Answer :
To find the missing number in the equation, we need to follow these steps:
Step 1: Multiply the matrix by 3 and 5 respectively.
Given matrix:
[tex]\[ \begin{bmatrix} -1 & 2 \\ 4 & -5 \end{bmatrix} \][/tex]
First, multiply by 3:
[tex]\[ 3 \times \begin{bmatrix} -1 & 2 \\ 4 & -5 \end{bmatrix} = \begin{bmatrix} 3 \times -1 & 3 \times 2 \\ 3 \times 4 & 3 \times -5 \end{bmatrix} = \begin{bmatrix} -3 & 6 \\ 12 & -15 \end{bmatrix} \][/tex]
Next, multiply by 5:
[tex]\[ 5 \times \begin{bmatrix} -1 & 2 \\ 4 & -5 \end{bmatrix} = \begin{bmatrix} 5 \times -1 & 5 \times 2 \\ 5 \times 4 & 5 \times -5 \end{bmatrix} = \begin{bmatrix} -5 & 10 \\ 20 & -25 \end{bmatrix} \][/tex]
Step 2: Add the resulting matrices.
Add:
[tex]\[ \begin{bmatrix} -3 & 6 \\ 12 & -15 \end{bmatrix} + \begin{bmatrix} -5 & 10 \\ 20 & -25 \end{bmatrix} = \begin{bmatrix} -3 + -5 & 6 + 10 \\ 12 + 20 & -15 + -25 \end{bmatrix} = \begin{bmatrix} -8 & 16 \\ 32 & -40 \end{bmatrix} \][/tex]
Step 3: Determine the scalar multiple.
The resulting matrix [tex]\(\begin{bmatrix} -8 & 16 \\ 32 & -40 \end{bmatrix} \)[/tex] is a scalar multiple of the original matrix [tex]\(\begin{bmatrix} -1 & 2 \\ 4 & -5 \end{bmatrix} \)[/tex].
To find the scalar, we compare the corresponding elements of the two matrices:
[tex]\[ \frac{-8}{-1} = 8, \quad \frac{16}{2} = 8, \quad \frac{32}{4} = 8, \quad \frac{-40}{-5} = 8 \][/tex]
In all cases, the scalar multiple is [tex]\( 8 \)[/tex].
Therefore, the missing number in the equation is [tex]\( 8 \)[/tex]:
[tex]\[ 3\left[\begin{array}{cc} -1 & 2 \\ 4 & -5 \end{array}\right] + 5\left[\begin{array}{cc} -1 & 2 \\ 4 & -5 \end{array}\right] = 8 \cdot \left[\begin{array}{cc} -1 & 2 \\ 4 & -5 \end{array}\right] \][/tex]
Step 1: Multiply the matrix by 3 and 5 respectively.
Given matrix:
[tex]\[ \begin{bmatrix} -1 & 2 \\ 4 & -5 \end{bmatrix} \][/tex]
First, multiply by 3:
[tex]\[ 3 \times \begin{bmatrix} -1 & 2 \\ 4 & -5 \end{bmatrix} = \begin{bmatrix} 3 \times -1 & 3 \times 2 \\ 3 \times 4 & 3 \times -5 \end{bmatrix} = \begin{bmatrix} -3 & 6 \\ 12 & -15 \end{bmatrix} \][/tex]
Next, multiply by 5:
[tex]\[ 5 \times \begin{bmatrix} -1 & 2 \\ 4 & -5 \end{bmatrix} = \begin{bmatrix} 5 \times -1 & 5 \times 2 \\ 5 \times 4 & 5 \times -5 \end{bmatrix} = \begin{bmatrix} -5 & 10 \\ 20 & -25 \end{bmatrix} \][/tex]
Step 2: Add the resulting matrices.
Add:
[tex]\[ \begin{bmatrix} -3 & 6 \\ 12 & -15 \end{bmatrix} + \begin{bmatrix} -5 & 10 \\ 20 & -25 \end{bmatrix} = \begin{bmatrix} -3 + -5 & 6 + 10 \\ 12 + 20 & -15 + -25 \end{bmatrix} = \begin{bmatrix} -8 & 16 \\ 32 & -40 \end{bmatrix} \][/tex]
Step 3: Determine the scalar multiple.
The resulting matrix [tex]\(\begin{bmatrix} -8 & 16 \\ 32 & -40 \end{bmatrix} \)[/tex] is a scalar multiple of the original matrix [tex]\(\begin{bmatrix} -1 & 2 \\ 4 & -5 \end{bmatrix} \)[/tex].
To find the scalar, we compare the corresponding elements of the two matrices:
[tex]\[ \frac{-8}{-1} = 8, \quad \frac{16}{2} = 8, \quad \frac{32}{4} = 8, \quad \frac{-40}{-5} = 8 \][/tex]
In all cases, the scalar multiple is [tex]\( 8 \)[/tex].
Therefore, the missing number in the equation is [tex]\( 8 \)[/tex]:
[tex]\[ 3\left[\begin{array}{cc} -1 & 2 \\ 4 & -5 \end{array}\right] + 5\left[\begin{array}{cc} -1 & 2 \\ 4 & -5 \end{array}\right] = 8 \cdot \left[\begin{array}{cc} -1 & 2 \\ 4 & -5 \end{array}\right] \][/tex]