Question

Let A, B and C be 3 square matrices of order 3 × 3. If $A^3 - 6A^2 + 7A + B = 0$ and $(adj(adj B)) = C$, where $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$, then $(\frac{\text{sum of the elements of C}}{10})$ is

• A. 2.4
• B. 4.8
• C. 3.2
• D. 6.4

Answer 1

Answer: (B)

4.8

Answer 2

The given matrix is $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$. The order of the matrix A is 3x3.

We are given the equation $A^3 - 6A^2 + 7A + B = 0$.

We first find the characteristic equation of matrix A using $|A - \lambda I| = 0$.

$A - \lambda I = \begin{bmatrix} 1-\lambda & 0 & 2 \\ 0 & 2-\lambda & 1 \\ 2 & 0 & 3-\lambda \end{bmatrix}$

$|A - \lambda I| = (1-\lambda) \begin{vmatrix} 2-\lambda & 1 \\ 0 & 3-\lambda \end{vmatrix} - 0 \begin{vmatrix} 0 & 1 \\ 2 & 3-\lambda \end{vmatrix} + 2 \begin{vmatrix} 0 & 2-\lambda \\ 2 & 0 \end{vmatrix}$

$|A - \lambda I| = (1-\lambda)[(2-\lambda)(3-\lambda) - 0] + 2[0 - 2(2-\lambda)]$

$|A - \lambda I| = (1-\lambda)(2-\lambda)(3-\lambda) - 4(2-\lambda)$

$|A - \lambda I| = (2-\lambda)[(1-\lambda)(3-\lambda) - 4]$

$|A - \lambda I| = (2-\lambda)[3 - \lambda - 3\lambda + \lambda^2 - 4]$

$|A - \lambda I| = (2-\lambda)[\lambda^2 - 4\lambda - 1]$

$|A - \lambda I| = 2\lambda^2 - 8\lambda - 2 - \lambda^3 + 4\lambda^2 + \lambda$

$|A - \lambda I| = -\lambda^3 + 6\lambda^2 - 7\lambda - 2$.

The characteristic equation is $-\lambda^3 + 6\lambda^2 - 7\lambda - 2 = 0$, or $\lambda^3 - 6\lambda^2 + 7\lambda + 2 = 0$.

By the Cayley-Hamilton theorem, every square matrix satisfies its own characteristic equation. Thus, $A^3 - 6A^2 + 7A + 2I = 0$, where I is the identity matrix of order 3.

We are given the equation $A^3 - 6A^2 + 7A + B = 0$.

Comparing this equation with the Cayley-Hamilton equation $A^3 - 6A^2 + 7A + 2I = 0$, we can see that $B = 2I$.

So, $B = 2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$.

We are given $C = adj(adj B)$.

For a square matrix M of order n, the property $adj(adj M) = |M|^{n-2} M$ holds.

In this case, M is B and n is 3.

So, $C = adj(adj B) = |B|^{3-2} B = |B| B$.

First, let's calculate the determinant of B.

$|B| = \begin{vmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{vmatrix}$. Since B is a diagonal matrix, its determinant is the product of its diagonal elements.

$|B| = 2 \times 2 \times 2 = 8$.

Now, we can find C.

$C = |B| B = 8 B = 8 \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} = \begin{bmatrix} 16 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 16 \end{bmatrix}$.

We need to find the sum of the elements of C.

Sum of elements of C = $16 + 0 + 0 + 0 + 16 + 0 + 0 + 0 + 16 = 16 \times 3 = 48$.

Finally, we need to calculate $(\frac{\text{sum of the elements of C}}{10})$.

$(\frac{\text{sum of the elements of C}}{10}) = \frac{48}{10} = 4.8$.