GO Classes Test Series 2023 | Linear Algebra | Test | Question: 9

Answer 1

Expanding along the third row, and then along the first row we obtain:
$$
\begin{aligned}
\operatorname{det}(A)=\left|\begin{array}{cccc}
1 & -1 & 0 & 3 \\
2 & 5 & 2 & 6 \\
0 & 1 & 0 & 0 \\
1 & 4 & 2 & 1
\end{array}\right|=-1\left|\begin{array}{ccc}
1 & 0 & 3 \\
2 & 2 & 6 \\
1 & 2 & 1
\end{array}\right|  =-1\left|\begin{array}{cc}
2 & 6 \\
2 & 1
\end{array}\right|-3\left|\begin{array}{cc}
2 & 2 \\
1 & 2
\end{array}\right| \\ = -(2-12)-3(4-2)=10-6=4.
\end{aligned}
$$

Comments

@GO Classes / @Lakshman Patel RJIT

There is a small correction needed in the explanation. First we are expanding along the third row and then along the first column

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@shishir__roy can we convert the matrix in row echelon form and det will be product of diagonal elements?
I am not getting where am I wrong? Can you please see this

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Bro,

$R_3 \leftarrow 7*R_3-5*R_2$ will change value of determinant by a factor of $7$.

Similar,

$R_4 \leftarrow 7*R_4+2*R_3$ will also change value of determinant by a factor of $7$.

You're getting this new determinant $=49*4$.

So, the original determinant should be $=\frac{49*4}{7*7} = 4$.

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@shishir__roy got it, thanks:)