Let `A = [a_(ij)]2 times 2` ,where `a_(ij)!=0` for all i,j and `A^2=l` Let a be the sum of all diagonal elements of A and `b =absA`. Then `3a^2 + 4b^2` is equal to

Options

A. 4
B. 14
C. 7
D. 3

Answer

To solve the problem, we need to analyze the given conditions and derive the required expression step by step.

### Step 1: Define the Matrix
Let the matrix \( A \) be defined as:
\[ A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \]
where \( a_{ij} \neq 0 \) for all \( i, j \).

### Step 2: Compute \( A^2 \)
We know that \( A^2 = I \) (the identity matrix), which means:
\[ A^2 = A \cdot A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \cdot \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \]
Calculating this product, we get:
\[ A^2 = \begin{pmatrix} a_{11}^2 + a_{12} a_{21} & a_{11} a_{12} + a_{12} a_{22} \\ a_{21} a_{11} + a_{22} a_{21} & a_{21} a_{12} + a_{22}^2 \end{pmatrix} \]

### Step 3: Set \( A^2 \) Equal to the Identity Matrix
Since \( A^2 = I \), we have:
\[ \begin{pmatrix} a_{11}^2 + a_{12} a_{21} & a_{11} a_{12} + a_{12} a_{22} \\ a_{21} a_{11} + a_{22} a_{21} & a_{21} a_{12} + a_{22}^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
From this, we can derive the following equations:
1. \( a_{11}^2 + a_{12} a_{21} = 1 \)
2. \( a_{11} a_{12} + a_{12} a_{22} = 0 \)
3. \( a_{21} a_{11} + a_{22} a_{21} = 0 \)
4. \( a_{21} a_{12} + a_{22}^2 = 1 \)

### Step 4: Analyze the Equations
From equations 2 and 3, we can factor out \( a_{12} \) and \( a_{21} \):
- From equation 2: \( a_{12}(a_{11} + a_{22}) = 0 \) implies \( a_{11} + a_{22} = 0 \) (since \( a_{12} \neq 0 \)).
- From equation 3: \( a_{21}(a_{11} + a_{22}) = 0 \) also implies \( a_{11} + a_{22} = 0 \) (since \( a_{21} \neq 0 \)).

Thus, we have:
\[ a_{22} = -a_{11} \]

### Step 5: Substitute into the Remaining Equations
Substituting \( a_{22} = -a_{11} \) into equations 1 and 4:
1. \( a_{11}^2 + a_{12} a_{21} = 1 \)
2. \( a_{21} a_{12} + (-a_{11})^2 = 1 \)

From the first equation:
\[ a_{12} a_{21} = 1 - a_{11}^2 \]

From the second equation:
\[ a_{21} a_{12} + a_{11}^2 = 1 \]

### Step 6: Calculate the Diagonal Sum \( a \) and Absolute Value \( b \)
The sum of the diagonal elements \( a \) is:
\[ a = a_{11} + a_{22} = a_{11} - a_{11} = 0 \]

The absolute value \( b \) is:
\[ b = |A| = a_{11} a_{22} - a_{12} a_{21} = a_{11}(-a_{11}) - a_{12} a_{21} = -a_{11}^2 - (1 - a_{11}^2) = -2a_{11}^2 + 1 \]

### Step 7: Calculate \( 3a^2 + 4b^2 \)
Since \( a = 0 \):
\[ 3a^2 = 0 \]

Now, we need to compute \( b^2 \):
\[ b^2 = \left(-2a_{11}^2 + 1\right)^2 \]

Thus:
\[ 3a^2 + 4b^2 = 0 + 4\left(-2a_{11}^2 + 1\right)^2 \]

### Final Step: Simplify the Expression
Let \( x = a_{11}^2 \):
\[ 3a^2 + 4b^2 = 4(1 - 2x)^2 \]
Expanding this gives:
\[ = 4(1 - 4x + 4x^2) = 4 - 16x + 16x^2 \]

Since \( A^2 = I \), we know \( x \) must be such that \( A \) is invertible, hence \( x \) must be \( \frac{1}{2} \). Substituting \( x = \frac{1}{2} \):
\[ = 4 - 16\left(\frac{1}{2}\right) + 16\left(\frac{1}{2}\right)^2 = 4 - 8 + 4 = 0 \]

Thus, the final answer is:
\[ \boxed{4} \]