Question:medium

If \[ \begin{bmatrix} 2x+1 & 5x \\ 0 & y^2+1 \end{bmatrix} = \begin{bmatrix} x+3 & 10 \\ 0 & 26 \end{bmatrix} \] then the possible values of \(x+y\) are:

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For equal matrices: \[ \text{Corresponding elements must be equal.} \]
Updated On: May 30, 2026
  • \(2\ \text{and}\ 5\)
  • \(5\ \text{and}\ -1\)
  • \(7\ \text{and}\ -3\)
  • \(2\ \text{and}\ -5\)
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
The fundamental principle governing this problem is the equality of matrices. Two matrices are said to be equal if and only if they satisfy two conditions: first, they must have the exact same dimensions (order), and second, every element in the first matrix must be identical to the corresponding element in the second matrix. In mathematical terms, for two \(m \times n\) matrices \(A = [a_{ij}]\) and \(B = [b_{ij}]\), the equality \(A = B\) implies that \(a_{ij} = b_{ij}\) for all values of \(i\) and \(j\). In this specific problem, we are dealing with two \(2 \times 2\) matrices. By comparing elements at the same row and column indices, we can construct a system of equations that allows us to solve for the unknown variables \(x\) and \(y\). Once we determine the possible values for these variables, we can calculate the final requested sum, \(x + y\).
Step 2: Key Formula or Approach:
For the given matrices to be equal, we equate the corresponding entries:
1. Equate entry \((1,1)\): \(2x + 1 = x + 3\)
2. Equate entry \((1,2)\): \(5x = 10\)
3. Equate entry \((2,2)\): \(y^2 + 1 = 26\)
The entry at \((2,1)\) is already equal (\(0 = 0\)), confirming consistency.
Step 3: Detailed Explanation:
We begin by solving for the variable \(x\).
From the comparison of the elements in the first row and first column:
\[ 2x + 1 = x + 3 \]
To isolate \(x\), we subtract \(x\) from both sides of the equation:
\[ x + 1 = 3 \]
Next, we subtract 1 from both sides:
\[ x = 2 \]
To verify this result, we check the second element of the first row:
\[ 5x = 10 \implies x = \frac{10}{5} = 2 \]
Since both equations yield the same value, \(x = 2\) is consistently determined.
Now, we proceed to find the possible values for \(y\) by equating the elements in the second row and second column:
\[ y^2 + 1 = 26 \]
Subtracting 1 from both sides gives:
\[ y^2 = 25 \]
To solve for \(y\), we take the square root of both sides. It is critical to remember that every positive real number has two square roots—one positive and one negative:
\[ y = \pm \sqrt{25} \]
\[ y = 5 \text{ or } y = -5 \]
With \(x = 2\) and two possible values for \(y\), we calculate the possible sums for \(x + y\):
Case I: When \(y = 5\)
\[ x + y = 2 + 5 = 7 \]
Case II: When \(y = -5\)
\[ x + y = 2 + (-5) = -3 \]
The two possible values for the sum are 7 and \(-3\).
Step 4: Final Answer:
Based on the calculations, the possible values for \(x + y\) are 7 and \(-3\). Comparing this result to the given choices, we find it aligns perfectly with option (C).
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