Q. 115.0( 3 Votes )

# Show that satisfies the equation A^{2} – 3A – 7I = 0 and hence find A^{–1}.

Answer :

We have the matrix A, such that

(i). We need to show that the matrix A satisfies the equation A^{2} – 3A – 7I = 0.

(ii). Also, we need to find A^{-1}.

(i). Take L.H.S: A^{2} – 3A – 7I

First, compute A^{2}.

A^{2} = A.A

In order to multiply two matrices, A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

Multiply 1^{st} row of matrix A by matching members of 1^{st} column of matrix A, then sum them up.

(5, 3).(5, -1) = (5 × 5) + (3 × -1)

⇒ (5, 3).(5, -1) = 25 + (-3)

⇒ (5, 3).(5, -1) = 25 – 3

⇒ (5, 3).(5, -1) = 22

Multiply 1^{st} row of matrix A by matching members of 2^{nd} column of matrix A, then sum them up.

(5, 3).(3, -2) = (5 × 3) + (3 × -2)

⇒ (5, 3).(3, -2) = 15 + (-6)

⇒ (5, 3).(3, -2) = 15 – 6

⇒ (5, 3).(3, -2) = 9

Multiply 2^{nd} row of matrix A by matching members of 1^{st} column of matrix A, then sum them up.

(-1, -2).(5, -1) = (-1 × 5) + (-2 × -1)

⇒ (-1, -2).(5, -1) = -5 + 2

⇒ (-1, -2).(5, -1) = -3

Multiply 2^{nd} row of matrix A by matching members of 2^{nd} column of matrix A, then sum them up.

(-1, -2).(3, -2) = (-1 × 3) + (-2 × -2)

⇒ (-1, -2).(3, -2) = -3 + 4

⇒ (-1, -2).(3, -2) = 1

Substitute values of A^{2} and A in A^{2} – 3A – 7I.

Also, since matrix A is of the order 2 × 2, then I will be the identity matrix of order 2 × 2 such that,

Clearly,

L.H.S = R.H.S

Thus, we have shown that matrix A satisfy A^{2} – 3A – 7I = 0.

(ii). Now, let us find A^{-1}.

We know that, inverse of matrix A is A^{-1} is true only when

A × A^{-1} = A^{-1} × A = I

Where, I = Identity matrix

We have,

A^{2} – 3A – 7I = 0

Multiply A^{-1} on both sides, we get

A^{-1}(A^{2} – 3A – 7I) = A^{-1} × 0

⇒ A^{-1}.A^{2} – A^{-1}.3A – A^{-1}.7I = 0

⇒ A^{-1}.A.A – 3A^{-1}.A – 7A^{-1}.I = 0

⇒ (A^{-1}A)A – 3(A^{-1}A) – 7(A^{-1}I) = 0

And as A^{-1}A = I and A^{-1}I = A^{-1}

⇒ IA – 3I – 7A^{-1} = 0

Since, IA = A

⇒ A – 3I – 7A^{-1} = 0

⇒ 7A^{-1} = A – 3I

[∵ ]

Thus, .

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