Question

If A, B, C are three matrices of order 2, such that $\qquad |A + B|- |C| = -10,$ $\qquad |B + C| - |A| = 25,$ $\qquad |C + A| - |B| = 15,$

then the value of $|A + B + C|$ is equal to ($|A|$ denotes the determinant of matrix A)

• A. 15
• B. 30
• C. 45
• D. 60

Answer 1

Answer: (B)

30

Answer 2

Let $A, B, C$ be three matrices of order 2. Let $|M|$ denote the determinant of a matrix $M$. For two matrices $M$ and $N$ of order 2, we have the property: $|M+N| = |M| + |N| + \det \begin{pmatrix} r_{M1} \\ r_{N2} \end{pmatrix} + \det \begin{pmatrix} r_{N1} \\ r_{M2} \end{pmatrix}$, where $r_{M1}, r_{M2}$ are the first and second rows of $M$, and $r_{N1}, r_{N2}$ are the first and second rows of $N$.

Let $X = |A|$, $Y = |B|$, $Z = |C|$. Let $d_{AB} = \det \begin{pmatrix} r_{A1} \\ r_{B2} \end{pmatrix} + \det \begin{pmatrix} r_{B1} \\ r_{A2} \end{pmatrix}$. Let $d_{BC} = \det \begin{pmatrix} r_{B1} \\ r_{C2} \end{pmatrix} + \det \begin{pmatrix} r_{C1} \\ r_{B2} \end{pmatrix}$. Let $d_{CA} = \det \begin{pmatrix} r_{C1} \\ r_{A2} \end{pmatrix} + \det \begin{pmatrix} r_{A1} \\ r_{C2} \end{pmatrix}$.

The given equations are:

1. $|A + B| - |C| = -10 \implies |A| + |B| + d_{AB} - |C| = -10 \implies X + Y + d_{AB} - Z = -10$
2. $|B + C| - |A| = 25 \implies |B| + |C| + d_{BC} - |A| = 25 \implies Y + Z + d_{BC} - X = 25$
3. $|C + A| - |B| = 15 \implies |C| + |A| + d_{CA} - |B| = 15 \implies Z + X + d_{CA} - Y = 15$

Adding the three equations: $(X + Y + d_{AB} - Z) + (Y + Z + d_{BC} - X) + (Z + X + d_{CA} - Y) = -10 + 25 + 15$ $(X - X + X) + (Y + Y - Y) + (-Z + Z + Z) + d_{AB} + d_{BC} + d_{CA} = 30$ $X + Y + Z + d_{AB} + d_{BC} + d_{CA} = 30$.

Now consider $|A+B+C|$. $A+B+C = \begin{pmatrix} r_{A1}+r_{B1}+r_{C1} \\ r_{A2}+r_{B2}+r_{C2} \end{pmatrix}$. Using the multilinearity property of the determinant with respect to rows: $|A+B+C| = \det \begin{pmatrix} r_{A1}+r_{B1}+r_{C1} \\ r_{A2}+r_{B2}+r_{C2} \end{pmatrix}$ $= \det \begin{pmatrix} r_{A1} \\ r_{A2}+r_{B2}+r_{C2} \end{pmatrix} + \det \begin{pmatrix} r_{B1} \\ r_{A2}+r_{B2}+r_{C2} \end{pmatrix} + \det \begin{pmatrix} r_{C1} \\ r_{A2}+r_{B2}+r_{C2} \end{pmatrix}$ $= \det \begin{pmatrix} r_{A1} \\ r_{A2} \end{pmatrix} + \det \begin{pmatrix} r_{A1} \\ r_{B2} \end{pmatrix} + \det \begin{pmatrix} r_{A1} \\ r_{C2} \end{pmatrix} + \det \begin{pmatrix} r_{B1} \\ r_{A2} \end{pmatrix} + \det \begin{pmatrix} r_{B1} \\ r_{B2} \end{pmatrix} + \det \begin{pmatrix} r_{B1} \\ r_{C2} \end{pmatrix} + \det \begin{pmatrix} r_{C1} \\ r_{A2} \end{pmatrix} + \det \begin{pmatrix} r_{C1} \\ r_{B2} \end{pmatrix} + \det \begin{pmatrix} r_{C1} \\ r_{C2} \end{pmatrix}$ $= |A| + \det \begin{pmatrix} r_{A1} \\ r_{B2} \end{pmatrix} + \det \begin{pmatrix} r_{A1} \\ r_{C2} \end{pmatrix} + \det \begin{pmatrix} r_{B1} \\ r_{A2} \end{pmatrix} + |B| + \det \begin{pmatrix} r_{B1} \\ r_{C2} \end{pmatrix} + \det \begin{pmatrix} r_{C1} \\ r_{A2} \end{pmatrix} + \det \begin{pmatrix} r_{C1} \\ r_{B2} \end{pmatrix} + |C|$ Rearranging the terms: $|A+B+C| = |A| + |B| + |C| + \left(\det \begin{pmatrix} r_{A1} \\ r_{B2} \end{pmatrix} + \det \begin{pmatrix} r_{B1} \\ r_{A2} \end{pmatrix}\right) + \left(\det \begin{pmatrix} r_{B1} \\ r_{C2} \end{pmatrix} + \det \begin{pmatrix} r_{C1} \\ r_{B2} \end{pmatrix}\right) + \left(\det \begin{pmatrix} r_{C1} \\ r_{A2} \end{pmatrix} + \det \begin{pmatrix} r_{A1} \\ r_{C2} \end{pmatrix}\right)$ $|A+B+C| = X + Y + Z + d_{AB} + d_{BC} + d_{CA}$.

Comparing this expression with the sum of the given equations, we see that: $|A+B+C| = (X + Y + Z + d_{AB} + d_{BC} + d_{CA}) = 30$.

The value of $|A + B + C|$ is 30.