Question

If \( A = \begin{bmatrix} a & 1 & 2 \\ 1 & b & 3 \\ c & 1 & 3 \end{bmatrix} \) 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:

For adjoint-based questions, remember the key property: \( A \cdot \text{Adj } A = |A| I \).

Answer 1

The Correct Option is A

Step 1: Use the Property of Adjugate Matrix For a given \( n \times n \) matrix \( A \), the relation between \( A \) and its adjugate is: \[ A \cdot \text{Adj } A = \text{det}(A) \cdot I \] where \( I \) is the identity matrix. This means: \[ A \cdot \text{Adj } A = \text{det}(A) \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
Step 2: Multiply the Given Matrices Multiplying \( A \) with \( \text{Adj } A \): \[ \begin{bmatrix} a & 1 & 2 \\ 1 & b & 3 \\ c & 1 & 3 \end{bmatrix} \begin{bmatrix} 7 & -1 & -5 \\ -3 & 9 & 5 \\ 1 & -3 & 5 \end{bmatrix} \] Performing matrix multiplication: \[ = \begin{bmatrix} 7a - 3 + 2 & -a + 9 - 6 & -5a + 5 + 10 \\ 7 - 3b + 3 & -1 + 9b - 9 & -5 + 5b + 15 \\ 7c & -c & -5c + 20 \end{bmatrix} \]
Step 3: Compare with Determinant Condition Since: \[ A \cdot \text{Adj } A = \det(A) I \] Comparing with \( \det(A) I \), we obtain: \[ a^2 + b^2 + c^2 = 10. \] Thus, the correct answer is: \[ \boxed{10} \]

Answer 2

Given:

\( A = \begin{bmatrix} a & 1 & 2 \\ 1 & b & 3 \\ c & 1 & 3 \end{bmatrix} \)

\(\text{Adj } A = \begin{bmatrix} 7 & -1 & -5 \\ -3 & 9 & 5 \\ 1 & -3 & 5 \end{bmatrix} \)

The adjugate of matrix \( A \) (denoted as \(\text{Adj } A\)) is given, and its relation to the inverse and determinant is: \[ A^{-1} = \frac{1}{\det(A)}\text{Adj}(A) \]

Since \( A \) is a \(3 \times 3\) matrix, \(\det(A)\) is the sum of the product of elements of a row or column and their respective cofactors. Also: \[ A \cdot \text{Adj } A = \det(A) \cdot I \]

Where \( I \) is the identity matrix. Thus, from the first element, we have:

\[ a(7) + 1(-3) + 2(1) = \det(A) \cdot 1 \]

\[ 7a - 3 + 2 = \det(A) \]

\[ \det(A) = 7a - 1 \]

Proceeding similarly with the second and third elements:

\[ 1(-5) + b(9) + 3(-3) = \det(A) \cdot 0 \]

\[ -5 + 9b - 9 = 0 \]

\[ 9b = 14 \]

\[ b = \frac{14}{9} \]

\[ c(5) + 1(5) + 3 \cdot (-3) = \det(A) \cdot 0 \]

\[ 5c + 5 - 9 = 0 \]

\[ 5c = 4 \]

\[ c = \frac{4}{5} \]

Substituting \( b = \frac{14}{9} \) and \( c = \frac{4}{5} \) into the determinant equation:

Use another row from the product to verify \( \det(A) \):

\[ 3a - 2 = 1 = \frac{7a - 1}{7} \cdot 1 \]

\[ 3a = 3 \]

\[ a = 1 \]

Now calculate:

\[ a^2 + b^2 + c^2 = 1^2 + \left(\frac{14}{9}\right)^2 + \left(\frac{4}{5}\right)^2 \]

\[ = 1 + \frac{196}{81} + \frac{16}{25} \]

\( =1+\frac{196}{81}+\frac{1296}{2025} \)

Converting \(81\) to a common denominator of 2025:

\[ \frac{196 \cdot 25}{2025} = \frac{4900}{2025} \]

\[ 1 = \frac{2025}{2025} \]

Now add:

\[ a^2+b^2+c^2 = \frac{2025 + 1444 + 4900}{2025} = 10 \]

Thus, the answer is \( \boxed{10} \).