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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).
in Linear Algebra by Veteran (52.9k points)
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2 Answers

+5 votes

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

by Boss (15.9k points)
0 votes

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

 

by (17 points)

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