Properties of Determinants

1 Mark Questions

2 7 65<br>

1. Write the value of 3 8 75 All India 2014C<br>

5 9 86

Given, 3 8 75<br>5 9 86 <math display="block">= 2 \begin{vmatrix} 8 & 75 \\ 9 & 86 \end{vmatrix} - 7 \begin{vmatrix} 3 & 75 \\ 5 & 86 \end{vmatrix} + 65 \begin{vmatrix} 3 & 8 \\ 5 & 9 \end{vmatrix}</math>

[expanding the determinant along <math>R_1</math>] <math>= 2 (688 - 675) - 7(258 - 375) + 65 (27 - 40)</math> <math>= 26 + 819 - 845</math> <math>= 845 - 845 = 0</math> (1)

1. Prove the following, using properties of determinants

<math>\begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix}</math><br>= <math>2(a+b+c)^3</math>

Delhi 2014

LHS = <math>\begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix}</math>

On applying <math>C_1 \rightarrow C_1 + C_2 + C_3</math>, we get

LHS = <math>\begin{vmatrix} 2(a+b+c) & a & b \\ 2(a+b+c) & b+c+2a & b \\ 2(a+b+c) & a & c+a+2b \end{vmatrix}</math>

On taking <math>2(a + b + c)</math> common from <math>C_1</math>, we get

LHS = <math>2(a+b+c)</math> | 1

(1/2)

On applying <math>R_2 \rightarrow R_2 - R_1</math> and <math>R_3 \rightarrow R_3 - R_1</math>, we get

LHS = <math>2(a + b + c)</math> <math>\begin{vmatrix} 1 & a & b \\ 0 & b + c + a & 0 \\ 0 & 0 & c + a + b \end{vmatrix}</math>

On taking <math>(a + b + c)</math> common from <math>R_2</math> and <math>R_3</math>, we get

LHS = <math>2(a + b + c)^3 \begin{vmatrix} 1 & a & b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}</math>

On expanding along <math>R_3</math>, we get

LHS = <math>2(a + b + c)^3</math> [(1) (1 – 0)] <math>= 2(a + b + c)^3 = RHS</math> Hence proved.

(1/2)

1. Using properties of determinants, prove that

<math>\begin{vmatrix} x^2 + 1 & xy & xz \\ xy & y^2 + 1 & yz \\ xz & yz & z^2 + 1 \end{vmatrix} = 1 + x^2 + y^2 + z^2.</math> Delhi 2014