Identity Matrix

Solution :

Given,

$$A=\begin{bmatrix}1 & 2 \\3 & 4\end{bmatrix}$$

First calculate

$$A^2$$

$$A^2=\begin{bmatrix}1 & 2 \\3 & 4\end{bmatrix}\begin{bmatrix}1 & 2 \\3 & 4\end{bmatrix}$$

$$=\begin{bmatrix}7 & 10 \\15 & 22\end{bmatrix}$$

Now,
$$5A=\begin{bmatrix}5 & 10 \\15 & 20\end{bmatrix}$$

Therefore,

$$A^2-5A=\begin{bmatrix}7 & 10 \\15 & 22\end{bmatrix} - \begin{bmatrix}5 & 10 \\15 & 20\end{bmatrix}$$

$$=\begin{bmatrix}2 & 0 \\0 & 2\end{bmatrix}$$

Given,

$$A^2-5A+kI=0$$

Therefore,

$$2I+kI=0$$

$$(2+k)I=0$$

$$2+k=0$$

$$k=-2$$

Final Answer :

$$-2$$

To find the least value of $$\det(A)$$, we use the properties of skew-symmetric matrices and substitute them into the updated matrix equation.

Step 1: Understand the property of matrix P

The given condition is:

$$P + P^T = 0 \implies P^T = -P$$

This means $$P$$ is a skew-symmetric matrix of order 2. Any real skew-symmetric matrix of order 2 can be expressed as:

$$P = \begin{pmatrix} 0 & k \\ -k & 0 \end{pmatrix}$$

where $$k$$ is a real number.

Step 2: Express matrix A in terms of P

From the second equation, isolate matrix $$A$$:

$$A + 2P = 3I$$

$$A = 3I - 2P$$

Step 3: Construct matrix A

Substitute the identity matrix $$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ and the matrix $$P$$ into the expression for $$A$$:

$$A = 3\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - 2\begin{pmatrix} 0 & k \\ -k & 0 \end{pmatrix}$$

$$A = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} - \begin{pmatrix} 0 & 2k \\ -2k & 0 \end{pmatrix}$$

$$A = \begin{pmatrix} 3 & -2k \\ 2k & 3 \end{pmatrix}$$

Step 4: Calculate the determinant of A

Evaluate $$\det(A)$$:

$$\det(A) = (3)(3) - (-2k)(2k)$$

$$\det(A) = 9 + 4k^2$$

Step 5: Find the minimum value

Since $$k$$ is a real number, its square is always non-negative ($$k^2 \ge 0$$), which means:

$$4k^2 \ge 0$$

Therefore:

$$\det(A) \ge 9$$

The least value of $$\det(A)$$ is $$9$$, which occurs when $$k = 0$$

Given the equation:

$$P^T = kP + (k^2 - 1)I \quad \dots \text{(1)}$$

Taking the transpose on both sides of equation (1):

$$(P^T)^T = (kP + (k^2 - 1)I)^T$$ $$P = kP^T + (k^2 - 1)I \quad \dots \text{(2)}$$

Substitute the value of $$P^T$$ from equation (1) into equation (2):

$$P = k[kP + (k^2 - 1)I] + (k^2 - 1)I$$ $$P = k^2 P + k(k^2 - 1)I + (k^2 - 1)I$$

Rearranging the terms to isolate $$P$$:

$$P - k^2 P = [k(k^2 - 1) + (k^2 - 1)]I$$ $$P(1 - k^2) = (k^2 - 1)(k + 1)I$$

To solve for $$P$$, we can rewrite $$(1 - k^2)$$ as $$-(k^2 - 1)$$:

$$-P(k^2 - 1) = (k^2 - 1)(k + 1)I$$

Dividing by $$(k^2 - 1)$$ (as $$k \neq 1$$ and $$P$$ is non-singular):

$$-P = (k + 1)I \implies P = -(k + 1)I$$

Since $$P$$ is a $$3 \times 3$$ matrix, we use the property $$|cA| = c^n |A|$$:

$$|P| = |-(k + 1)I|$$ $$|P| = [-(k + 1)]^3 |I|$$ $$|P| = -(k + 1)^3$$

Using the determinant property of adjoints for an $$n \times n$$ matrix, $$|\text{adj}(P)| = |P|^{n-1}$$. For $$n = 3$$:

$$|\text{adj}(P)| = |P|^2$$ $$|\text{adj}(P)| = [-(k + 1)^3]^2$$ $$|\text{adj}(P)| = (k + 1)^6$$

Since $$(k + 1)^6$$ is the same as $$(1 + k)^6$$, we reach our final result.

Correct Option: A