beljuA

2020-12-03

If A and B are both $n\times n$ matrices (of the same size), then

det(A+B)=det(A)+det(B)

True or False?

det(A+B)=det(A)+det(B)

True or False?

stuth1

Skilled2020-12-04Added 97 answers

Step 1

In general, the given identity det(A+B)=det(A)+det(B) does not hold true. To prove the given identity is wrong, it is enough to give a counter example. Consider the matrices$A=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\text{and}B=\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]$

Evaluate det(A) as follows.

$det(A)=\left|\begin{array}{cc}1& 0\\ 0& 0\end{array}\right|=0$

Evaluate det(B) as follows.

$det(B)=\left|\begin{array}{cc}0& 0\\ 0& 1\end{array}\right|=0$

Step 2

Evaluate det(A)+det(B) as follows.

det(A)+det(B)=0+0=0

Thus, det(A)+det(B)=0

Evaluate A+B as follows.

$A+B=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]+\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]=$

$=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Evaluate det(A+B) as follows.

$det(A+B)=\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|$

=1

Thus, det(A+B)=1

Hence it is proved that$det(A)+det(B)\ne det(A+B)$

Therefore, the given statement is FALSE.

In general, the given identity det(A+B)=det(A)+det(B) does not hold true. To prove the given identity is wrong, it is enough to give a counter example. Consider the matrices

Evaluate det(A) as follows.

Evaluate det(B) as follows.

Step 2

Evaluate det(A)+det(B) as follows.

det(A)+det(B)=0+0=0

Thus, det(A)+det(B)=0

Evaluate A+B as follows.

Evaluate det(A+B) as follows.

=1

Thus, det(A+B)=1

Hence it is proved that

Therefore, the given statement is FALSE.

Jeffrey Jordon

Expert2022-01-29Added 2605 answers

Answer is given below (on video)

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