If A = {bmatrix} 1 & 5 \\ 4 & 2 {bmatrix}, then find A² - 4A - 12I .

Correct Answer is : [5−5−44]{bmatrix} 5 & -5 \\ -4 & 4 {bmatrix}[5−4−54]

{bmatrix} -3 & 2 \\ 2 & -3 {bmatrix}

{bmatrix} 0 & 4 \\ 4 & 3 {bmatrix}

{bmatrix} 5 & -5 \\ -4 & 4 {bmatrix}

{bmatrix} 13 & 4 \\ 4 & 5 {bmatrix}

Correct Answer is : [5−5−44]{bmatrix} 5 & -5 \\ -4 & 4 {bmatrix}[5−4−54]

Test Index

UPSC NDA Math Model Paper 2

Question : 62 of 120

Marks: +1, -0

A = \begin{bmatrix} 1 & 5 \\ 4 & 2 \end{bmatrix},

A^{2} - 4A - 12I .

Solution:

Let us find the characteristic polynomial p(λ) of the given matrix

A = \begin{bmatrix} 1 & 5 \\ 4 & 2 \end{bmatrix}

p(\lambda) = \left| \lambda I - A \right|

\Rightarrow p(\lambda) = \begin{vmatrix} \lambda-1 & 5 \\ 4 & \lambda-2 \end{vmatrix}

\Rightarrow p(\lambda) = (\lambda-1)(\lambda-2) - (4)(5)

\Rightarrow p(\lambda) = \lambda^{2} - 2\lambda - \lambda + 2 - 20

\Rightarrow p(\lambda) = \lambda^{2} - 3\lambda - 18

According to the Cayley - Hamilton theorem,

p(A) = 0

\Rightarrow A^{2} - 3A - 18I = 0

Now, the expression given to evaluate is

A^{2} - 4A - 12I,

which can be written as:

(A^{2} - 3A - 18I) - A + 6I

= 0 - \begin{bmatrix} 1 & 5 \\ 4 & 2 \end{bmatrix} + \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}

= \begin{bmatrix} 5 & -5 \\ -4 & 4 \end{bmatrix}

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