IF A = ((2,-3,5),(3,2,-4),(1,1,-2)), find A^(-1).

Question

If A \(\begin{pmatrix} 2 & -3 & 5 \\ 3 &2&-4 \\ 1 &1 & -2 \end{pmatrix}\), find A^-1

If A = ((2,-3,5),(3,2,-4),(1,1,-2)), find A^-1

Using A ^-1, solve the following system of equations: 

2x – 3y + 5z = 11; 

3x + 2y – 4z = - 5; 

x + y – 2z = - 3.

Answer 1

Given,

\(\begin{pmatrix} 2 & -3 & 5 \\ 3 &2&-4 \\ 1 &1 & -2 \end{pmatrix}\)

A ^-1 = \(\frac{1}{|A|}\)adj (A)

The determinant of matrix A is

|A| = \(\begin{pmatrix} 2 & -3 & 5 \\ 3 &2&-4 \\ 1 &1 & -2 \end{pmatrix}\)

= 2( 2 × - 2 – ( - 4) ×1) + 3(3× - 2 – ( - 4) ×1) + 5(3×1 – 2 × 1) 

= 2( - 4 + 4 ) + 3( - 6 + 4 ) + 5( 3 – 2 ) 

= 2(0) + 3( - 2) + 5(1) 

= - 6 + 5 

= - 1 

|A| ≠ 0 

∴ A - 1 is possible.

A ^-1 = \(\frac{1}{|A|}\)adj (A)

Given set of lines are : - 

2x – 3y + 5z = 11 

3x + 2y – 4z = - 5 

x + y – 2z = - 3 

Converting following equations in matrix form, 

AX = B

Where A

 

Pre - multiplying by A^-1 

A ^-1AX = A ^-1B

IX = A^-1B 

X = A^-1B

∴ x = 1 , y = 2 , z = 3