Question

If A, B, C are three matrices of order 2, such that

$|A| = -2, |B| = 2, |C| = 10,$

find the value of $k = |A + B - C| + |B + C - A| + |C + A - B| + |A + B + C|.$ (|A| denotes the determinant of matrix A)

• A. 40
• B. 30
• C. 20
• D. 10

Answer 1

Answer: (A)

40

Answer 2

Let A, B, C be $2 \times 2$ diagonal matrices of the form $A = \begin{pmatrix} a_1 & 0 \\ 0 & d \end{pmatrix}$, $B = \begin{pmatrix} b_1 & 0 \\ 0 & d \end{pmatrix}$, $C = \begin{pmatrix} c_1 & 0 \\ 0 & d \end{pmatrix}$. Given $|A| = a_1d = -2$, $|B| = b_1d = 2$, $|C| = c_1d = 10$.

We can calculate the determinants of the sums and differences of these matrices:

1. $A + B - C = \begin{pmatrix} a_1+b_1-c_1 & 0 \\ 0 & d+d-d \end{pmatrix} = \begin{pmatrix} a_1+b_1-c_1 & 0 \\ 0 & d \end{pmatrix}$ $|A + B - C| = (a_1+b_1-c_1)d = a_1d + b_1d - c_1d = |A| + |B| - |C| = -2 + 2 - 10 = -10$.

2. $B + C - A = \begin{pmatrix} b_1+c_1-a_1 & 0 \\ 0 & d+d-d \end{pmatrix} = \begin{pmatrix} b_1+c_1-a_1 & 0 \\ 0 & d \end{pmatrix}$ $|B + C - A| = (b_1+c_1-a_1)d = b_1d + c_1d - a_1d = |B| + |C| - |A| = 2 + 10 - (-2) = 14$.

3. $C + A - B = \begin{pmatrix} c_1+a_1-b_1 & 0 \\ 0 & d+d-d \end{pmatrix} = \begin{pmatrix} c_1+a_1-b_1 & 0 \\ 0 & d \end{pmatrix}$ $|C + A - B| = (c_1+a_1-b_1)d = c_1d + a_1d - b_1d = |C| + |A| - |B| = 10 + (-2) - 2 = 6$.

4. $A + B + C = \begin{pmatrix} a_1+b_1+c_1 & 0 \\ 0 & d+d+d \end{pmatrix} = \begin{pmatrix} a_1+b_1+c_1 & 0 \\ 0 & 3d \end{pmatrix}$ $|A + B + C| = (a_1+b_1+c_1)(3d) = 3(a_1d + b_1d + c_1d) = 3(|A| + |B| + |C|) = 3(-2 + 2 + 10) = 3(10) = 30$.

Now, we sum these determinants to find $k$: $k = |A + B - C| + |B + C - A| + |C + A - B| + |A + B + C|$ $k = (-10) + (14) + (6) + (30)$ $k = 40$.

This result is independent of the choice of $d$ (as long as $d \neq 0$, which is implied by $|A|, |B|, |C|$ being non-zero). This specific construction suggests that the value of $k$ is indeed constant for any matrices A, B, C satisfying the conditions, provided they can be represented in this form or that the general case simplifies to this. The problem implies a unique answer, and this method provides one.