What is the value of ≤ft|{array}{lll}{a}₂1} & {a}₃1} & {a}₁1} \\ {a}₂3} & {a}₃3} & {a}₁3} \\ {a}₂2} & {a}₃2} & {a}₁2}{array}| ?

Correct Answer is : −Δ-Δ−Δ

Correct Answer is : −Δ-Δ−Δ

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UPSC NDA 2 2022 Mathematics Solved Paper

Question Numbers: 28-30

Let Δ be the determinant of a matrix A, where A =

\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right]

and C11, C12, C13 be the cofactors of a11, a12, a13 respectively.

Question : 30 of 120

Marks: +1, -0

What is the value of

\left|\begin{array}{lll}\mathrm{a}_{21} & \mathrm{a}_{31} & \mathrm{a}_{11} \\ \mathrm{a}_{23} & \mathrm{a}_{33} & \mathrm{a}_{13} \\ \mathrm{a}_{22} & \mathrm{a}_{32} & \mathrm{a}_{12}\end{array}\right|

Solution:

Concept:

Interchanging two rows or two columns of a determinant changes its sign.

The determinant of a matrix equals the determinant of its transpose.

Explanation:

Let the original determinant be

\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}

First, take the transpose:

\begin{vmatrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \end{vmatrix} = \Delta

Now interchange column 1 and column 3: the determinant becomes

-\Delta

The matrix becomes

\begin{vmatrix} a_{31} & a_{21} & a_{11} \\ a_{32} & a_{22} & a_{12} \\ a_{33} & a_{23} & a_{13} \end{vmatrix} = -\Delta

Next, interchange column 1 and column 2 in this new matrix:

-\Delta

changes sign to

\Delta

We get

\begin{vmatrix} a_{21} & a_{31} & a_{11} \\ a_{22} & a_{32} & a_{12} \\ a_{23} & a_{33} & a_{13} \end{vmatrix} = \Delta

Finally, interchange row 2 and row 3:

\Delta

changes sign to

-\Delta

Thus

\begin{vmatrix} a_{21} & a_{31} & a_{11} \\ a_{23} & a_{33} & a_{13} \\ a_{22} & a_{32} & a_{12} \end{vmatrix} = -\Delta

Answer:

-\Delta

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