Let the given matrix
we have to find A-1.
Now A-1 will exist only of |A| ≠ 0
Let us first find |A|
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⇒ |A| = (-3) (-3) – (5)(2)
⇒ |A| = 9 – 10 = -1
Hence |A| ≠ 0 and A-1 exist
Now A-1 is given by
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Now adjoint(A) = CT where C is the cofactor matrix and CT is the transpose of cofactor matrix
The cofactor is given by
⇒ Cij = (-1) i+jMij …(a)
Where M represents minor
Mij is the determinant of matrix leaving the ith row and jth column
The cofactor matrix C will be
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Now transpose of C that is CT
For CT we will interchange the rows and columns
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Using (a)
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From matrix A the minors are
M11 = -3, M12 = 5, M21 = 2 and M22 = -3
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Now inverse of A is
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Now we have to solve the matrix equation
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As we know that AA-1 = I hence multiply equation by A-1 from right side where I is identity matrix
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Hence ![]()
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