If A = [(3,1)(7,5)], find x and y such that A^2 + xI = yA. Hence, find A^-1.

Question

If A = \(\begin{bmatrix} 3 & 1 \\[0.3em] 7 & 5 \\[0.3em] \end{bmatrix},\) find x and y such that A² + xI = yA. Hence, find A^-1.

A = [(3,1)(7,5)

Answer 1

Given: A = \(\begin{bmatrix} 3 & 1 \\[0.3em] 7& 5 \\[0.3em] \end{bmatrix}\)

To find: value of x and y

Given equation: A² + xI = yA

Firstly, we find the A²

Putting the values in given equation

A² + xI = yA

On Comparing, we get 

16 + x = 3y …(i) 

y = 8 …(ii) 

56 = 7y …(iii) 

32 + x = 5y …(iv) 

Putting the value of y = 8 in eq. (i), we get 

16 + x = 3(8) 

⇒ 16 + x = 24 

⇒ x = 8 

Hence, the value of x = 8 and y = 8 

So, the given equation become A² + 8I = 8A 

Now, we have to find A^-1 

Finding A^-1 using given equation 

A² + 8I = 8A 

Post multiplying by A^-1 both sides, we get 

(A² + 8I)A^-1 = 8AA^-1 

⇒ A².A^-1 + 8I.A^-1 = 8AA^-1 

⇒ A.(AA^-1) + 8A^-1 = 8I [AA^-1 = I] 

⇒ A(I) + 8A^-1 = 8I 

⇒ A + 8A^-1 = 8I 

⇒ 8A^-1 = – A + 8I

x = 8, y = 8 and A^-1 = \(\frac{1}{8}.\begin{bmatrix} 5 & -1 \\[0.3em] -7 & 3 \\[0.3em] \end{bmatrix}\)