Question

Question for Brainly Teachers​

Answer 1

Step-by-step explanation:

Q3)

$\left|\begin{array}{ccc}</p><p>1+i&1-i&i\\1-i&i&1+i\\i&1+i&1-i\end{array}\right|$

R1->R1+R2+R3

$\left|\begin{array}{ccc}</p><p>1+i+1-i+i&1-i+i+1+i&i+1+i+1-i\\1-i&i&1+i\\i&1+i&1-i\end{array}\right|$

$\left|\begin{array}{ccc}</p><p>2+i&2+i&2+i\\1-i&i&1+i\\i&1+i&1-i\end{array}\right|$

Take (2+i) common from R1

$(2+i)\left|\begin{array}{ccc}</p><p>1&1&1\\1-i&i&1+i\\i&1+i&1-i\end{array}\right|$

C1->C1-C3

C2->C2-C3

$(2+i)\left|\begin{array}{ccc}</p><p>1-1&1-1&1\\1-i-1-i&i-1-i&1+i\\i-1+i&1+i-1+i&1-i\end{array}\right|$

$(2+i)\left|\begin{array}{ccc}</p><p>0&0&1\\-2i&-1&1+i\\2i-1&2i&1-i\end{array}\right|$

Expand the determinant along R1

$= (2 + i)[- 2i(2i) + (2i - 1)]\\ \\ = (2 + i)( - 4 {i}^{2} + 2i - 1) \\ \\ = (2 + i)( - 4( - 1) + 2i - 1) \\ \\ = (2 + i)(3 + 2i) \\ \\ = 6 + 4i + 3i + 2 {i}^{2} \\ \\ = 6 + 7i - 2 \\ \\ \because \: {i}^{2} = - 1 \\ \\ \left|\begin{array}{ccc}</p><p>1+i&1-i&i\\1-i&i&1+i\\i&1+i&1-i\end{array}\right| = 4 + 7i$

Hope it helps you.

Answer 2

Step-by-step explanation:

Q3)

$$\begin{lgathered}\left|\begin{array}{ccc} 1+i&1-i&i\\1-i&i&1+i\\i&1+i&1-i\end{array}\right|\end{lgathered}$$

R1->R1+R2+R3

$$\begin{lgathered}\left|\begin{array}{ccc} 1+i+1-i+i&1-i+i+1+i&i+1+i+1-i\\1-i&i&1+i\\i&1+i&1-i\end{array}\right|\end{lgathered}$$

$$\begin{lgathered}\left|\begin{array}{ccc} 2+i&2+i&2+i\\1-i&i&1+i\\i&1+i&1-i\end{array}\right|\end{lgathered}$$

Take (2+i) common from R1

$$\begin{lgathered}(2+i)\left|\begin{array}{ccc} 1&1&1\\1-i&i&1+i\\i&1+i&1-i\end{array}\right|\end{lgathered}$$

C1->C1-C3

C2->C2-C3

$$\begin{lgathered}(2+i)\left|\begin{array}{ccc} 1-1&1-1&1\\1-i-1-i&i-1-i&1+i\\i-1+i&1+i-1+i&1-i\end{array}\right|\end{lgathered}$$

$$\begin{lgathered}(2+i)\left|\begin{array}{ccc} 0&0&1\\-2i&-1&1+i\\2i-1&2i&1-i\end{array}\right|\end{lgathered}$$

Expand the determinant along R1

$$\begin{lgathered}= (2 + i)[- 2i(2i) + (2i - 1)]\\ \\ = (2 + i)( - 4 {i}^{2} + 2i - 1) \\ \\ = (2 + i)( - 4( - 1) + 2i - 1) \\ \\ = (2 + i)(3 + 2i) \\ \\ = 6 + 4i + 3i + 2 {i}^{2} \\ \\ = 6 + 7i - 2 \\ \\ \because \: {i}^{2} = - 1 \\ \\ \left|\begin{array}{ccc} 1+i&1-i&i\\1-i&i&1+i\\i&1+i&1-i\end{array}\right| = 4 + 7i \\ \\\end{lgathered}$$

Hope it helps you.

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