1.

Evaluate \(\begin{vmatrix}2 &5 &8 \\ 6&10 &9\\ 4& 6& 1 \end{vmatrix}\)1. 302. 153. 244. 21

Answer» Correct Answer - Option 1 : 30

Concept:

Let A is any square matrix, A = \(\begin{vmatrix} a_{11} &a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix}\) 

|A| = a11(a22 a33 - a23 a32) - a12(a21 a33 - a23 a31) + a13(a21 a32 - a22 a31)

 

 

Calculation:

|A| = \(\begin{vmatrix} 2 & 5 & 8 \\ 6 & 10 & 9 \\ 4 &6& 1 \end{vmatrix}\) 

R2 = R2 - R1

⇒ |A| = \(\begin{vmatrix} 2 & 5 & 8 \\ 4 & 5 & 1 \\ 4 &6& 1 \end{vmatrix}\)

R2 = R2 - R3

⇒ |A| = \(\begin{vmatrix} 2 & 5 & 8 \\ 0 & -1 & 0 \\ 4 &6& 1 \end{vmatrix}\)

Expanding along  R1

⇒ |A| = 2(-1) - 5(0) + 8(-(-4))

⇒ |A| = -2 + 32

⇒ |A| = 30

 

Any scalar multiplied to a row or a column can be taken out, e.g. 

\(\begin{vmatrix} ka_{11} &ka_{12} & ka_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix}\) = k\(\begin{vmatrix} a_{11} &a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix}\) = k[a11(a22 a33 - a23 a32) - a12(a21 a33 - a23 a31) + a13(a21 a32 - a22 a31)]

Addition or subtraction in a row or column can be written as:

\(\begin{vmatrix} a_{11}+z &a_{12} & a_{13}\\ a_{21}+y & a_{22} & a_{23}\\ a_{31}+x & a_{32} & a_{33} \end{vmatrix}\) = \(\begin{vmatrix} a_{11} &a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix}\)\(\begin{vmatrix} z &a_{12} & a_{13}\\ y & a_{22} & a_{23}\\ x & a_{32} & a_{33} \end{vmatrix}\)

Area of a triangle with points (a, b), (x, y) and (p, q) = \({1\over2}\begin{vmatrix} a &b & 1\\ x & y& 1\\ p & q & 1 \end{vmatrix}\)

Value of determinant does not change with the row or column additions or subtractions within themselves.

If Two rows or columns are same or multiple of each other then the value of the determinant is zero.

The determinant can be possible only for a square matrix. 



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