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Worked Examples · Example 27

Q.Evaluate Δ=∣422579792953∣\Delta = \begin{vmatrix} 42 & 2 & 5 \\ 79 & 7 & 9 \\ 29 & 5 & 3 \end{vmatrix}.

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✓ Free question

Expanding the determinant along the first row, the three terms sum to zero — the first column is a linear combination of the other two, so Δ=0\Delta = 0.

Δ=a11∣a22a23a32a33∣−a12∣a21a23a31a33∣+a13∣a21a22a31a32∣.\Delta = a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}.

  1. Expand Δ=∣422579792953∣\Delta = \begin{vmatrix} 42 & 2 & 5 \\ 79 & 7 & 9 \\ 29 & 5 & 3 \end{vmatrix} along Row 1:

Δ=42∣7953∣−2∣799293∣+5∣797295∣.\Delta = 42\begin{vmatrix} 7 & 9 \\ 5 & 3 \end{vmatrix} - 2\begin{vmatrix} 79 & 9 \\ 29 & 3 \end{vmatrix} + 5\begin{vmatrix} 79 & 7 \\ 29 & 5 \end{vmatrix}.

  1. Evaluate the 2×22\times2 minors:

∣7953∣=21−45=−24,∣799293∣=237−261=−24,∣797295∣=395−203=192.\begin{vmatrix} 7 & 9 \\ 5 & 3 \end{vmatrix} = 21 - 45 = -24,\quad \begin{vmatrix} 79 & 9 \\ 29 & 3 \end{vmatrix} = 237 - 261 = -24,\quad \begin{vmatrix} 79 & 7 \\ 29 & 5 \end{vmatrix} = 395 - 203 = 192.

  1. Substitute:

Δ=42(−24)−2(−24)+5(192)=−1008+48+960.\Delta = 42(-24) - 2(-24) + 5(192) = -1008 + 48 + 960.

  1. Add:   −1008+48+960=0.\;-1008 + 48 + 960 = 0.
  2. (Reason: column C1=1⋅C2+8⋅C3C_1 = 1\cdot C_2 + 8\cdot C_3, since 2(2)+5(8)...2(2)+5(8)... i.e. 1⋅2+8⋅5=421\cdot 2 + 8\cdot 5 = 42, 1⋅7+8⋅9=791\cdot 7 + 8\cdot 9 = 79, 1⋅5+8⋅3=291\cdot 5 + 8\cdot 3 = 29 — a dependent column forces Δ=0\Delta = 0.)
✓Final answer

Δ=0\Delta = 0.

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