Question

If \( a, b, c \in \mathbb{R} \), \( a < b < c \), \( A = \begin{bmatrix} -c & 0 & c 0 & -b & b a & a & 0 \end{bmatrix} \), \( 12A^{-1} = \begin{bmatrix} -b & c & bc b & -c & bc b & c & bc \end{bmatrix} \) and Trace of \( A = -5 \), then \( a^2 + b^2 + c^2 = \)

• A. 29
• B. 21
• C. 26
• D. 14

Hint:

To avoid calculating the full adjoint, compare just one or two corresponding elements. For example, the element at (3,3) in the given \( 12A^{-1} \) is \( bc \). In the computed \( 12A^{-1} \), it is \( 12 \times (1/2a) = 6/a \). Setting \( 6/a = bc \) quickly connects the variables.

Answer 1

The Correct Option is D

Concept:& nbsp;

This problem uses the properties of the inverse of a matrix and the trace of a matrix.

• Inverse of a matrix: \[ A^{-1}=\frac{1}{|A|}\,\operatorname{adj}(A). \]

• Trace of a matrix: The trace of a matrix is the sum of its diagonal elements.

Step 1: Find \(\operatorname{adj}(A)\) and \(|A|\).

The determinant of the matrix is

\[ |A| = -c(-ab)-0+c\bigl(0-(-ab)\bigr) = abc+abc = 2abc. \]

The cofactor matrix is

\[ C= \begin{bmatrix} -ab & amp; ab & amp; ab\\ ac & amp; -ac & amp; ac\\ bc & amp; bc & amp; bc \end{bmatrix}. \]

Therefore,

\[ \operatorname{adj}(A) = C^T = \begin{bmatrix} -ab & amp; ac & amp; bc\\ ab & amp; -ac & amp; bc\\ ab & amp; ac & amp; bc \end{bmatrix}. \]

Hence,

\[ A^{-1} = \frac{1}{2abc} \begin{bmatrix} -ab & amp; ac & amp; bc\\ ab & amp; -ac & amp; bc\\ ab & amp; ac & amp; bc \end{bmatrix} = \begin{bmatrix} -\frac{1}{2c} & amp; \frac{1}{2b} & amp; \frac{1}{2a}\\ \frac{1}{2c} & amp; -\frac{1}{2b} & amp; \frac{1}{2a}\\ \frac{1}{2c} & amp; \frac{1}{2b} & amp; \frac{1}{2a} \end{bmatrix}. \]

Step 2: Compare the given matrices to find \(a\), \(b\), and \(c\).

Given,

\[ 12A^{-1} = \begin{bmatrix} -\frac{6}{c} & amp; \frac{6}{b} & amp; \frac{6}{a}\\ \frac{6}{c} & amp; -\frac{6}{b} & amp; \frac{6}{a}\\ \frac{6}{c} & amp; \frac{6}{b} & amp; \frac{6}{a} \end{bmatrix} = \begin{bmatrix} -b & amp; c & amp; bc\\ b & amp; -c & amp; bc\\ b & amp; c & amp; bc \end{bmatrix}. \]

Comparing corresponding entries,

\[ -\frac{6}{c}=-b \quad\Rightarrow\quad bc=6. \] \[ \frac{6}{a}=bc \quad\Rightarrow\quad \frac{6}{a}=6 \quad\Rightarrow\quad a=1. \]

Step 3: Use the trace condition to determine \(b\) and \(c\).

Since the trace of \(A\) is \(-5\),

\[ -c-b+0=-5 \quad\Rightarrow\quad b+c=5. \]

Now,

\[ b+c=5, \qquad bc=6. \]

Solving

\[ x^2-5x+6=0, \]

gives the roots

\[ x=2,\;3. \]

Since \[ a

Therefore,

\[ a^2+b^2+c^2 = 1^2+2^2+3^2 = 1+4+9 = 14. \]