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Algebra · real student question

Compute the determinant of the 4x4 matrix with rows (19,5,7,8), (18,4,4,4), (9,1,2,1) and (9,2,3,4).

Question

Compute

∣195781844491219234∣\begin{vmatrix}19&5&7&8\\18&4&4&4\\9&1&2&1\\9&2&3&4\end{vmatrix}

Step-by-step solution

  1. Use row 3 to attack the first column. Row 33 begins with 99, and the other first-column entries are 1919, 1818 and 99. Subtracting multiples of row 33 (an operation that preserves the determinant) gives

    R1←R1−2R3,R2←R2−2R3,R4←R4−R3R_1\leftarrow R_1-2R_3,\quad R_2\leftarrow R_2-2R_3,\quad R_4\leftarrow R_4-R_3

    ∣1336020291210113∣\begin{vmatrix}1&3&3&6\\0&2&0&2\\9&1&2&1\\0&1&1&3\end{vmatrix}

    Note 19−18=119-18=1, so the first column is not fully cleared — one more step is needed.

  2. Finish clearing the column with the new leading 1.

    R3←R3−9R1⟹(0, 1−27, 2−27, 1−54)=(0,−26,−25,−53)R_3\leftarrow R_3-9R_1\quad\Longrightarrow\quad (0,\ 1-27,\ 2-27,\ 1-54)=(0,-26,-25,-53)

    ∣133602020−26−25−530113∣\begin{vmatrix}1&3&3&6\\0&2&0&2\\0&-26&-25&-53\\0&1&1&3\end{vmatrix}

  3. Expand along the first column. Only the leading 11 remains, with cofactor sign ++:

    det⁡=1⋅∣202−26−25−53113∣\det=1\cdot\begin{vmatrix}2&0&2\\-26&-25&-53\\1&1&3\end{vmatrix}

  4. Simplify and evaluate the 3×3. Factor 22 out of the first row:

    2∣101−26−25−53113∣=2[1(−75+53)−0+1(−26+25)]=2[−22−1]=2(−23)=−462\begin{vmatrix}1&0&1\\-26&-25&-53\\1&1&3\end{vmatrix}=2\Big[1(-75+53)-0+1(-26+25)\Big]=2\big[-22-1\big]=2(-23)=-46

    −46\boxed{-46}

  5. Cross-check the size against the sister problem. The very similar matrix with rows (18,5,7,8),(18,3,4,4),(9,1,1,1),(9,2,3,4)(18,5,7,8),(18,3,4,4),(9,1,1,1),(9,2,3,4) has determinant 99; changing four entries here has moved it to −46-46, including a sign flip. Because determinants are alternating multilinear functions, no smooth relationship between the two values should be expected — each has to be computed independently.

Answer

−46-46

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