If P = B + C = [ 2 5 0 5 2 3 0 3 12 ] and Q = B - C = [ 0 -1 2 1 0 1 -2 -1 0 ], then Tr (BC) =

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q4 – AP-EAMCET 2026

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Home Maths MatricesQuestion #280

P = B + C = [\begin{matrix} 2 & 5 & 0 \\ 5 & 2 & 3 \\ 0 & 3 & 12 \end{matrix}]

and

Q = B - C = [\begin{matrix} 0 & - 1 & 2 \\ 1 & 0 & 1 \\ - 2 & - 1 & 0 \end{matrix}]

, then

T r (B C) =

Clue 1 of 3: Matrix Algebra

Recognize that the given equations are linear combinations of matrices B and C, which can be solved to find B and C in terms of P and Q.

🧠 Full Solution Path

Step 1: Express B and C in terms of P and Q, then form BC✦ Active

Given the matrix equations $P = B + C$ and $Q = B - C$ . We can solve these linear equations for matrices B and C:

(B + C) + (B - C) = P + Q \Rightarrow 2 B = P + Q \Rightarrow B = \frac{1}{2} (P + Q)

(B + C) - (B - C) = P - Q \Rightarrow 2 C = P - Q \Rightarrow C = \frac{1}{2} (P - Q)

Now, we can express the product $B C$ in terms of $P$ and $Q$ :

B C = (\frac{1}{2} (P + Q)) (\frac{1}{2} (P - Q)) = \frac{1}{4} (P + Q) (P - Q)

Expanding the product $(P + Q) (P - Q)$ , we get:

B C = \frac{1}{4} (P^{2} - P Q + Q P - Q^{2})

💡 Teacher's Secret Hint

Remember that for matrices, $(P + Q) (P - Q)$ is not generally $P^{2} - Q^{2}$ unless the matrices $P$ and $Q$ commute (i.e., $P Q = Q P$ ). So, the intermediate terms $- P Q + Q P$ must be included.

Step 2: Apply Trace Properties to Simplify the Expression○ Expand

We need to find $T r (B C)$ . We can use the linearity of the trace operator ( $T r (A + B) = T r (A) + T r (B)$ and $T r (k A) = k T r (A)$ ) and the cyclic property ( $T r (A B) = T r (B A)$ ):

T r (B C) = T r (\frac{1}{4} (P^{2} - P Q + Q P - Q^{2}))

T r (B C) = \frac{1}{4} (T r (P^{2}) - T r (P Q) + T r (Q P) - T r (Q^{2}))

Since $T r (P Q) = T r (Q P)$ , the terms $- T r (P Q)$ and $+ T r (Q P)$ cancel each other out, simplifying the expression significantly:

T r (B C) = \frac{1}{4} (T r (P^{2}) - T r (Q^{2}))

💡 Teacher's Secret Hint

This key property, $T r (A B) = T r (B A)$ , is very powerful in simplifying matrix trace problems. It allows us to avoid calculating the full product $P Q$ and $Q P$ .

Step 3: Calculate

T r (P^{2})

○ Expand

First, we calculate the trace of $P^{2}$ . We only need the diagonal elements of $P^{2}$ for this. The matrix $P$ is given by:

P = [\begin{matrix} 2 & 5 & 0 \\ 5 & 2 & 3 \\ 0 & 3 & 12 \end{matrix}]

The diagonal elements of $P^{2}$ are:

p_{11}^{2} = (2) (2) + (5) (5) + (0) (0) = 4 + 25 + 0 = 29

p_{22}^{2} = (5) (5) + (2) (2) + (3) (3) = 25 + 4 + 9 = 38

p_{33}^{2} = (0) (0) + (3) (3) + (12) (12) = 0 + 9 + 144 = 153

The trace of $P^{2}$ is the sum of these diagonal elements:

T r (P^{2}) = 29 + 38 + 153 = 220

💡 Teacher's Secret Hint

To find $T r (A^{2})$ , you only need to compute the dot product of each row vector with its corresponding column vector for the diagonal elements. You don't need to compute the entire $A^{2}$ matrix.

Step 4: Calculate

T r (Q^{2})

○ Expand

Next, we calculate the trace of $Q^{2}$ . The matrix $Q$ is given by:

Q = [\begin{matrix} 0 & - 1 & 2 \\ 1 & 0 & 1 \\ - 2 & - 1 & 0 \end{matrix}]

The diagonal elements of $Q^{2}$ are:

q_{11}^{2} = (0) (0) + (- 1) (1) + (2) (- 2) = 0 - 1 - 4 = - 5

q_{22}^{2} = (1) (- 1) + (0) (0) + (1) (- 1) = - 1 + 0 - 1 = - 2

q_{33}^{2} = (- 2) (2) + (- 1) (1) + (0) (0) = - 4 - 1 + 0 = - 5

The trace of $Q^{2}$ is the sum of these diagonal elements:

T r (Q^{2}) = - 5 - 2 - 5 = - 12

💡 Teacher's Secret Hint

Pay close attention to negative signs during multiplication, as a small error can lead to an incorrect final answer.

Step 5: Calculate

T r (B C)

○ Expand

Finally, substitute the calculated values of $T r (P^{2})$ and $T r (Q^{2})$ into the simplified formula for $T r (B C)$ :

T r (B C) = \frac{1}{4} (T r (P^{2}) - T r (Q^{2}))

T r (B C) = \frac{1}{4} (220 - (- 12))

T r (B C) = \frac{1}{4} (220 + 12)

T r (B C) = \frac{1}{4} (232)

T r (B C) = 58

Thus, the trace of $B C$ is $58$ .

💡 Teacher's Secret Hint

Double-check your arithmetic, especially when dealing with subtraction of negative numbers. A common mistake is $220 - (- 12)$ becoming $220 - 12$ instead of $220 + 12$ .

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Maths Question 4 – AP-EAMCET 2026

Clue 1 of 3: Matrix Algebra

🧠 Full Solution Path