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Question: Let $A = \begin{bmatrix} 1 & a & -1 \\ b & 1 & 2 \\ -2 & 1 & c \end{bmatrix}$ and $adj(A) = \begin{b...

Let A=[1a−1b12−21c]A = \begin{bmatrix} 1 & a & -1 \\ b & 1 & 2 \\ -2 & 1 & c \end{bmatrix} and adj(A)=[−2−11−4−2−22−11]adj(A) = \begin{bmatrix} -2 & -1 & 1 \\ -4 & -2 & -2 \\ 2 & -1 & 1 \end{bmatrix}, where a, b and c are real numbers. Then, which of the following option(s) is/are correct?

A

det(A) = 4

B

a + b + c = 0

C

A−1=[1214−1411212−1214−14]A^{-1} = \begin{bmatrix} \frac{1}{2} & \frac{1}{4} & -\frac{1}{4} \\ 1 & \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{4} & -\frac{1}{4} \end{bmatrix}

D

adj(A−1)=[−140140−14−1212−140]adj(A^{-1}) = \begin{bmatrix} -\frac{1}{4} & 0 & \frac{1}{4} \\ 0 & -\frac{1}{4} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{1}{4} & 0 \end{bmatrix}

Answer

B, C, D

Explanation

Solution

Given the matrix A=[1a−1b12−21c]A = \begin{bmatrix} 1 & a & -1 \\ b & 1 & 2 \\ -2 & 1 & c \end{bmatrix} and its adjoint adj(A)=[−2−11−4−2−22−11]adj(A) = \begin{bmatrix} -2 & -1 & 1 \\ -4 & -2 & -2 \\ 2 & -1 & 1 \end{bmatrix}.

We use the fundamental property A⋅adj(A)=det(A)⋅IA \cdot adj(A) = det(A) \cdot I, where II is the identity matrix.

  1. Calculate A⋅adj(A)A \cdot adj(A): A⋅adj(A)=[1a−1b12−21c][−2−11−4−2−22−11]A \cdot adj(A) = \begin{bmatrix} 1 & a & -1 \\ b & 1 & 2 \\ -2 & 1 & c \end{bmatrix} \begin{bmatrix} -2 & -1 & 1 \\ -4 & -2 & -2 \\ 2 & -1 & 1 \end{bmatrix}

    The elements of the product matrix are: (A⋅adj(A))11=(1)(−2)+(a)(−4)+(−1)(2)=−2−4a−2=−4−4a(A \cdot adj(A))_{11} = (1)(-2) + (a)(-4) + (-1)(2) = -2 - 4a - 2 = -4 - 4a (A⋅adj(A))12=(1)(−1)+(a)(−2)+(−1)(−1)=−1−2a+1=−2a(A \cdot adj(A))_{12} = (1)(-1) + (a)(-2) + (-1)(-1) = -1 - 2a + 1 = -2a (A⋅adj(A))13=(1)(1)+(a)(−2)+(−1)(1)=1−2a−1=−2a(A \cdot adj(A))_{13} = (1)(1) + (a)(-2) + (-1)(1) = 1 - 2a - 1 = -2a (A⋅adj(A))21=(b)(−2)+(1)(−4)+(2)(2)=−2b−4+4=−2b(A \cdot adj(A))_{21} = (b)(-2) + (1)(-4) + (2)(2) = -2b - 4 + 4 = -2b (A⋅adj(A))22=(b)(−1)+(1)(−2)+(2)(−1)=−b−2−2=−b−4(A \cdot adj(A))_{22} = (b)(-1) + (1)(-2) + (2)(-1) = -b - 2 - 2 = -b - 4 (A⋅adj(A))23=(b)(1)+(1)(−2)+(2)(1)=b−2+2=b(A \cdot adj(A))_{23} = (b)(1) + (1)(-2) + (2)(1) = b - 2 + 2 = b (A⋅adj(A))31=(−2)(−2)+(1)(−4)+(c)(2)=4−4+2c=2c(A \cdot adj(A))_{31} = (-2)(-2) + (1)(-4) + (c)(2) = 4 - 4 + 2c = 2c (A⋅adj(A))32=(−2)(−1)+(1)(−2)+(c)(−1)=2−2−c=−c(A \cdot adj(A))_{32} = (-2)(-1) + (1)(-2) + (c)(-1) = 2 - 2 - c = -c (A⋅adj(A))33=(−2)(1)+(1)(−2)+(c)(1)=−2−2+c=−4+c(A \cdot adj(A))_{33} = (-2)(1) + (1)(-2) + (c)(1) = -2 - 2 + c = -4 + c

