find A–1 using A–1 solve2x – 3y + 5z = 113x + 2y – 4z = – 5x + y – 2z =

Engineering

Mathematics

Introduction to Matrix

Transpose and Adjoint of a Matrix

Inverse of a Matrix

Question

$A = [\begin{array}{ccc} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{array}]$ find A^–1 using A^–1 solve
2x – 3y + 5z = 11
3x + 2y – 4z = – 5
x + y – 2z = – 3

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∣ A ∣ = [\begin{array}{ccc} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{array}] = 2 (0) + 3 (- 6 + 4) + 5 (3 - 2) = - 6 + 5 = - 1 \neq 0

co factor of

A = [\begin{array}{ccc} 0 & 2 & 1 \\ - 1 & - 9 & - 5 \\ 2 & 23 & 13 \end{array}]

Adj . A = [\begin{array}{ccc} 0 & 2 & 1 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13 \end{array}]

A^{- 1} = \frac{Adj . A}{∣ A ∣} = \frac{1}{- 1} [\begin{array}{ccc} 0 & 2 & 1 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13 \end{array}]

the system of equation can be written as

[\begin{array}{ccc} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 11 \\ - 5 \\ - 3 \end{array}] \Rightarrow AX = B \Rightarrow X = A^{- 1} B

[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{ccc} 0 & 2 & 1 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13 \end{array}] [\begin{array}{c} 11 \\ - 5 \\ - 3 \end{array}] = [\begin{array}{c} 1 \\ 2 \\ 3 \end{array}]

x = 1, y = 2, z = 3