If A = `[(1, 0, 1),(0, 2, 3),(1, 2, 1)] "and B" = [(1, 2, 3),(1, 1, 5),(2, 4, 7)]`, then find a matrix X such that XA = B.

#### Solution 1

A = `[(1, 0, 1),(0, 2, 3),(1, 2, 1)]` and B = `[(1, 2, 3),(1, 1, 5),(2, 4, 7)]`

XA = B

Post multiplying by A^{–1}, we get

XAA^{–1} = BA^{–1}

∴ X = BA^{–1} ...(i)

|A| = `|(1, 0, 1),(0, 2, 3),(1, 2, 1)|`

= 1(2 – 6) – 0 + 1(0 – 2)

= – 4 – 2

= – 6 ≠ 0

∴ A^{–1} exists.

A_{11} = (– 1)^{1+1} M_{11} = `1 |(2, 3),(2, 1)|` = 1(2 – 6) = – 4

A_{12} = (– 1)^{1+2} M_{12} = `-1 |(0, 3),(1, 1)|` = –1(0 – 3) = 3

A_{13} = (– 1)^{1+3} M_{13} = `1 |(0, 2),(1, 2)|` = 1(0 – 2) = – 2

A_{21} = (– 1)^{2+1} M_{21} = `-1 |(0, 1),(2, 1)|` = –1(0 – 2) = 2

A_{22} = (– 1)^{2+2} M_{22} = `1 |(1, 1),(1, 1)|` = 1(1 – 1) = 0

A_{23} = (– 1)^{2+3} M_{23 }= `-1 |(1, 0),(1, 2)|` = –1(2 – 0) = – 2

A_{31} = (– 1)^{3+1} M_{31 }= `1 |(0, 1),(2, 3)|` = 1(0 – 2) = – 2

A_{32} = (– 1)^{3+2} M_{32 }= `-1 |(1, 1),(0, 3)|` = –1(3 – 0) = – 3

A_{33} = (– 1)^{3+3} M_{33 }= `1 |(1, 0),(0, 2)|` = 1(2 – 0) = 2

∴ The matrix of the co-factors is

[A_{ij}]_{3×3} = `[("A"_11, "A"_12, "A"_13),("A"_21, "A"_22, "A"_23),("A"_31, "A"_32, "A"_33)]`

= `[(-4, 3, -2),(2, 0, -2),(-2, -3, 2)]`

Now, adj A = `["A"_"ij"]_(3 xx 3)^"T"`

= `[(-4, 2, -2),(3, 0, -3),(-2, -2, 2)]`

∴ A^{–1} = `1/|"A"| ("adj A")`

= `-1/6[(-4, 2, -2),(3, 0, -3),(-2, -2, 2)]`

X = BA^{–1 } ...[From (i)]

∴ X = `[(1, 2, 3),(1, 1, 5),(2, 4, 7)] {(1/6) [(-4, 2, -2),(3, 0, -3),(-2, -2, 2)]}`

= `(1)/(6) [(1, 2, 3),(1, 1, 5),(2, 4, 7)] [(4, -2, 2),(-3, 0, 3),(2, 2, -2)]`

= `(1)/(6) [(4 - 6 + 6, -2 + 0 + 6, 2 + 6 - 6),(4 - 3 + 10, -2 + 0 + 10, 2 + 3 - 10),(8 - 12 + 14, -4 + 0 + 14, 4 + 12 - 14)]`

∴ X = `(1)/(6)[(4, 4, 2),(11, 8, -5),(10, 10, 2)]`

#### Solution 2

Consider XA = B

∴ `X[(1, 0, 1),(0, 2, 3),(1, 2, 1)] = [(1, 2, 3),(1, 1, 5),(2, 4, 7)]`

By C_{3} – C_{1}, we get

`[(1, 0, 0),(0, 2, 3),(1, 2, 0)] = [(1, 2, 2),(1, 1, 4),(2, 4, 5)]`

By `(1/2)C_2`, we get

`[(1, 0, 0),(0, 1, 3),(1, 1, 0)] = [(1, 1, 2),(1, 1/2, 4),(2, 2, 5)]`

By C_{3} – 3C_{2}, we get

`[(1, 0, 0),(0, 1, 0),(1, 1, -3)] = [(1, 1, -1),(1, 1/2, 5/2),(2, 2, -1)]`

By `(-1/3)C_3`, we get

`[(1, 0, 0),(0, 1, 0),(1, 1, 1)] = [(1, 1, 1/3),(1, 1/2, -5/6),(2, 2, 1/3)]`

By C_{1} – C_{3} and C_{2} – C_{3}, we get

`[(1, 0, 0),(0, 1, 0),(0, 0, 1)] = [(2/3, 2/3, 1/3),(11/6, 4/3, -5/6),(5/3, 5/3, 1/3)]`

∴ X = `1/6[(4, 4, 2),(11, 8, -5),(10, 10, 2)]`