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Byju's Answer

Standard XII

Mathematics

Evaluation of a Determinant

Consider the ...

Question

Consider the determinant $Δ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣$
$M_{i j} =$ Minor of the element of $i^{t h}$ row & $j^{t h}$ column.
$C_{i j} =$ Cofactor of element of $i^{t h}$ row & $j^{t h}$ column.
Value of $b_{1} . C_{31} + b_{2} . C_{32} + b_{3} . C_{33}$ is

0

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Δ

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2 Δ

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Δ^{2}

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Solution

The correct option is A $0$
Value of $b_{1} . C_{31} + b_{2} . C_{32} + b_{3} . C_{33}$

$= b_{1} . {(- 1)}^{3 + 1} ∣ ∣ ∣ \begin{matrix} a_{2} & a_{3} b_{2} & b_{3} \end{matrix} ∣ ∣ ∣ + b_{3} . {(- 1)}^{3 + 2} ∣ ∣ ∣ \begin{matrix} a_{1} & a_{3} b_{1} & b_{3} \end{matrix} ∣ ∣ ∣ + b_{3} . {(- 1)}^{3 + 3} ∣ ∣ ∣ \begin{matrix} a_{1} & a_{3} b_{1} & b_{2} \end{matrix} ∣ ∣ ∣$

$= b_{1} (b_{2} a_{3} - a_{2} b_{3}) - b_{2} (b_{1} a_{3} - a_{1} b_{3}) + b_{3} (b_{1} a_{3} - a_{1} b_{2})$

$= 0$

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Similar questions

Consider the determinant

Δ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣

M_{i j} =

Minor of the element of

i^{t h}

row &

j^{t h}

column.

C_{i j} =

Cofactor of element of

i^{t h}

row &

j^{t h}

column.

a_{2} . C_{12} + b_{2} . C_{22} + c_{2} . C_{32}

is equal to

Consider the determinant

Δ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣

M_{i j} =

Minor of the element of

i^{t h}

row &

j^{t h}

column.

C_{i j} =

Cofactor of element of

i^{t h}

row &

j^{t h}

column.

a_{3} M_{13} - b_{3} M_{23} + c_{3} M_{33}

is equal to

Consider the determinant

Δ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣

M_{i j} =

Minor of the element of

i^{t h}

row &

j^{t h}

column.

C_{i j} =

Cofactor of element of

i^{t h}

row &

j^{t h}

column.

If all the elements of the determinant are multiplied by 2, then the value of new determinant is

Q. If

Δ = ∣ ∣ ∣ ∣ \begin{matrix} a_{11} & a_{12} & a_{13} a_{21} & a_{22} & a_{23} a_{31} & a_{32} & a_{33} \end{matrix} ∣ ∣ ∣ ∣

and

c_{i j} = {(- 1)}^{i + j}

(determinant obtained by deleting ith row and jth column), then

∣ ∣ ∣ ∣ \begin{matrix} c_{11} & c_{12} & c_{13} c_{21} & c_{22} & c_{23} c_{31} & c_{32} & c_{33} \end{matrix} ∣ ∣ ∣ ∣ = Δ^{2}

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x^{2} x & x^{2} & 1 x^{2} & 1 & x \end{matrix} ∣ ∣ ∣ ∣ ∣ = 7

and

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{3} - 1 & 0 & x - x^{4} 0 & x - x^{4} & x^{3} - 1 x - x^{4} & x^{3} - 1 & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣

, then

If A1, B1, C1 … are the cofactors of the elements a1, b1, c1... of the matrix

$⎡ ⎢ ⎣ \begin{matrix} a 1 & b 1 & c 1 a 2 & b 2 & c 2 a 3 & b 3 & c 3 \end{matrix} ⎤ ⎥ ⎦$ then $∣ ∣ ∣ \begin{matrix} B 2 & C 2 B 3 & C 3 \end{matrix} ∣ ∣ ∣ =$

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Consider the determinant Δ=∣∣ ∣∣a1a2a3b1b2b3c1c2c3∣∣ ∣∣Mij= Minor of the element of ith row & jth column.Cij= Cofactor of element of ith row & jth column.Value of b1.C31+b2.C32+b3.C33 is

The correct option is A 0Value of b1.C31+b2.C32+b3.C33=b1.(−1)3+1∣∣∣a2a3b2b3∣∣∣+b3.(−1)3+2∣∣∣a1a3b1b3∣∣∣+b3.(−1)3+3∣∣∣a1a3b1b2∣∣∣=b1(b2a3−a2b3)−b2(b1a3−a1b3)+b3(b1a3−a1b2)=0