Key Concept: The Binomial Theorem for Matrices

The problem requires us to find the sum of all elements of a matrix $B$, which is given in a series form. The first crucial step is to recognize that this series resembles a binomial expansion. For any two commuting matrices $X$ and $Y$ (or in this case, a scalar identity matrix $I$ and any matrix $A$), the binomial expansion for $(X-Y)^n$ is given by: $(X-Y)^n = {}^n{C_0}X^n - {}^n{C_1}X^{n-1}Y + {}^n{C_2}X^{n-2}Y^2 - \,\,.....\,\, + (-1)^n {}^n{C_n}Y^n$ Since $I$ commutes with any matrix, we can apply this theorem directly.

Comparing the given expression for $B$: $B = I - {}^5{C_1}(\text{adj }A) + {}^5{C_2}{(\text{adj }A)^2} - {}^5{C_3}{(\text{adj }A)^3} + {}^5{C_4}{(\text{adj }A)^4} - {}^5{C_5}{(\text{adj }A)^5}$ with the binomial expansion, we can identify:

• $n = 5$
• $X = I$ (the identity matrix)
• $Y = \text{adj }A$ (the adjoint of matrix $A$)

Therefore, the matrix $B$ can be significantly simplified to: $B = (I - \text{adj }A)^5$

Now, we proceed with calculating $\text{adj }A$ and then $B$.

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Step 1: Calculate the Adjoint of Matrix A ($\text{adj }A$)

Given matrix A = \left( {\matrix{ 2 & { - 1} \cr 0 & 2 \cr } } \right). For a $2 \times 2$ matrix M = \left( {\matrix{ a & b \cr c & d \cr } } \right), its adjoint is given by \text{adj }M = \left( {\matrix{ d & { - b} \cr { - c} & a \cr } } \right). Applying this formula to matrix $A$: \text{adj }A = \left( {\matrix{ 2 & { - ( - 1)} \cr { - 0} & 2 \cr } } \right) = \left( {\matrix{ 2 & 1 \cr 0 & 2 \cr } } \right)

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Step 2: Calculate the Matrix $(I - \text{adj }A)$

Now we substitute the calculated $\text{adj }A$ into the simplified expression for $B$. First, let's find the matrix $(I - \text{adj }A)$: I - \text{adj }A = \left( {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right) - \left( {\matrix{ 2 & 1 \cr 0 & 2 \cr } } \right) Performing the matrix subtraction: I - \text{adj }A = \left( {\matrix{ 1-2 & 0-1 \cr 0-0 & 1-2 \cr } } \right) = \left( {\matrix{ -1 & -1 \cr 0 & -1 \cr } } \right)

Let's call this resulting matrix $M$, so M = \left( {\matrix{ -1 & -1 \cr 0 & -1 \cr } } \right). Our goal is to find $B = M^5$.

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Step 3: Calculate $M^5$

We need to compute the 5th power of matrix M = \left( {\matrix{ -1 & -1 \cr 0 & -1 \cr } } \right). Let's calculate the first few powers of $M$ to identify a pattern: M^1 = \left( {\matrix{ -1 & -1 \cr 0 & -1 \cr } } \right) M^2 = M \cdot M = \left( {\matrix{ -1 & -1 \cr 0 & -1 \cr } } \right) \left( {\matrix{ -1 & -1 \cr 0 & -1 \cr } } \right) = \left( {\matrix{ (-1)(-1)+(-1)(0) & (-1)(-1)+(-1)(-1) \cr (0)(-1)+(-1)(0) & (0)(-1)+(-1)(-1) \cr } } \right) M^2 = \left( {\matrix{ 1 & 1+1 \cr 0 & 1 \cr } } \right) = \left( {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right) M^3 = M^2 \cdot M = \left( {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right) \left( {\matrix{ -1 & -1 \cr 0 & -1 \cr } } \right) = \left( {\matrix{ (1)(-1)+(2)(0) & (1)(-1)+(2)(-1) \cr (0)(-1)+(1)(0) & (0)(-1)+(1)(-1) \cr } } \right) M^3 = \left( {\matrix{ -1 & -1-2 \cr 0 & -1 \cr } } \right) = \left( {\matrix{ -1 & -3 \cr 0 & -1 \cr } } \right)

Observing the pattern:

