Answer :

Let the given matrix we have to find A^{-1}.

Now A^{-1} will exist only of |A| ≠ 0

Let us first find |A|

⇒ |A| = (-3) (-3) – (5)(2)

⇒ |A| = 9 – 10 = -1

Hence |A| ≠ 0 and A^{-1} exist

Now A^{-1} is given by

Now adjoint(A) = C^{T} where C is the cofactor matrix and C^{T} is the transpose of cofactor matrix

The cofactor is given by

⇒ C_{ij} = (-1) ^{i+j}M_{ij} …(a)

Where M represents minor

M_{ij} is the determinant of matrix leaving the i^{th} row and j^{th} column

The cofactor matrix C will be

Now transpose of C that is C^{T}

For C^{T} we will interchange the rows and columns

Using (a)

From matrix A the minors are

M_{11} = -3, M_{12} = 5, M_{21} = 2 and M_{22} = -3

Now inverse of A is

Now we have to solve the matrix equation

As we know that AA^{-1} = I hence multiply equation by A^{-1} from right side where I is identity matrix

Hence

Rate this question :

<span lang="EN-USMathematics - Exemplar

<span lang="EN-USMathematics - Exemplar

Using matrices soMathematics - Board Papers

On her birthday SMathematics - Board Papers

Using matrices soMathematics - Board Papers

<span lang="EN-USRS Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal - Mathematics

Using elementary RS Aggarwal - Mathematics

Using elementary RS Aggarwal - Mathematics