Solveeit Logo
Login

Question

Question: If $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is a 2x2 matrix such that $A = \begin{bmatrix}...

If A=[abcd]A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} is a 2x2 matrix such that A=[1−12−1]A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} and A2=[1−1−10]A^2 = \begin{bmatrix} 1 & -1 \\ -1 & 0 \end{bmatrix}. Then the value of a + 2b + c + d is

Answer

0

Explanation

Solution

The problem asks for the value of the expression a+2b+c+da + 2b + c + d, given a 2x2 matrix A=[abcd]A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}.

The problem provides two pieces of information about matrix A:

  1. A=[1−12−1]A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}
  2. A2=[1−1−10]A^2 = \begin{bmatrix} 1 & -1 \\ -1 & 0 \end{bmatrix}

From the first piece of information, where AA is explicitly defined, we can directly determine the values of a,b,c,a, b, c, and dd by comparing the elements of the given matrix with the general form [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}:

a=1a = 1

b=−1b = -1

c=2c = 2

d=−1d = -1

Now, let's evaluate the expression a+2b+c+da + 2b + c + d using these values:

a+2b+c+d=1+2(−1)+2+(−1)a + 2b + c + d = 1 + 2(-1) + 2 + (-1)

=1−2+2−1= 1 - 2 + 2 - 1

=(1+2)−(2+1)= (1 + 2) - (2 + 1)

=3−3= 3 - 3

=0= 0

Note on the contradictory information:

The second piece of information, A2=[1−1−10]A^2 = \begin{bmatrix} 1 & -1 \\ -1 & 0 \end{bmatrix}, appears to be contradictory to the first definition of AA. Let's verify A2A^2 using the values of AA from the first definition:

A=[1−12−1]A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}

A2=A⋅A=[1−12−1][1−12−1]A^2 = A \cdot A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}

A2=[(1)(1)+(−1)(2)(1)(−1)+(−1)(−1)(2)(1)+(−1)(2)(2)(−1)+(−1)(−1)]A^2 = \begin{bmatrix} (1)(1) + (-1)(2) & (1)(-1) + (-1)(-1) \\ (2)(1) + (-1)(2) & (2)(-1) + (-1)(-1) \end{bmatrix}

A2=[1−2−1+12−2−2+1]A^2 = \begin{bmatrix} 1 - 2 & -1 + 1 \\ 2 - 2 & -2 + 1 \end{bmatrix}

A2=[−100−1]A^2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}

This calculated A2A^2 is [−100−1]\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}, which is not equal to the given A2=[1−1−10]A^2 = \begin{bmatrix} 1 & -1 \\ -1 & 0 \end{bmatrix}.

In such cases, when a variable is explicitly defined, that definition takes precedence. The additional, contradictory information is usually a distractor or an error in the question's premise. Therefore, we rely on the direct definition of AA to find a,b,c,da, b, c, d.