The system of linear equations x + y = 2 2 x + 2 y = 3 has

more than one but finitely many solutions

Find the inverse (if exist) of the matrix 2 4 2 3 .

Let A = 2 4 2 3 Now, we have, | A | = 2 4 2 3 = 2 · 3 - 4 · 2 = 6 - 8 = - 2 ≠ 0 Thus, matrix A is non-singular matrix and hence its unique inverse A - 1 exists. Now, Adj A = 3 - 4 - 2 2 [We can write the adjoint matrix by interchanging the main diagonal entries and reversing the sign of the other two elements] So, using A - 1 = 1 | A | Adj A = 1 - 2 3 - 4 - 2 2 = - 3 2 2 1 - 1 Hence, the inverse of the given matrix is - 3 2 2 1 - 1 .

Find the inverse (if exist) of the matrix cosec θ cot θ cot θ cosec θ .

Let A = cosec θ cot θ cot θ cosec θ Now, we have, | A | = cosec θ cot θ cot θ cosec θ = cosec θ · cosec θ - cot θ · cot θ = cosec 2 θ - cot 2 θ = 1 ≠ 0 Thus, matrix A is non-singular matrix and hence its unique inverse A - 1 exists. Now, Adj A = cosec θ - cot θ - cot θ cosec θ [We can write the adjoint matrix by interchanging the main diagonal entries and reversing the sign of the other two elements] So, using A - 1 = 1 | A | Adj A = 1 1 cosec θ - cot θ - cot θ cosec θ = cosec θ - cot θ - cot θ cosec θ Hence, the inverse of the given matrix is cosec θ - cot θ - cot θ cosec θ .

Find the inverse (if exist) of the matrix cos α sin α − sin α cos α .

Let A = cos α sin α - sin α cos α Now, we have, | A | = cos α sin α - sin α cos α = cos α · cos α + sin α · sin α = cos 2 α + sin 2 α = 1 ≠ 0 Thus, matrix A is non-singular matrix and hence its unique inverse A - 1 exists. Now, Adj A = cos α - sin α sin α cos α [We can write the adjoint matrix by interchanging the main diagonal entries and reversing the sign of the other two elements] So, using A - 1 = 1 | A | Adj A = 1 1 cos α - sin α sin α cos α = cos α - sin α sin α cos α Hence, the inverse of the given matrix is cos α - sin α sin α cos α .

Find the inverse (if exist) of the matrix sec θ + tan θ 1 1 sec θ − tan θ

Let A = sec θ + tan θ 1 1 sec θ - tan θ Now, we have, | A | = sec θ + tan θ 1 1 sec θ - tan θ = sec θ + tan θ · sec θ - tan θ - 1 · 1 = sec 2 θ - tan 2 θ - 1 = 1 - 1 = 0 Thus, matrix A is singular matrix and hence its unique inverse A - 1 does not exist.

Given that A 2 − 5 A + 7 I 2 = 0 for A = 2 − 1 1 3 , use it to show that A − 1 = 1 7 3 1 − 1 2 .

Given that, A = 2 - 1 1 3 Now, A 2 = A · A = 2 - 1 1 3 2 - 1 1 3 = 4 - 1 - 2 - 3 2 + 3 - 1 + 9 = 3 - 5 5 8 Considering A 2 - 5 A + 7 I 2 = 3 - 5 5 8 - 5 2 - 1 1 3 + 7 1 0 0 1 [Where I 2 = 1 0 0 1 ] = 3 - 5 5 8 - 10 - 5 5 15 + 7 0 0 7 = 3 - 10 + 7 - 5 + 5 + 0 5 - 5 + 0 8 - 15 + 7 = 0 0 0 0 Hence, proved. Now, A 2 - 5 A + 7 I 2 = 0 or A A - 5 I 2 = - 7 I 2 or A - 1 7 A - 5 I 2 = I 2 Hence, by definition of inverse of a matrix, we have A - 1 = - 1 7 A - 5 I 2 = - 1 7 2 - 1 1 3 - 5 0 0 5 [Pre multiplying with A - 1 and then use A - 1 I = A - 1 ] = - 1 7 - 3 - 1 1 - 2 = 1 7 3 1 - 1 2 Hence, proved.

Give that the matrix A = 2 3 1 2 satisfies the equation A 2 − 4 A + I 2 = 0 , find A − 1 .

Given matrix, A = 2 3 1 2 Now, A 2 - 4 A + I 2 = O or A A - 4 I 2 = - I 2 or A - A - I 2 = I 2 Hence, by definition of inverse of a matrix, we have A - 1 = - A - 4 I 2 = - 2 3 1 2 - 4 0 0 4 [Pre multiplying with A - 1 and using A - 1 I = A - 1 ] = - 2 - 4 3 - 0 1 - 0 2 - 4 = 2 - 3 - 1 2 Thus, A - 1 = 2 - 3 - 1 2 .