Question

If $A = \begin{bmatrix} 1 & 1 & 0 \\ 2 & 2 & 1 \\ 0 & 1 & -1 \end{bmatrix}$ and $A^3 - 2A^2 + \lambda A + I = \mathbf{O}$ (Null matrix), then-

• A. $\lambda = 4$
• B. $\lambda = -4$
• C. $|A^2 + A^{-1}| = 56$
• D. $|A^2 + A^{-1}| = 14$

Answer 1

Answer: (B), (C)

(B), (C)

Answer 2

The given matrix is $A = \begin{bmatrix} 1 & 1 & 0 \\ 2 & 2 & 1 \\ 0 & 1 & -1 \end{bmatrix}$. The given matrix equation is $A^3 - 2A^2 + \lambda A + I = \mathbf{O}$.

According to the Cayley-Hamilton theorem, every square matrix satisfies its own characteristic equation. The characteristic equation of matrix A is given by $|A - xI| = 0$. $A - xI = \begin{bmatrix} 1-x & 1 & 0 \\ 2 & 2-x & 1 \\ 0 & 1 & -1-x \end{bmatrix}$

The determinant $|A - xI|$ is: $|A - xI| = (1-x)[(2-x)(-1-x) - 1(1)] - 1[2(-1-x) - 1(0)] + 0[\dots]$ $= (1-x)[-2 - 2x + x + x^2 - 1] - 1[-2 - 2x]$ $= (1-x)[x^2 - x - 3] + 2 + 2x$ $= x^2 - x - 3 - x^3 + x^2 + 3x + 2 + 2x$ $= -x^3 + 2x^2 + 4x - 1$

The characteristic equation is $-x^3 + 2x^2 + 4x - 1 = 0$, or $x^3 - 2x^2 - 4x + 1 = 0$.

By the Cayley-Hamilton theorem, the matrix A satisfies this equation: $A^3 - 2A^2 - 4A + I = \mathbf{O}$

We are given the equation $A^3 - 2A^2 + \lambda A + I = \mathbf{O}$. Comparing the two equations, we have: $(A^3 - 2A^2 - 4A + I) - (A^3 - 2A^2 + \lambda A + I) = \mathbf{O} - \mathbf{O}$ $(-4A) - (\lambda A) = \mathbf{O}$ $(-4 - \lambda)A = \mathbf{O}$

To find $\lambda$, we need to check if A is the null matrix. A is clearly not the null matrix. We can also check the determinant of A to see if it is invertible. $|A| = \begin{vmatrix} 1 & 1 & 0 \\ 2 & 2 & 1 \\ 0 & 1 & -1 \end{vmatrix} = 1(2 \times -1 - 1 \times 1) - 1(2 \times -1 - 1 \times 0) + 0 = 1(-2-1) - 1(-2) = -3 + 2 = -1$. Since $|A| = -1 \neq 0$, A is invertible.

Since $(-4 - \lambda)A = \mathbf{O}$ and A is invertible, we can multiply by $A^{-1}$ from the right: $(-4 - \lambda)A A^{-1} = \mathbf{O} A^{-1}$ $(-4 - \lambda)I = \mathbf{O}$ $-4 - \lambda = 0$ $\lambda = -4$.

Thus, option (B) is correct, and option (A) is incorrect.

Now let's evaluate $|A^2 + A^{-1}|$. From the characteristic equation $A^3 - 2A^2 - 4A + I = \mathbf{O}$, we can find an expression for $A^{-1}$. Since A is invertible, multiply the equation by $A^{-1}$: $A^{-1}(A^3 - 2A^2 - 4A + I) = A^{-1}\mathbf{O}$ $A^{-1}A^3 - 2A^{-1}A^2 - 4A^{-1}A + A^{-1}I = \mathbf{O}$ $A^2 - 2A - 4I + A^{-1} = \mathbf{O}$ $A^{-1} = -A^2 + 2A + 4I$

Now, consider the expression $A^2 + A^{-1}$: $A^2 + A^{-1} = A^2 + (-A^2 + 2A + 4I) = 2A + 4I$

We need to calculate the determinant of $2A + 4I$. $2A = 2 \begin{bmatrix} 1 & 1 & 0 \\ 2 & 2 & 1 \\ 0 & 1 & -1 \end{bmatrix} = \begin{bmatrix} 2 & 2 & 0 \\ 4 & 4 & 2 \\ 0 & 2 & -2 \end{bmatrix}$ $4I = 4 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}$ $2A + 4I = \begin{bmatrix} 2 & 2 & 0 \\ 4 & 4 & 2 \\ 0 & 2 & -2 \end{bmatrix} + \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} 2+4 & 2+0 & 0+0 \\ 4+0 & 4+4 & 2+0 \\ 0+0 & 2+0 & -2+4 \end{bmatrix} = \begin{bmatrix} 6 & 2 & 0 \\ 4 & 8 & 2 \\ 0 & 2 & 2 \end{bmatrix}$

Now, calculate the determinant of this matrix: $|2A + 4I| = \begin{vmatrix} 6 & 2 & 0 \\ 4 & 8 & 2 \\ 0 & 2 & 2 \end{vmatrix}$ Expand along the first row: $= 6 \begin{vmatrix} 8 & 2 \\ 2 & 2 \end{vmatrix} - 2 \begin{vmatrix} 4 & 2 \\ 0 & 2 \end{vmatrix} + 0 \begin{vmatrix} 4 & 8 \\ 0 & 2 \end{vmatrix}$ $= 6((8)(2) - (2)(2)) - 2((4)(2) - (0)(2)) + 0$ $= 6(16 - 4) - 2(8 - 0)$ $= 6(12) - 2(8)$ $= 72 - 16 = 56$

So, $|A^2 + A^{-1}| = 56$. Option (C) $|A^2 + A^{-1}| = 56$ is correct, and option (D) $|A^2 + A^{-1}| = 14$ is incorrect.

The question asks "then-", followed by options, suggesting that multiple options might be correct. Based on our calculations, options (B) and (C) are correct.