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Mathematics Question on Determinants

Let
A=[2−1 02]\begin{bmatrix} 2 & -1 \\\ 0 & 2 \end{bmatrix}
If B = I – 5 C 1(adj A) + 5 C 2(adj A)2 – …. – 5 C 5(adj A)5, then the sum of all elements of the matrix B is

A

-5

B

-6

C

-7

D

-8

Answer

-7

Explanation

Solution

The correct answer is (C) : -7
Given A = [2−1 02]\begin{bmatrix} 2 & -1 \\\ 0 & 2 \end{bmatrix}
And B = I - 5C1(adj A) + 5C2(adj A)2 .... 5C5(adj A)5
⇒ B = ( I - adj A)5
Now , adj A = [21 02]\begin{bmatrix} 2 & 1 \\\ 0 & 2 \end{bmatrix}
∴B=∴ B = ([10 01]−[21 02])5\left( \begin{bmatrix} 1 & 0 \\\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 1 \\\ 0 & 2 \end{bmatrix} \right) ^5
⇒ B = [−1−1 01]5=−[11 01]5\begin{bmatrix} -1 & -1 \\\ 0 & 1 \end{bmatrix}^5 = - \begin{bmatrix} 1 & 1 \\\ 0 & 1 \end{bmatrix}^5
Now [11 01]2=[11 01][11 01]=[12 01]\begin{bmatrix} 1 & 1 \\\ 0 & 1 \end{bmatrix}^2 = \begin{bmatrix} 1 & 1 \\\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\\ 0 & 1 \end{bmatrix}
⇒ [11 01]3=[12 01][11 01]=[13 01]\begin{bmatrix} 1 & 1 \\\ 0 & 1 \end{bmatrix}^3 = \begin{bmatrix} 1 & 2 \\\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\\ 0 & 1 \end{bmatrix}
Similarly , [12 01]5=[15 01]\begin{bmatrix} 1 & 2 \\\ 0 & 1 \end{bmatrix}^5 = \begin{bmatrix} 1 & 5 \\\ 0 & 1 \end{bmatrix}
So, B = −[15 01]=[−1−5 0−1]- \begin{bmatrix} 1 & 5 \\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -1 & -5 \\\ 0 & -1 \end{bmatrix}
Therefore , Modulus of sum of all elements of Matrix B :
B=∣−1−5−1∣=−7B = | -1-5-1| = -7