Question

7 number what is the answer of adjoins A

Answer 1

Question:

$If\:A=\left[\begin{array}{ccc}1&-1&2\\2&3&5\\-2&0&1\end{array}\right], find\:adjoint\:of\:A$

Answer:

$adj\:A =\left[\begin{array}{ccc}3&1&-11\\-12&5&-1\\6&2&5\end{array}\right]$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

$A=\left[\begin{array}{ccc}1&-1&2\\2&3&5\\-2&0&1\end{array}\right]$

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Adjoint of the matrix

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ First find the minor and cofactor of the matrix

  M₁₁ = 3 - 0 = 3                        A₁₁ = ( - 1 )¹⁺¹ ( 3 ) =3

  M₁₂ = 2 + 10 = 12                   A₁₂ = (-1)¹⁺² ( 12 ) = -12

  M₁₃ = 0 + 6 = 6                      A₁₃ = (-1)¹⁺³ (6) = 6

  M₂₁ = -1 - 0 = -1                      A₂₁ = (-1)²⁺¹ (-1) = 1

  M₂₂ = 1 + 4 = 5                      A₂₂ = (-1)²⁺² (5) = 5

  M₂₃ = 0 - 2 = - 2                     A₂₃ = (-1) ²⁺³ (-2) = 2

  M₃₁ = -5 - 6 = -11                    A₃₁ = (-1) ³⁺¹ (-11) = -11

  M₃₂ = 5 - 4 =1                        A₃₂ = (-1)³⁺² (1) = -1

  M₃₃ = 3 + 2 = 5                     A₃₃ = (-1)³⁺³ (5 ) = 5

$adj\:A = Transpose\:of\:\left[\begin{array}{ccc}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{array}\right]$

$adj\:A=\left[\begin{array}{ccc}A_{11}&A_{21}&A_{31}\\A_{12}&A_{22}&A_{32}\\A_{13}&A_{23}&A_{33}\end{array}\right]$

→ Substitute the values,

  $adj\:A =\left[\begin{array}{ccc}3&1&-11\\-12&5&-1\\6&2&5\end{array}\right]$

$\Large{\underline{\underline{\bf{Notes:}}}}$

$adj\:A = Transpose\:of\:\left[\begin{array}{ccc}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{array}\right]$