# Thread: Find the value of 'q' in the matrix

1. ## Find the value of 'q' in the matrix

Question : If P= $\displaystyle \begin{bmatrix}1 & -1 \\ 2 & -1\end{bmatrix}$ and Q = $\displaystyle \begin{bmatrix} 1 & 1 \\ q & -1\end{bmatrix}$ and $\displaystyle (P+Q)^2 = P^2 + Q^2$ , determine the value of q

2. Originally Posted by zorro
Question : If P= $\displaystyle \begin{bmatrix}1 & -1 \\ 2 & -1\end{bmatrix}$ and Q = $\displaystyle \begin{bmatrix} 1 & 1 \\ q & -1\end{bmatrix}$ and $\displaystyle (P+Q)^2 = P^2 + Q^2$ , determine the value of q
What you have boils down to finding the value of q that satisfies $\displaystyle PQ + QP = 0 \Rightarrow PQ = -QP$. So do the necessary multiplications (it's assumed you can do this without trouble) and then equate the appropriate matrix elements to get an equation in q.

3. ## Please let me know if the value of q = 0

Originally Posted by mr fantastic
What you have boils down to finding the value of q that satisfies $\displaystyle PQ + QP = 0 \Rightarrow PQ = -QP$. So do the necessary multiplications (it's assumed you can do this without trouble) and then equate the appropriate matrix elements to get an equation in q.

Please let me know if the value of q = 0 or not

4. Originally Posted by zorro
Please let me know if the value of q = 0 or not
Have you substituted the value of q and checked that it works? If you post your work I will review it.

5. $\displaystyle (P+Q)^2 = P^2 + Q^2\Leftrightarrow$ $\displaystyle 2PQ=0\Leftrightarrow P=0$ or $\displaystyle Q=0$

6. Originally Posted by Raoh
$\displaystyle (P+Q)^2 = P^2 + Q^2\Leftrightarrow$ $\displaystyle 2PQ=0\Leftrightarrow P=0$ or $\displaystyle Q=0$
NO! matrices do not satisfy the cancellation law. It is quite possible to have PQ= 0 without either P or Q equal to 0.

For example, if $\displaystyle P= \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}$ and $\displaystyle Q= \begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix}$ then neither is the 0 matrix but $\displaystyle PQ= \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$.

7. Matrices aren't always commutative, as well... So $\displaystyle PQ + QP = 0 \Leftrightarrow 2PQ = 0$ isn't neccesarily correct. Just do what Mr.F said and you should be fine.

8. Originally Posted by HallsofIvy
NO! matrices do not satisfy the cancellation law. It is quite possible to have PQ= 0 without either P or Q equal to 0.

For example, if $\displaystyle P= \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}$ and $\displaystyle Q= \begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix}$ then neither is the 0 matrix but $\displaystyle PQ= \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$.
thank you

9. zorro, just go ahead and do the computation.

What is P+ Q? What is (P+Q)^2?

What are P^2 and Q^2? What is P^2+ Q^2?

Set (P+ Q)^2 equal to P^2+ Q^2 and set terms equal component by component. You will have four equations in q. Do they all give the same value for q?

10. ## Is this correct?

Originally Posted by mr fantastic
Have you substituted the value of q and checked that it works? If you post your work I will review it.

$\displaystyle P \ = \ \begin{bmatrix}1 & -1 \\ 2 & -1 \end{bmatrix}$

$\displaystyle Q \ = \ \begin{bmatrix}1 & 1 \\ q & -1 \end{bmatrix}$

$\displaystyle \therefore (P + Q) \ = \ \begin{bmatrix} 1 & 0 \\ (2 + q) & -2 \end{bmatrix}$

$\displaystyle (P + Q)^2 \ = \ \begin{bmatrix} 1 & 0 \\ (q + 2)^2 & 4 \end{bmatrix} \ = \ 4 - (q + 2)^2 \ = \ [- q^2 - 4q] \longmapsto eq\ 1$

$\displaystyle P^2 \ = \ \begin{bmatrix} 1 & 1 \\ 4 & 1 \end{bmatrix}$

$\displaystyle Q^2 \ = \ \begin{bmatrix} 1 & 1 \\ q^2 & 1 \end{bmatrix}$

$\displaystyle P^2 + Q^2 \ = \ \begin{bmatrix} 2 & 2 \\ (4 + q^2) & 2 \end{bmatrix} \ = \ 4 - 2(4 + q^2) \ = \ [-4 - 2q^2] \longmapsto eq \ 2$

Since

$\displaystyle (P + Q)^2 \ = \ (P^2 + Q^2)$ From eq 1 and eq 2

$\displaystyle -q^2 - 4q \ = \ -4 - 2q^2$

$\displaystyle -q^2 + 2q^2 - 4q \ = \ -4$

$\displaystyle q^2 - 4q + 4 \ = \ 0$

$\displaystyle (q - 2)(q - 2)$

$\displaystyle \therefore$ $\displaystyle q = 2$

Is this correct ????????

