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Evaluate the determinant \(\left|\begin{array}{rrr} {2} & {-1} & {-2} \\ {0} & {2} & {-1} \\ {3} & {-5} & {0} \end{array}\right|\) 
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Evaluate the determinant \(\left|\begin{array}{rrr} {2} & {-1} & {-2} \\ {0} & {2} & {-1} \\ {3} & {-5} & {0} \end{array}\right|\) 

Answer

ATQ,
We know that a determinant of a 3 \(\times\) 3 matrix is calculated as
\(|A|=\left|\begin{array}{ccc} {i} & {j} & {k} \\ {a} & {b} & {c} \\ {d} & {e} & {f} \end{array}\right|=i\left|\begin{array}{cc} {b} & {c} \\ {e} & {f} \end{array}\right|-j\left|\begin{array}{cc} {a} & {c} \\ {d} & {f} \end{array}\right|+k\left|\begin{array}{cc} {a} & {b} \\ {d} & {e} \end{array}\right|\)
\(2\left|\begin{array}{cc} {2} & {-1} \\ {-5} & {0} \end{array}\right|-(-1)\left|\begin{array}{cc} {0} & {-1} \\ {3} & {0} \end{array}\right|+(-2)\left|\begin{array}{cc} {0} & {2} \\ {3} & {-5} \end{array}\right|\)
= 2[0 – (-1)(-5)] + 1[0 – (-1)(3)] – 2[0 – 3(2)]
= 2[0 – 5] + 1[0 + 3] – 2[-6]
= 2[-5] + 1[3] -2[-6]
= -10 + 3 + 12
= 5
The determinant of the above matrix is 5.
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