JEE Main 2021 — Mathematics Algebra

Let $A=[\begin{matrix}{a}_{11} & {a}_{12} & {a}_{13} \\ {a}_{21} & {a}_{22} & {a}_{23} \\ {a}_{31} & {a}_{32} & {a}_{33}\end{matrix}]$A=[a11​a21​a31​​a12​a22​a32​​a13​a23​a33​​]$\Rightarrow 2A=[\begin{matrix}2{a}_{11} & 2{a}_{12} & 2{a}_{13} \\ 2{a}_{21} & 2{a}_{22} & 2{a}_{23} \\ 2{a}_{31} & 2{a}_{32} & 2{a}_{33}\end{matrix}]$⇒2A=[2a11​2a21​2a31​​2a12​2a22​2a32​​2a13​2a23​2a33​​]

$${R}_{2}\rightarrow 2{R}_{2}+5{R}_{3}$$R2​→2R2​+5R3​

$$B=[\begin{matrix}2{a}_{11} & 2{a}_{12} & 2{a}_{13} \\ 4{a}_{21}+10{a}_{31} & 4{a}_{22}+10{a}_{32} & 4{a}_{23}+10{a}_{33} \\ 2{a}_{31} & 2{a}_{32} & 2{a}_{33}\end{matrix}]$$B=[2a11​4a21​+10a31​2a31​​2a12​4a22​+10a32​2a32​​2a13​4a23​+10a33​2a33​​]

$$|B|=|\begin{matrix}2{a}_{11} & 2{a}_{12} & 2{a}_{13} \\ 4{a}_{21}+10{a}_{31} & 4{a}_{22}+10{a}_{32} & 4{a}_{23}+10{a}_{33} \\ 2{a}_{31} & 2{a}_{32} & 2{a}_{33}\end{matrix}|$$∣B∣=∣2a11​4a21​+10a31​2a31​​2a12​4a22​+10a32​2a32​​2a13​4a23​+10a33​2a33​​∣

$$|B|=|\begin{matrix}2{a}_{11} & 2{a}_{12} & 2{a}_{13} \\ 4{a}_{21} & 4{a}_{22} & 4{a}_{23} \\ 2{a}_{31} & 2{a}_{32} & 2{a}_{33}\end{matrix}|+|\begin{matrix}2{a}_{11} & 2{a}_{12} & 2{a}_{13} \\ 10{a}_{31} & 10{a}_{32} & 10{a}_{33} \\ 2{a}_{31} & 2{a}_{32} & 2{a}_{33}\end{matrix}|$$∣B∣=∣2a11​4a21​2a31​​2a12​4a22​2a32​​2a13​4a23​2a33​​∣+∣2a11​10a31​2a31​​2a12​10a32​2a32​​2a13​10a33​2a33​​∣

$$|B|=2\times 4\times 2|\begin{matrix}{a}_{11} & {a}_{12} & {a}_{13} \\ {a}_{21} & {a}_{22} & {a}_{23} \\ {a}_{31} & {a}_{32} & {a}_{33}\end{matrix}|+2\times 10\times 2|\begin{matrix}{a}_{11} & {a}_{12} & {a}_{13} \\ {a}_{31} & {a}_{32} & {a}_{33} \\ {a}_{31} & {a}_{32} & {a}_{33}\end{matrix}|$$∣B∣=2×4×2∣a11​a21​a31​​a12​a22​a32​​a13​a23​a33​​∣+2×10×2∣a11​a31​a31​​a12​a32​a32​​a13​a33​a33​​∣

$$|B|=16|\begin{matrix}{a}_{11} & {a}_{12} & {a}_{13} \\ {a}_{21} & {a}_{22} & {a}_{23} \\ {a}_{31} & {a}_{32} & {a}_{33}\end{matrix}|+0$$∣B∣=16∣a11​a21​a31​​a12​a22​a32​​a13​a23​a33​​∣+0

$$|B|=16|A|$$∣B∣=16∣A∣

Value of the second determinant is zero. Since Row $2$2 and Row $3$3 are identical.

$$|B|=16\times 4=64(\because |A|=4)$$∣B∣=16×4=64(∵∣A∣=4)