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MrRissso [65]
4 years ago
8

The given matrix equation is false in general. Explain why the equation is false. Assume that all matrices are n ✕ n. (A + B)2 =

A2 + 2AB + B2 (A + B)2 = A2 + AB + BA + B2. This is only equal to A2 + 2AB + B2 when AB = −BA, which in general is not true. (A + B)2 = A2 + AB − AB + B2 = A2 + B2. This is only equal to A2 + 2AB + B2 when A = 0nn or B = 0nn, which in general is not true. (A + B)2 = A2 − BA + BA + B2 = A2 + B2. This is only equal to A2 + 2AB + B2 when A = 0nn or B = 0nn, which in general is not true. (A + B)2 = A2 + AB + BA + B2. This is only equal to A2 + 2AB + B2 when AB = BA, which in general is not true. Give a 2 ✕ 2 example showing the equation is false. A = 1 2 3 4
Mathematics
1 answer:
DENIUS [597]4 years ago
8 0

Answer:

Fourth option is the correct one.

Step-by-step explanation:

Given:

A = \left[\begin{array}{ccc}1&2\\3&4\end{array}\right]  

Lets Assume

B = \left[\begin{array}{ccc}4&0\\0&1\end{array}\right]

Now,

if we multiply both in sequence AB, we get

AB = \left[\begin{array}{ccc}1&2\\3&4\end{array}\right] \left[\begin{array}{ccc}4&0\\0&1\end{array}\right]

AB = \left[\begin{array}{ccc}1.4+2.0&1.0+2.1\\3.4+4.0&3.0+4.1\end{array}\right]

AB = \left[\begin{array}{ccc}4&2\\12&4\end{array}\right]

Now if we multiply both in sequence BA, we get

BA = \left[\begin{array}{ccc}4&0\\0&1\end{array}\right] \left[\begin{array}{ccc}1&2\\3&4\end{array}\right]

      =  \left[\begin{array}{ccc}4.1+0.3&4.2+0.4\\0.1+1.3&0.2+1.4\end{array}\right]

BA = \left[\begin{array}{ccc}4&8\\3&4\end{array}\right]

Hence AB ≠ BA

Now according to equation,

(A+B)² = A² + B² + AB + BA = A² + B² + 2AB only when AB = BA which is not the case as seen in example above

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