Mathematics GATE 2018 EE: 44

Question

Let $A$ = $\begin{bmatrix} 1 & 0 & -1 \\-1 & 2 &  0 \\  0 &  0 & -2 \end{bmatrix}$  and $B$ = $A^{3} - A^{2} -4A +5I$  where
$I$ is the $3\times 3$ identity matrix. The determinant of  $B$  is ________ (up to $1$ decimal place).

Answer 1

Answer should be 1

As we know that Determinant of a matrix = Product of the eigen values of that matrix

So ,to find the determinant of matrix B , first we have to find the eigen values of B and to find the eigen values of B , first we have to find the eigen values of A.

So, characteristic equation should be | A- λI |  = 0

So , After expanding ,(1-λ)(2-λ)(2+λ) = 0

So , λ = 1,2,-2

Since , B = A³ - A² -4A +5I

So , 1st Eigen value of B = 1³ - 1² - 4*1 + 5 = 1

2nd Eigen value of B = 2³ - 2² - 4*2 + 5 =1

3rd Eigen value of B = (-2)³ - (-2)² -4*(-2) + 5 = 1

Since Eigen values of B = 1,1,1 , So , Det(B) =1*1*1 = 1

Answer 2

characteristic equation of A be | A- λI |  = 0

Now, Eigenvalues of A are  λ = 1 ,2,-2

Therefore, Characterstic Eqn is (λ-1)(λ-2)(λ+2)=0

Expanding It we get       λ³ - λ² -4λ +4I = 0

According to Cayley-Hamilton theorem   " Every Square Matrix Satisfies It 's Characteristic Eqn".

therefore, A³ - A² -4A +4I = O (Null Matrix)

B = A³ - A² -4A +5I = (A³ - A² -4A +4I) +I

=> B= O+I ( we know A³ - A² -4A +4I =O)

=> B =I

Det(B) = Det(I) =1

 

Answer 3

First here find the eigens value of A . If we find the eigens value of then the characteristics equation will become  A3 - A2 -4A +4I=0 now given equation given that  B = A3 - A2 -4A +5I

  B = A3 - A2 -4A +4I+I

B=0+I=I so determinate of B=1.