Question

Use expansion by cofactors to find the determinant of the matrix.wxyz18−212427−3218−2410−354032−22

Answer 1

To find the determinant of the given matrix using expansion by cofactors, we can start by selecting any row or column. Let's choose the first row for this example.

The formula for expanding the determinant by cofactors along the first row is:

`\displaystyle \text{det}(A)=a_(11)C_(11)-a_(12)C_(12)+a_(13)C_(13)-a_(14)C_(14),`

where
`\displaystyle a_(ij)` represents the elements of the matrix and
`\displaystyle C_(ij)` represents the cofactors.

Given matrix
`\displaystyle A`:

`\displaystyle \begin{bmatrix} w & x & y & z\\ 18 &-21 &24 &27\\ -32 &18 &-24 &10\\ -35 &40 &32 &-22 \end{bmatrix} .`

Expanding along the first row, we have:

`\displaystyle \text{det}(A)=wC_(11)-xC_(12)+yC_(13)-zC_(14),`

where
`\displaystyle C_(ij)` is the cofactor of
`\displaystyle a_(ij)`.

The cofactor of
`\displaystyle a_(11)` is given by the determinant of the 3×3 matrix obtained by removing the first row and first column:

`\displaystyle C_(11)=\begin{vmatrix} -21 &24 &27\\ 18 &-24 &10\\ 40 &32 &-22 \end{vmatrix} .`

The cofactor of
`\displaystyle a_(12)` is given by the determinant of the 3×3 matrix obtained by removing the first row and second column:

`\displaystyle C_(12)=-\begin{vmatrix} 18 &24 &27\\ -32 &-24 &10\\ -35 &32 &-22 \end{vmatrix} .`

The cofactor of
`\displaystyle a_(13)` is given by the determinant of the 3×3 matrix obtained by removing the first row and third column:

`\displaystyle C_(13)=\begin{vmatrix} 18 &-21 &27\\ -32 &18 &10\\ -35 &40 &-22 \end{vmatrix} .`

The cofactor of
`\displaystyle a_(14)` is given by the determinant of the 3×3 matrix obtained by removing the first row and fourth column:

`\displaystyle C_(14)=-\begin{vmatrix} 18 &-21 &24\\ -32 &18 &-24\\ -35 &40 &32 \end{vmatrix} .`

Calculating the determinants of the corresponding matrices, we find:

`\displaystyle C_(11)=\begin{vmatrix} -21 &24 &27\\ 18 &-24 &10\\ 40 &32 &-22 \end{vmatrix} =2184,`

`\displaystyle C_(12)=-\begin{vmatrix} 18 &24 &27\\ -32 &-24 &10\\ -35 &32 &-22 \end{vmatrix} =6480,`

`\displaystyle C_(13)=\begin{vmatrix} 18 &-21 &27\\ -32 &18 &10\\ -35 & 40 &-22 \end{vmatrix} =3240,`

`\displaystyle C_(14)=-\begin{vmatrix} 18 &-21 &24\\ -32 &18 &-24\\ -35 &40 &32 \end{vmatrix} =4320.`

Now, substituting these values back into the expansion formula, we have:

`\displaystyle \text{det}(A)=w(2184)-x(6480)+y(3240)-z(4320).`

Hence, the determinant of the given matrix is:

`\displaystyle \text{det}(A)=2184w-6480x+3240y-4320z.`

`\huge{\mathfrak{\colorbox{black}{\textcolor{lime}{I\:hope\:this\:helps\:!\:\:}}}}`

♥️
`\large{\underline{\textcolor{red}{\mathcal{SUMIT\:\:ROY\:\:(:\:\:}}}}`