If ` [ (1, 2),(-2,-b)]+ [(a,4),(3,2)]=[(5,6),(1,0)]` , then ` a^(2) + b^(2) ` is equal to

Options

A. 10
B. 12
C. 20
D. 22

Answer

To solve the equation given in the question, we will follow these steps:

**Step 1: Write down the matrices.**
We have:
\[ \begin{pmatrix} 1 & 2 \\ -2 & -b \end{pmatrix} + \begin{pmatrix} a & 4 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} 5 & 6 \\ 1 & 0 \end{pmatrix} \]

**Step 2: Add the matrices on the left-hand side.**
We add the corresponding elements of the matrices:
- First element: \(1 + a\)
- Second element: \(2 + 4 = 6\)
- Third element: \(-2 + 3 = 1\)
- Fourth element: \(-b + 2\)

Thus, the left-hand side becomes:
\[ \begin{pmatrix} 1 + a & 6 \\ 1 & 2 - b \end{pmatrix} \]

**Step 3: Set the left-hand side equal to the right-hand side.**
Now we equate the left-hand side to the right-hand side:
\[ \begin{pmatrix} 1 + a & 6 \\ 1 & 2 - b \end{pmatrix} = \begin{pmatrix} 5 & 6 \\ 1 & 0 \end{pmatrix} \]

**Step 4: Create equations from the matrix equality.**
From the matrix equality, we can derive the following equations:
1. \(1 + a = 5\)
2. \(2 - b = 0\)

**Step 5: Solve for \(a\) and \(b\).**
From the first equation:
\[ 1 + a = 5 \implies a = 5 - 1 = 4 \]

From the second equation:
\[ 2 - b = 0 \implies b = 2 \]

**Step 6: Calculate \(a^2 + b^2\).**
Now we need to find \(a^2 + b^2\):
\[ a^2 + b^2 = 4^2 + 2^2 = 16 + 4 = 20 \]

Thus, the final answer is:
\[ \boxed{20} \]

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