Question:
Let $ A = \begin{bmatrix} 2 & 1 & 3 & -1 \\1 & -2 & 2 & -3 \end{bmatrix}, B = \begin{bmatrix} 2 & 1 & 0 & 3 \\1 & -1 & 2 & 3 \end{bmatrix} $, and the equation $ 2A + 3B - 5C = 0 $. Find the matrix $ C $.
Let $ A = \begin{bmatrix} 2 & 1 & 3 & -1 \\1 & -2 & 2 & -3 \end{bmatrix}, B = \begin{bmatrix} 2 & 1 & 0 & 3 \\1 & -1 & 2 & 3 \end{bmatrix} $, and the equation $ 2A + 3B - 5C = 0 $. Find the matrix $ C $.
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When working with matrix equations, ensure to perform the operations element-wise. Remember that matrix addition, scalar multiplication, and other operations follow the standard linear algebra rules.
Updated On: May 9, 2025
\( \begin{bmatrix} 2 & 1 & 6/5 & 7/5 \\1 & 7/5 & 2/5 & 3/5 \end{bmatrix} \)
\( \begin{bmatrix} -2 & 1 & 6/5 & 7/5 \\1 & -7/5 & 2/5 & 3/5 \end{bmatrix} \)
\( \begin{bmatrix} -2 & 1 & 6/5 & 7/5 \\1 & 7/5 & 2/5 & 3/5 \end{bmatrix} \)
\( \begin{bmatrix} 2 & 1 & 6/5 & 7/5 \\1 & -7/5 & 2 & 3/5 \end{bmatrix} \)
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The Correct Option is D
Solution and Explanation
Step 1: Use the given equation \( 2A + 3B - 5C = 0 \).
Rearranging, we get: \[ 5C = 2A + 3B \] \[ C = \frac{1}{5} (2A + 3B) \] Step 2: Calculate \( 2A \) and \( 3B \). First, calculate \( 2A \): \[ 2A = 2 \times \begin{bmatrix} 2 & 1 & 3 & -1 \\ 1 & -2 & 2 & -3 \end{bmatrix} = \begin{bmatrix} 4 & 2 & 6 & -2 \\ 2 & -4 & 4 & -6 \end{bmatrix} \] Next, calculate \( 3B \): \[ 3B = 3 \times \begin{bmatrix} 2 & 1 & 0 & 3 \\ 1 & -1 & 2 & 3 \end{bmatrix} = \begin{bmatrix} 6 & 3 & 0 & 9 \\ 3 & -3 & 6 & 9 \end{bmatrix} \] Step 3: Add \( 2A \) and \( 3B \). \[ 2A + 3B = \begin{bmatrix} 4 & 2 & 6 & -2 \\ 2 & -4 & 4 & -6 \end{bmatrix} + \begin{bmatrix} 6 & 3 & 0 & 9 \\ 3 & -3 & 6 & 9 \end{bmatrix} = \begin{bmatrix} 10 & 5 & 6 & 7 \\ 5 & -7 & 10 & 3 \end{bmatrix} \] Step 4: Divide by 5 to find \( C \). \[ C = \frac{1}{5} \begin{bmatrix} 10 & 5 & 6 & 7 \\ 5 & -7 & 10 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 6/5 & 7/5 \\ 1 & -7/5 & 2 & 3/5 \end{bmatrix} \] Thus, the correct matrix \( C \) is: \[ \boxed{\begin{bmatrix} 2 & 1 & 6/5 & 7/5 \\ 1 & -7/5 & 2 & 3/5 \end{bmatrix}} \]
Rearranging, we get: \[ 5C = 2A + 3B \] \[ C = \frac{1}{5} (2A + 3B) \] Step 2: Calculate \( 2A \) and \( 3B \). First, calculate \( 2A \): \[ 2A = 2 \times \begin{bmatrix} 2 & 1 & 3 & -1 \\ 1 & -2 & 2 & -3 \end{bmatrix} = \begin{bmatrix} 4 & 2 & 6 & -2 \\ 2 & -4 & 4 & -6 \end{bmatrix} \] Next, calculate \( 3B \): \[ 3B = 3 \times \begin{bmatrix} 2 & 1 & 0 & 3 \\ 1 & -1 & 2 & 3 \end{bmatrix} = \begin{bmatrix} 6 & 3 & 0 & 9 \\ 3 & -3 & 6 & 9 \end{bmatrix} \] Step 3: Add \( 2A \) and \( 3B \). \[ 2A + 3B = \begin{bmatrix} 4 & 2 & 6 & -2 \\ 2 & -4 & 4 & -6 \end{bmatrix} + \begin{bmatrix} 6 & 3 & 0 & 9 \\ 3 & -3 & 6 & 9 \end{bmatrix} = \begin{bmatrix} 10 & 5 & 6 & 7 \\ 5 & -7 & 10 & 3 \end{bmatrix} \] Step 4: Divide by 5 to find \( C \). \[ C = \frac{1}{5} \begin{bmatrix} 10 & 5 & 6 & 7 \\ 5 & -7 & 10 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 6/5 & 7/5 \\ 1 & -7/5 & 2 & 3/5 \end{bmatrix} \] Thus, the correct matrix \( C \) is: \[ \boxed{\begin{bmatrix} 2 & 1 & 6/5 & 7/5 \\ 1 & -7/5 & 2 & 3/5 \end{bmatrix}} \]
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