Question

If \( A = \begin{bmatrix} a & 1 & 2 1 & 2 & b c & 1 & 3 \end{bmatrix} \) and \( \text{adj } A = \begin{bmatrix} 7 & -1 & -5 -3 & 9 & 5 1 & -3 & 5 \end{bmatrix} \), then \( a^2 + b^2 + c^2 = \)

• A. 10
• B. 14
• C. 11
• D. 29

Hint:

To find missing variables in matrix \( A \) when \( \text{adj } A \) is provided, use the fact that the dot product of any row of \( A \) with a non-corresponding column of \( \text{adj } A \) is zero.

Answer 1

The Correct Option is A

Concept: The adjoint of a matrix \( A \), denoted as \( \text{adj } A \), is the transpose of the cofactor matrix. A key property used to find unknown elements in matrix \( A \) when its adjoint is given is:
• \( A \cdot (\text{adj } A) = |A| \cdot I \), where \( |A| \) is the determinant and \( I \) is the identity matrix.
• Alternatively, the element \( A_{ij} \) is related to the cofactors. Specifically, the transpose of the matrix of cofactors equals the adjoint. Thus, \( C_{ij} = (\text{adj } A)_{ji} \).

Step 1: Using the property \( A \cdot (\text{adj } A) = |A|I \).
Let's consider the product of the first row of \( A \) and the first column of \( \text{adj } A \): \[ (a)(7) + (1)(-3) + (2)(1) = |A| \quad \Rightarrow \quad 7a - 3 + 2 = |A| \quad \Rightarrow \quad |A| = 7a - 1 \quad \cdots (1) \] Now consider the product of the first row of \( A \) and the second column of \( \text{adj } A \) (which must be 0): \[ (a)(-1) + (1)(9) + (2)(-3) = 0 \quad \Rightarrow \quad -a + 9 - 6 = 0 \quad \Rightarrow \quad -a + 3 = 0 \quad \Rightarrow \quad \mathbf{a = 3} \]

Step 2: Finding unknowns \( b \) and \( c \) using orthogonality properties.
Multiply the second row of \( A \) with the second column of \( \text{adj } A \): \[ (1)(-1) + (2)(9) + (b)(-3) = |A| \quad \Rightarrow \quad -1 + 18 - 3b = |A| \quad \Rightarrow \quad 17 - 3b = |A| \] From Step 1, \( |A| = 7(3) - 1 = 20 \). So: \[ 17 - 3b = 20 \quad \Rightarrow \quad -3b = 3 \quad \Rightarrow \quad \mathbf{b = -1} \] Multiply the third row of \( A \) with the third column of \( \text{adj } A \): \[ (c)(-5) + (1)(5) + (3)(5) = |A| \quad \Rightarrow \quad -5c + 5 + 15 = 20 \] \[ -5c + 20 = 20 \quad \Rightarrow \quad -5c = 0 \quad \Rightarrow \quad \mathbf{c = 0} \]

Step 3: Calculating the final value.
We have \( a = 3 \), \( b = -1 \), and \( c = 0 \). \[ a^2 + b^2 + c^2 = (3)^2 + (-1)^2 + (0)^2 \] \[ = 9 + 1 + 0 = 10 \]