Question

If \(A^{-1}=\begin{bmatrix} 1 & 0 & -4 \\ 0 & -1 & 2 \\ -1 & 2 & 1 \end{bmatrix}=\frac{\text{adj}(A)}{k}\), then \(k\) is a root of the equation...

• A. \(x^{2}+1=0\)
• B. \(x^{2}-1=0\)
• C. \(x^{2}-2=0\)
• D. \(x^{2}-x+1=0\)

Hint:

Since the inverse equals adjoint over determinant, k equals the determinant of A, which is the reciprocal of the determinant of the inverse.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
We know \(A^{-1}=\dfrac{\text{adj}(A)}{|A|}\). So \(k=|A|\).

Step 2: Use the Determinant Property
\(|A^{-1}|=\dfrac{1}{|A|}\), so \(k=\dfrac{1}{|A^{-1}|}\).

Step 3: Calculation
\(|A^{-1}|=1\,[(-1)(1)-(2)(2)]-0+(-4)\,[0\cdot 2-(-1)(-1)]\)
\(=1(-5)+(-4)(-1)=-1\).

Step 4: Find k
\(k=\dfrac{1}{-1}=-1\).
Check the options: \(x^{2}-1=0\) has the root \(-1\).

Final Answer:
k equals minus one, a root of x squared minus one equal to zero. \[ \boxed{\text{(B) }x^{2}-1=0} \]