Chapter Determinants Section DM Exercise M11

Find a value of k so that the matrix \(A\) has det(A) = 0, or explain why it is not possible.

\[A=\begin{bmatrix} 1 & 2 & 1 \\ 2 & 0 & 1 \\ 2 & 3 & k \end{bmatrix}\]

Process:

  • Steps:

    • First break down the 3x3 matrix into a 2x2 matrix of values not in the same row or column. (See example below)
    • Take the number in row 1, column 1
    • Multiply it by the 2x2 matrix not in the row and column of the number in row 1, column 1

Example:

\[{ A= }{ X }_{ 11 }\begin{bmatrix} { X }_{ 22 } & { X }_{ 23 } \\ { X }_{ 32 } & { X }_{ 33 } \end{bmatrix}-{ X }_{ 12 }\begin{bmatrix} { X }_{ 21 } & { X }_{ 23 } \\ { X }_{ 31 } & { X }_{ 33 } \end{bmatrix}-{ X }_{ 13 }\begin{bmatrix} { X }_{ 21 } & { X }_{ 22 } \\ { X }_{ 31 } & { X }_{ 32 } \end{bmatrix}\] The Subscript refers to the Column and row of the first Matrix. \({ X }_{ 32 }\) in the first matrix is value \(3\) (Third row; second column)

\[1\begin{bmatrix} 0 & 1 \\ 3 & k \end{bmatrix}-2\begin{bmatrix} 2 & 1 \\ 2 & k \end{bmatrix}-1\begin{bmatrix} 2 & 0 \\ 2 & 3 \end{bmatrix}\]

\[1(0(k)\quad -3(1))\quad -2(2(k)\quad -1(2))\quad -1(2(3)\quad -2(0))\] \[-3-2(2k-2)+1(6-0)\] \[-3-4k+4+6\] \[(-3 + 4 + 6) - 4k\]

\[7 - 4k\]

Since, det(A) = 0 therefore \(7-4k=0\)

This leads to: \[-4k=-7\] \[k=7/4\]

Solution

det(A) = 0 if \(k=7/4\).