Question at position 11 If 4[1x−20]+2[−20y0]=[0000]4 \begin{bmatrix} 1 & x \ -2 & 0 \end{bmatrix} +
2 \begin{bmatrix} -2 & 0 \ y & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} , thenx = 4 and y = -5.There are no values for x and y which satisfy the equation.x = 4 and y = 0.x = 0 and y = 4.x = 0 and y = 0.

Question

Question at position 11 If 4[1x−20]+2[−20y0]=[0000]4 \begin{bmatrix} 1 & x  \ -2 & 0 \end{bmatrix} +
2 \begin{bmatrix} -2 & 0 \ y & 0  \end{bmatrix} =  \begin{bmatrix} 0 & 0  \ 0 & 0  \end{bmatrix} , thenx = 4 and y = -5.There are no values for x and y which satisfy the equation.x = 4 and y = 0.x = 0 and y = 4.x = 0 and y = 0.

BlackTom AI · Accepted Answer

Start by restating what the problem is asking: we have a matrix equation where a linear combination of two 2x2 matrices must equal the zero matrix, and we need to determine the values of x and y that satisfy it.
Option 1: 'x = 4 and y = -5.' To check this, plug these into the constructed matrix expression step by step. Compute 4 times the first matrix: 4[[1, x],[-2,0]] becomes [[4, 4x],[-8,0]]; with x = 4 that gives [[4,16],[-8,0]]. Next, compute 2 times the second matrix: 2[[-2,0],[y,0]] becomes [[-4,0],[2y,0]]; with y = -5 this yields [[-4,0],[-10,0]]. Adding the two results: [[4-4, 16+0],[-8-10, 0+0]] = [[0,16],[-18,0]]. This is not the zero matrix, since the top-right entry is 16 and the bottom-left is -18, so this option does not satisfy the equation. 
Option 2: 'There are no values for x and y which satisfy the equation.' After performing the addition of the two scaled matrices, it is possible to set up equations from the corresponding entries to enforce zero values. Since at least one entry (the top-right) can be forced to zero by choosing x appropriately, and another entry (the bottom-left) can be forced to zero by choosing y appropriately, this option claims impossibility. The explicit entry equations show that zeros can be achieved, so this blanket impossibility statement is unlikely to be correct. 
Option 3: 'x = 4 and y = 0.' Compute with x = 4: the top-right entry becomes 4x = 16, which is not zero, so the top-right entry fails to be zero. Therefore this pair does not satisfy the zero-matrix condition. 
Option 4: 'x = 0 and y = 4.' With x = 0, the top-right entry is 4x = 0, satisfying that position. With y = 4, the bottom-left entry is -8 + 2y = -8 + 8 = 0, satisfying that position as well. Since both nonzero entries can be made zero simultaneously, this pair satisfies the matrix equation. 
Option 5: 'x = 0 and y = 0.' Here, the top-right entry is 4x = 0, which is good, but the bottom-left entry becomes -8 + 2y = -8 + 0 = -8, which is not zero. Thus this pair fails to yield the zero matrix. 
In summary, each option changes particular entries of the resulting matrix. The key is to enforce zero on each nonzero position: the top-right entry gives the equation 4x = 0, so x must be 0; the bottom-left entry gives -8 + 2y = 0, so y must be 4. Only the option that matches x = 0 and y = 4 satisfies both conditions. The remaining options either misstate the solvability or propose values that do not zero out the necessary entries, leading to a nonzero matrix in at least one position.

类似问题

如果 \[ \left[\begin{matrix}1 & 1\\3 & -3\end{matrix}\right ] X + \left [\begin{matrix}0 & 0\\4 & -4\end{matrix}\right ] = \left [\begin{matrix}-4 & -3\\1 & 0\end{matrix}\right ] X， \] 以下哪一个矩阵是 \（X\）？

如果， [ 1 1 3 − 3 ] 𝑋 + [ 0 0 4 − 4 ] = [ − 4 − 3 1 0 ] 𝑋 ，以下哪一个矩阵是 \（X\）？

If [ − 5 4 0 0 ] 𝑋 + [ 1 1 − 5 − 4 ] = [ 4 3 3 4 ] 𝑋 , which of the following matrices is 𝑋 ?

If [ − 1 1 5 5 ] 𝑋 + [ 4 − 1 2 − 5 ] = [ 1 0 1 − 4 ] 𝑋 , which of the following matrices is 𝑋 ?

If [−4 3 −3 1]X+[2 −3 1 −5]=[−1 3 −2 4]X, which of the following matrices is X?

If \[ \left [\begin{matrix}0 & 0\\0 & -2\end{matrix}\right ] X + \left [\begin{matrix}-2 & 0\\2 & -3\end{matrix}\right ] = \left [\begin{matrix}-3 & 1\\-3 & 0\end{matrix}\right ] X, \] which of the following matrices is \(X\)?

If [ − 4 3 − 3 1 ] 𝑋 + [ 2 − 3 1 − 5 ] = [ − 1 3 − 2 4 ] 𝑋 , which of the following matrices is 𝑋 ?

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