Question

given ( a = \begin{bmatrix} -1 & -2 & 3 \\ 4 & 8 & -6 \\ 0 & 1 & 5 end{bmatrix} ) and ( b = \begin{bmatrix} -5 & -1 & 6 \\ 4 & 1 & 2 \\ 0 & -3 & 2 end{bmatrix} ). what is ( x ) if ( b - x = a )?
( \begin{bmatrix} 4 & -1 & -3 \\ 0 & 7 & -8 \\ 0 & 4 & 3 end{bmatrix} ) (selected with a blue dot)
( \begin{bmatrix} -4 & 1 & 3 \\ 0 & -7 & 8 \\ 0 & -4 & -3 end{bmatrix} ) (unselected with a circle)

Explanation:

Step1: Rearrange the equation

From \( B - X = A \), we can rearrange it to solve for \( X \). Subtract \( A \) from both sides and add \( X \) to both sides, we get \( X = B - A \).

Step2: Subtract the matrices

To subtract two matrices, we subtract their corresponding elements.

For the first row, first column: \( -5 - (-1) = -5 + 1 = -4 \)? Wait, no, wait. Wait, \( X = B - A \), so each element \( X_{ij} = B_{ij} - A_{ij} \).

Let's compute each element:

First row:

• Column 1: \( B_{11} - A_{11} = -5 - (-1) = -5 + 1 = -4 \)? Wait, no, wait the first option has 4 in first row first column. Wait, maybe I made a mistake. Wait, the equation is \( B - X = A \), so \( X = B - A \). Wait, let's check the first option. Wait, maybe I messed up the rearrangement. Let's re-arrange: \( B - X = A \) => \( -X = A - B \) => \( X = B - A \). Yes, that's correct.

Wait, let's compute \( B - A \):

Matrix \( A \):
\[

$$\begin{bmatrix} -1 & -2 & 3 \\ 4 & 8 & -6 \\ 0 & 1 & 5 \end{bmatrix}$$

\]

Matrix \( B \):
\[

$$\begin{bmatrix} -5 & -1 & 6 \\ 4 & 1 & 2 \\ 0 & -3 & 2 \end{bmatrix}$$

\]

Now, subtract \( A \) from \( B \) (element-wise):

First row:

• Column 1: \( -5 - (-1) = -5 + 1 = -4 \)? Wait, no, the first option's first row first column is 4. Wait, maybe the equation is \( B - X = A \), so \( X = B - A \). Wait, let's check the second option. Wait, maybe I made a mistake in the sign. Wait, let's re-arrange the equation again. \( B - X = A \) => \( B - A = X \). Wait, let's compute each element:

First row, first column: \( B_{11} - A_{11} = -5 - (-1) = -4 \)

First row, second column: \( B_{12} - A_{12} = -1 - (-2) = -1 + 2 = 1 \)

First row, third column: \( B_{13} - A_{13} = 6 - 3 = 3 \)

Second row, first column: \( B_{21} - A_{21} = 4 - 4 = 0 \)

Second row, second column: \( B_{22} - A_{22} = 1 - 8 = -7 \)

Second row, third column: \( B_{23} - A_{23} = 2 - (-6) = 2 + 6 = 8 \)

Third row, first column: \( B_{31} - A_{31} = 0 - 0 = 0 \)

Third row, second column: \( B_{32} - A_{32} = -3 - 1 = -4 \)

Third row, third column: \( B_{33} - A_{33} = 2 - 5 = -3 \)

So the matrix \( X \) is:
\[

$$\begin{bmatrix} -4 & 1 & 3 \\ 0 & -7 & 8 \\ 0 & -4 & -3 \end{bmatrix}$$

\]
Which matches the second option. Wait, but the first option was selected. Wait, maybe I made a mistake in the rearrangement. Wait, let's check the equation again. \( B - X = A \) => \( X = B - A \). Wait, let's verify with the first option. Let's take the first option matrix as \( X \) and check if \( B - X = A \).

First option \( X \):
\[

$$\begin{bmatrix} 4 & -1 & -3 \\ 0 & 7 & -8 \\ 0 & 4 & 3 \end{bmatrix}$$

\]

Compute \( B - X \):

First row, first column: \( -5 - 4 = -9 \), but \( A_{11} = -1 \). Not equal. So first option is wrong.

Second option \( X \):
\[

$$\begin{bmatrix} -4 & 1 & 3 \\ 0 & -7 & 8 \\ 0 & -4 & -3 \end{bmatrix}$$

\]

Compute \( B - X \):

First row, first column: \( -5 - (-4) = -5 + 4 = -1 \) (matches \( A_{11} = -1 \))

First row, second column: \( -1 - 1 = -2 \) (matches \( A_{12} = -2 \))

First row, third column: \( 6 - 3 = 3 \) (matches \( A_{13} = 3 \))

Second row, first column: \( 4 - 0 = 4 \) (matches \( A_{21} = 4 \))

Second row, second column: \( 1 - (-7) = 1 + 7 = 8 \) (matches \( A_{22} = 8 \))

Second row, third column: \( 2 - 8 = -6 \) (matches \( A_{23} = -6 \))

Third row, first column: \( 0 - 0 = 0 \) (matches \( A_{31} = 0 \))

Third row, second column: \( -3 - (-4) = -3 + 4 = 1 \) (matches \( A_{32} = 1 \))

Third row, third column: \( 2 - (-3) = 2 + 3 = 5 \) (matches \( A_{33} = 5 \))

Yes, so the correct \( X \) is the second option.

Answer:

The correct option is the second matrix: \(

$$\begin{bmatrix}-4 & 1 & 3 \\ 0 & -7 & 8 \\ 0 & -4 & -3\end{bmatrix}$$

\) (the one with the radio button not selected initially). So the answer is the second option, which can be represented as:

\(\boldsymbol{

$$\begin{bmatrix}-4 & 1 & 3 \\ 0 & -7 & 8 \\ 0 & -4 & -3\end{bmatrix}$$

}\) (or the option with this matrix, likely the second choice in the list)