Let A = begin{bmatrix} 1 + x^2 - y^2 - z^2 & | vedclass

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(B) To find the determinant of matrix $A$,we can use the substitution method for an objective approach.Let $y = 0$ and $z = 0$.Then the matrix $A$ bec

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$(1 + x^2 + y^2 + z^2)^3$

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(B) To find the determinant of matrix $A$,we can use the substitution method for an objective approach.Let $y = 0$ and $z = 0$.Then the matrix $A$ becomes:$A = \begin{bmatrix} 1 + x^2 & 0 & 0 \ 0 & 1 - x^2 & 0 \ 0 & 0 & 1 - x^2 \end{bmatrix}$.The determinant of this diagonal matrix is $\det(A) = (1 + x^2)(1 - x^2)(1 - x^2) = (1 + x^2)(1 - x^2)^2$.Wait,let us re-evaluate the matrix with $y=0, z=0$ carefully:$A_{11} = 1 + x^2 - 0 - 0 = 1 + x^2$$A_{12} = 2(0 + 0) = 0$$A_{13} = 2(0 - 0) = 0$$A_{21} = 2(0 - 0) = 0$$A_{22} = 1 + 0 - 0 - x^2 = 1 - x^2$$A_{23} = 2(0 + x) = 2x$$A_{31} = 2(0 + 0) = 0$$A_{32} = 2(0 - x) = -2x$$A_{33} = 1 + 0 - x^2 - 0 = 1 - x^2$So,$\det(A) = (1 + x^2) \cdot [(1 - x^2)^2 - (-2x)(2x)] = (1 + x^2) \cdot [1 - 2x^2 + x^4 + 4x^2] = (1 + x^2)(1 + 2x^2 + x^4) = (1 + x^2)(1 + x^2)^2 = (1 + x^2)^3$.Comparing this with the given options by substituting $y=0, z=0$:$A: (1 + 0 + 0 + 0)^3 = 1$$B: (1 + x^2 + 0 + 0)^3 = (1 + x^2)^3$$C: (0 + 0 + 0)^3 = 0$$D: (1 + x^3 + 0 + 0)^2 = (1 + x^3)^2$Thus,the correct option is $B$.

Let $A = \begin{bmatrix} 1 + x^2 - y^2 - z^2 & 2(xy + z) & 2(zx - y) \\ 2(xy - z) & 1 + y^2 - z^2 - x^2 & 2(yz + x) \\ 2(zx + y) & 2(yz - x) & 1 + z^2 - x^2 - y^2 \end{bmatrix}$. Then $\det(A)$ is equal to:

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