11. Originally Posted by zorro
$\displaystyle P \ = \ \begin{bmatrix}1 & -1 \\ 2 & -1 \end{bmatrix}$

$\displaystyle Q \ = \ \begin{bmatrix}1 & 1 \\ q & -1 \end{bmatrix}$

$\displaystyle \therefore (P + Q) \ = \ \begin{bmatrix} {\color{red}1} & 0 \\ (2 + q) & -2 \end{bmatrix}$ Mr F says: That red 1 should be a 2.

$\displaystyle (P + Q)^2 \ = \ \begin{bmatrix} 1 & 0 \\ (q + 2)^2 & 4 \end{bmatrix}$ Mr F says: This is completely wrong. Please go back and review how to multiply two matrices (which is what you're doing when you take the square of a matrix).

$\displaystyle \ = \ 4 - (q + 2)^2 \ = \ [- q^2 - 4q] \longmapsto eq\ 1$ Mr F says: What is this meant to be? What is its relevance?

$\displaystyle P^2 \ = \ \begin{bmatrix} 1 & 1 \\ 4 & 1 \end{bmatrix}$

$\displaystyle Q^2 \ = \ \begin{bmatrix} 1 & 1 \\ q^2 & 1 \end{bmatrix}$

Mr F says: The above is completely wrong. Please go back and review how to multiply two matrices (which is what you're doing when you take the square of a matrix).

$\displaystyle P^2 + Q^2 \ = \ \begin{bmatrix} 2 & 2 \\ (4 + q^2) & 2 \end{bmatrix} \ = \ 4 - 2(4 + q^2) \ = \ [-4 - 2q^2] \longmapsto eq \ 2$

Since

$\displaystyle (P + Q)^2 \ = \ (P^2 + Q^2)$ From eq 1 and eq 2

$\displaystyle -q^2 - 4q \ = \ -4 - 2q^2$

$\displaystyle -q^2 + 2q^2 - 4q \ = \ -4$

$\displaystyle q^2 - 4q + 4 \ = \ 0$

$\displaystyle (q - 2)(q - 2)$

$\displaystyle \therefore$ $\displaystyle q = 2$

Is this correct ????????
There are many mistakes you need to fix. Because of these mistakes, all working that follows from them will be wrong.

12. ## I have reworked the problem could u please check

Originally Posted by mr fantastic
There are many mistakes you need to fix. Because of these mistakes, all working that follows from them will be wrong.

Is it right now

$\displaystyle (P + Q)^2$ = $\displaystyle \begin{bmatrix} 2 & 0 \\ (2 +q) & -2 \end{bmatrix}$ . $\displaystyle \begin{bmatrix} 2 & 0 \\ (2 +q) & -2 \end{bmatrix}$ = $\displaystyle \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}$

$\displaystyle P^2$ = $\displaystyle \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}$ . $\displaystyle \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}$ = $\displaystyle \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$ $\displaystyle ; \qquad \qquad Q^2$ = $\displaystyle \begin{bmatrix} 1 & 1 \\ q & -1 \end{bmatrix}$ . $\displaystyle \begin{bmatrix} 1 & 1 \\ q & -1 \end{bmatrix}$ = $\displaystyle \begin{bmatrix} (1-q) & 0 \\ 0 & (1+q) \end{bmatrix}$

$\displaystyle (P^2 + Q^2)$ = $\displaystyle \begin{bmatrix} -q & 0 \\ 0 & q \end{bmatrix}$

$\displaystyle (P^2 + Q^2) = P^2 + Q^2$

$\displaystyle 16 = 1+1-q^2$

$\displaystyle 16 = 2 - q^2$

$\displaystyle q^2 = 2 - 16$

$\displaystyle q^2 = -14$

$\displaystyle q = - \sqrt{14}$

Is this correct ????

13. Originally Posted by zorro
Is it right now

$\displaystyle (P + Q)^2$ = $\displaystyle \begin{bmatrix} 2 & 0 \\ (2 +q) & -2 \end{bmatrix}$ . $\displaystyle \begin{bmatrix} 2 & 0 \\ (2 +q) & -2 \end{bmatrix}$ = $\displaystyle \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}$

$\displaystyle P^2$ = $\displaystyle \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}$ . $\displaystyle \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}$ = $\displaystyle \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$ $\displaystyle ; \qquad \qquad Q^2$ = $\displaystyle \begin{bmatrix} 1 & 1 \\ q & -1 \end{bmatrix}$ . $\displaystyle \begin{bmatrix} 1 & 1 \\ q & -1 \end{bmatrix}$ = $\displaystyle \begin{bmatrix} (1 {\color{red}+}q) & 0 \\ 0 & (1+q) \end{bmatrix}$

$\displaystyle (P^2 + Q^2)$ = $\displaystyle \begin{bmatrix} -q & 0 \\ 0 & q \end{bmatrix}$

$\displaystyle (P^2 + Q^2) = P^2 + Q^2$

$\displaystyle 16 = 1+1-q^2$

$\displaystyle 16 = 2 - q^2$

$\displaystyle q^2 = 2 - 16$

$\displaystyle q^2 = -14$

$\displaystyle q = - \sqrt{14}$

Is this correct ????
Look carefully - there is a correction in red to your expression for Q^2. That will change everything that follows.