• For $M^1$: \left( {\matrix{ (-1)^1 & -1 \cr 0 & (-1)^1 \cr } } \right)
• For $M^2$: \left( {\matrix{ (-1)^2 & 2 \cr 0 & (-1)^2 \cr } } \right)
• For $M^3$: \left( {\matrix{ (-1)^3 & -3 \cr 0 & (-1)^3 \cr } } \right)

It appears that for an integer $k \ge 1$: M^k = \left( {\matrix{ (-1)^k & k(-1)^{k-1} \cr 0 & (-1)^k \cr } } \right) Alternatively, this can be written as: M^k = (-1)^k \left( {\matrix{ 1 & -k \cr 0 & 1 \cr } } \right) Let's verify this form with $k=2$: (-1)^2 \left( {\matrix{ 1 & -2 \cr 0 & 1 \cr } } \right) = \left( {\matrix{ 1 & -2 \cr 0 & 1 \cr } } \right). This doesn't match M^2 = \left( {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right). Let's re-examine the first pattern: M^k = \left( {\matrix{ (-1)^k & k(-1)^{k+1} \cr 0 & (-1)^k \cr } } \right) for the top right element. No, the M^k = \left( {\matrix{ (-1)^k & -k(-1)^k \cr 0 & (-1)^k \cr } } \right) would be the correct one from the pattern if we express $-1$ as $-1 \times 1$, $2$ as $-1 \times (-2)$, $-3$ as $-1 \times 3$. It is M^k = \left( {\matrix{ (-1)^k & -k(-1)^{k-1} \cr 0 & (-1)^k \cr } } \right). For $k=1$: \left( {\matrix{ -1 & -1(-1)^0 \cr 0 & -1 \cr } } \right) = \left( {\matrix{ -1 & -1 \cr 0 & -1 \cr } } \right). Correct. For $k=2$: \left( {\matrix{ (-1)^2 & -2(-1)^1 \cr 0 & (-1)^2 \cr } } \right) = \left( {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right). Correct. For $k=3$: \left( {\matrix{ (-1)^3 & -3(-1)^2 \cr 0 & (-1)^3 \cr } } \right) = \left( {\matrix{ -1 & -3 \cr 0 & -1 \cr } } \right). Correct.

Using this pattern for $k=5$: B = M^5 = \left( {\matrix{ (-1)^5 & -5(-1)^{5-1} \cr 0 & (-1)^5 \cr } } \right) B = \left( {\matrix{ -1 & -5(-1)^4 \cr 0 & -1 \cr } } \right) = \left( {\matrix{ -1 & -5(1) \cr 0 & -1 \cr } } \right) B = \left( {\matrix{ -1 & -5 \cr 0 & -1 \cr } } \right)

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Step 4: Sum of all elements of Matrix B

The matrix $B$ is \left( {\matrix{ -1 & -5 \cr 0 & -1 \cr } } \right). The sum of all elements is $(-1) + (-5) + 0 + (-1)$. Sum $= -1 - 5 + 0 - 1 = -7$.

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Tips and Common Mistakes:

1. Recognizing the Binomial Expansion: This is the most critical step. If you miss this, you might attempt to calculate $(\text{adj }A)^2$, $(\text{adj }A)^3$, etc., and then combine them, which would be very tedious and error-prone.
2. Adjoint of a 2x2 Matrix: Remember the formula: for \left( {\matrix{ a & b \cr c & d \cr } } \right), the adjoint is \left( {\matrix{ d & { - b} \cr { - c} & a \cr } } \right). Pay attention to signs.
3. Matrix Power Calculation: For matrices with a specific structure (like upper triangular matrices with identical diagonal elements), often a pattern emerges for their powers. Calculating the first few powers and looking for a pattern can save a lot of computational effort. Don't blindly multiply 5 times if a pattern is evident.
4. Careful Arithmetic: Matrix operations, especially subtraction and multiplication, require precision. A small error can propagate.

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Summary and Key Takeaway:

This problem beautifully combines knowledge of the Binomial Theorem with matrix operations. The key to solving it efficiently was recognizing the given series as a binomial expansion, which drastically simplified the expression for matrix $B$. Subsequently, the problem reduced to calculating a power of a matrix, which was made easier by identifying a pattern in its successive powers. This highlights the importance of pattern recognition and formula application in competitive mathematics.

The final answer is $\boxed{-7}$.