    So, A⋅adj(A)=[−4−4a−2a−2a−2b−b−4b2c−c−4+c]A \cdot adj(A) = \begin{bmatrix} -4 - 4a & -2a & -2a \\ -2b & -b - 4 & b \\ 2c & -c & -4 + c \end{bmatrix}

  2. Determine det(A)det(A) and values of a,b,ca, b, c:

    Comparing A⋅adj(A)A \cdot adj(A) with det(A)⋅I=[det(A)000det(A)000det(A)]det(A) \cdot I = \begin{bmatrix} det(A) & 0 & 0 \\ 0 & det(A) & 0 \\ 0 & 0 & det(A) \end{bmatrix}:

    From element (1,2)(1,2): −2a=0  ⟹  a=0-2a = 0 \implies a = 0. From element (1,1)(1,1): −4−4a=det(A)-4 - 4a = det(A). Substituting a=0a=0, we get −4−4(0)=det(A)  ⟹  det(A)=−4-4 - 4(0) = det(A) \implies det(A) = -4. This means option A (det(A)=4det(A) = 4) is incorrect.

    Now, substitute a=0a=0 and det(A)=−4det(A)=-4 into the product matrix and equate it to det(A)⋅Idet(A) \cdot I: [−400−2b−b−4b2c−c−4+c]=[−4000−4000−4]\begin{bmatrix} -4 & 0 & 0 \\ -2b & -b - 4 & b \\ 2c & -c & -4 + c \end{bmatrix} = \begin{bmatrix} -4 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & -4 \end{bmatrix}

    From element (2,1)(2,1): −2b=0  ⟹  b=0-2b = 0 \implies b = 0. From element (3,1)(3,1): 2c=0  ⟹  c=02c = 0 \implies c = 0.

    Let's verify with other diagonal elements: −b−4=−0−4=−4-b - 4 = -0 - 4 = -4 (matches det(A)det(A)). −4+c=−4+0=−4-4 + c = -4 + 0 = -4 (matches det(A)det(A)). All values are consistent. So, a=0,b=0,c=0a=0, b=0, c=0.

  3. Check Option B: a+b+c=0a+b+c = 0

    a+b+c=0+0+0=0a+b+c = 0+0+0 = 0. Therefore, option B is correct.

  4. Check Option C: A−1A^{-1}

    We know A−1=1det(A)adj(A)A^{-1} = \frac{1}{det(A)} adj(A). A−1=1−4[−2−11−4−2−22−11]=[−2−4−1−41−4−4−4−2−4−2−42−4−1−41−4]A^{-1} = \frac{1}{-4} \begin{bmatrix} -2 & -1 & 1 \\ -4 & -2 & -2 \\ 2 & -1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{-2}{-4} & \frac{-1}{-4} & \frac{1}{-4} \\ \frac{-4}{-4} & \frac{-2}{-4} & \frac{-2}{-4} \\ \frac{2}{-4} & \frac{-1}{-4} & \frac{1}{-4} \end{bmatrix} A−1=[1214−1411212−1214−14]A^{-1} = \begin{bmatrix} \frac{1}{2} & \frac{1}{4} & -\frac{1}{4} \\ 1 & \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{4} & -\frac{1}{4} \end{bmatrix} Therefore, option C is correct.

  5. Check Option D: adj(A−1)adj(A^{-1})

    We use the property adj(A−1)=1det(A)Aadj(A^{-1}) = \frac{1}{det(A)} A. First, write down matrix A with the found values a=0,b=0,c=0a=0, b=0, c=0: A=[10−1012−210]A = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ -2 & 1 & 0 \end{bmatrix} adj(A−1)=1−4[10−1012−210]=[−140140−14−2424−140]adj(A^{-1}) = \frac{1}{-4} \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ -2 & 1 & 0 \end{bmatrix} = \begin{bmatrix} -\frac{1}{4} & 0 & \frac{1}{4} \\ 0 & -\frac{1}{4} & -\frac{2}{4} \\ \frac{2}{4} & -\frac{1}{4} & 0 \end{bmatrix} adj(A−1)=[−140140−14−1212−140]adj(A^{-1}) = \begin{bmatrix} -\frac{1}{4} & 0 & \frac{1}{4} \\ 0 & -\frac{1}{4} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{1}{4} & 0 \end{bmatrix} Therefore, option D is correct.

The final answer is B,C,D\boxed{B, C, D}