Question:
Solve the given matrix equation for \(x\), given \(x > 0\):
\[ x\begin{bmatrix} 2x & 2 \\ 3 & x \end{bmatrix} + 2\begin{bmatrix} 8 & 5x \\ 4 & 0 \end{bmatrix} = 2\begin{bmatrix} x^2+8 & 24 \\ 10 & 8 \end{bmatrix} \]
Solve the given matrix equation for \(x\), given \(x > 0\):
\[ x\begin{bmatrix} 2x & 2 \\ 3 & x \end{bmatrix} + 2\begin{bmatrix} 8 & 5x \\ 4 & 0 \end{bmatrix} = 2\begin{bmatrix} x^2+8 & 24 \\ 10 & 8 \end{bmatrix} \]
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Multiply the scalars into the matrices, add, then equate matching entries. The entry 12x = 48 gives x = 4.
Updated On: Oct 22, 2026
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Solution and Explanation
Step 1: Multiply the scalars into the matrices.
A scalar multiplies every entry of a matrix. So we get the matrices below.
\[ x\begin{bmatrix} 2x & 2 \\ 3 & x \end{bmatrix} = \begin{bmatrix} 2x^2 & 2x \\ 3x & x^2 \end{bmatrix}, \quad 2\begin{bmatrix} 8 & 5x \\ 4 & 0 \end{bmatrix} = \begin{bmatrix} 16 & 10x \\ 8 & 0 \end{bmatrix} \]
\[ 2\begin{bmatrix} x^2+8 & 24 \\ 10 & 8 \end{bmatrix} = \begin{bmatrix} 2x^2+16 & 48 \\ 20 & 16 \end{bmatrix} \]
Step 2: Add the two matrices on the left side.
Matrices of the same order are added entry by entry.
\[ \begin{bmatrix} 2x^2+16 & 12x \\ 3x+8 & x^2 \end{bmatrix} = \begin{bmatrix} 2x^2+16 & 48 \\ 20 & 16 \end{bmatrix} \]
Step 3: Equate the corresponding entries.
Two matrices are equal only when all matching entries are equal.
The first entries give \(2x^2+16 = 2x^2+16\), which is true for every \(x\).
The second entry of row 1 gives \(12x = 48\), so \(x = 4\).
The first entry of row 2 gives \(3x + 8 = 20\), so \(3x = 12\) and \(x = 4\).
The last entry gives \(x^2 = 16\), so \(x = \pm 4\).
Step 4: Apply the condition x > 0.
The value \(x = -4\) fails the other equations and the given condition.
Only \(x = 4\) satisfies all the equations at once.
Final Answer:
The value of \(x\) is 4. \[ \boxed{x = 4} \]
A scalar multiplies every entry of a matrix. So we get the matrices below.
\[ x\begin{bmatrix} 2x & 2 \\ 3 & x \end{bmatrix} = \begin{bmatrix} 2x^2 & 2x \\ 3x & x^2 \end{bmatrix}, \quad 2\begin{bmatrix} 8 & 5x \\ 4 & 0 \end{bmatrix} = \begin{bmatrix} 16 & 10x \\ 8 & 0 \end{bmatrix} \]
\[ 2\begin{bmatrix} x^2+8 & 24 \\ 10 & 8 \end{bmatrix} = \begin{bmatrix} 2x^2+16 & 48 \\ 20 & 16 \end{bmatrix} \]
Step 2: Add the two matrices on the left side.
Matrices of the same order are added entry by entry.
\[ \begin{bmatrix} 2x^2+16 & 12x \\ 3x+8 & x^2 \end{bmatrix} = \begin{bmatrix} 2x^2+16 & 48 \\ 20 & 16 \end{bmatrix} \]
Step 3: Equate the corresponding entries.
Two matrices are equal only when all matching entries are equal.
The first entries give \(2x^2+16 = 2x^2+16\), which is true for every \(x\).
The second entry of row 1 gives \(12x = 48\), so \(x = 4\).
The first entry of row 2 gives \(3x + 8 = 20\), so \(3x = 12\) and \(x = 4\).
The last entry gives \(x^2 = 16\), so \(x = \pm 4\).
Step 4: Apply the condition x > 0.
The value \(x = -4\) fails the other equations and the given condition.
Only \(x = 4\) satisfies all the equations at once.
Final Answer:
The value of \(x\) is 4. \[ \boxed{x = 4} \]
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