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How to find determinant of a 4th order matrix

How to find determinant of a 4th order matrix

Grade:12th Pass

1 Answers

ruchi yadav
askIITians Faculty 27 Points
10 years ago
Any matrixAand its transpose have the same determinant, meaning

\begin{displaymath}\det A = \det A^T.\end{displaymath}

2.
The determinant of a triangular matrix is the product of the entries on the diagonal.
3.
If we interchange two rows, the determinant of the new matrix is the opposite of the old one.
4.
If we multiply one row with a constant, the determinant of the new matrix is the determinant of the old one multiplied by the constant.
5.
If we add one row to another one multiplied by a constant, the determinant of the new matrix is the same as the old one.
6.
We have

\begin{displaymath}\det(AB) = \det(A) \det(B).\end{displaymath}

In particular, ifAis invertible (which happens if and only if$\det(A) \neq 0$), then

\begin{displaymath}\det(A^{-1}) = \frac{1}{\det(A)}.\end{displaymath}

So let us see how this works in case of a matrix of order 4.

Example.Evaluate

\begin{displaymath}\left\vert\begin{array}{cccc} 1&2&3&4\\ 5&6&7&8\\ 2&6&4&8\\ 3&1&1&2\\ \end{array}\right\vert.\end{displaymath}

We have

\begin{displaymath}\left\vert\begin{array}{cccc} 1&2&3&4\\ 5&6&7&8\\ 2&6&4&8\\... ...3&4\\ 5&6&7&8\\ 1&3&2&4\\ 3&1&1&2\\ \end{array}\right\vert.\end{displaymath}

If we subtract every row multiplied by the appropriate number from the first row, we get

\begin{displaymath}\left\vert\begin{array}{cccc} 1&2&3&4\\ 5&6&7&8\\ 1&3&2&4\\... ...-4&-8&-12\\ 0&1&-1&0\\ 0&-5&-8&-10\\ \end{array}\right\vert.\end{displaymath}

We do not touch the first row and work with the other rows. We interchange the second with the third to get

\begin{displaymath}\left\vert\begin{array}{rrrr} 1&2&3&4\\ 0&-4&-8&-12\\ 0&1&-... ...1&-1&0\\ 0&-4&-8&-12\\ 0&-5&-8&-10\\ \end{array}\right\vert.\end{displaymath}

If we subtract every row multiplied by the appropriate number from the second row, we get

\begin{displaymath}\left\vert\begin{array}{rrrr} 1&2&3&4\\ 0&1&-1&0\\ 0&-4&-8&... ...1&-1&0\\ 0&0&-12&-12\\ 0&0&-13&-10\\ \end{array}\right\vert.\end{displaymath}

Using previous properties, we have

\begin{displaymath}\left\vert\begin{array}{rrrr} 1&2&3&4\\ 0&1&-1&0\\ 0&0&-12&... ... 0&1&-1&0\\ 0&0&1&1\\ 0&0&-13&-10\\ \end{array}\right\vert.\end{displaymath}

If we multiply the third row by 13 and add it to the fourth, we get

\begin{displaymath}\left\vert\begin{array}{rrrr} 1&2&3&4\\ 0&1&-1&0\\ 0&0&1&1\... ...3&4\\ 0&1&-1&0\\ 0&0&1&1\\ 0&0&0&3\\ \end{array}\right\vert\end{displaymath}

which is equal to 3. Putting all the numbers together, we get

\begin{displaymath}\left\vert\begin{array}{cccc} 1&2&3&4\\ 5&6&7&8\\ 2&6&4&8\\... ...\end{array}\right\vert = 2 \cdot (-1) \cdot (-12) \cdot 3 = 72.\end{displaymath}



Thank You
Ruchi
Askiitians Faculty

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