`3(11f+2i) - 2(16f+3i) = 12 - 10`

`33f + 6i - 32f - 6i = 2`

`f = 2 =gt c = 4 - 1 = 3`

`12+6+2i = 0 =gt 2i = -18 =gt i = -9`

**Hence, evaluating the matrix B under given conditions yields `B = ((3,-4,3),(1,-2,2),(-7,11,-9)).` **