Given the matrix
\[\mathbf{B} = \left[\begin{array} {rrr} 4-x & -4 & -4 \\ 2 & -2-x & -4 \\ 3 & -3 & -4-x \end{array}\right] \]
find all values of x for which det(B)=0
Sol: Perform matrix row and columnn operations to find the determinant.
Step:1 C1 -> C1+C2.
\[\mathbf{} \left[\begin{array} {rrr} -x & -4 & -4 \\ -x & -2-x & -4 \\ 0 & -3 & -4-x \end{array}\right] \]
Step :2 R2 -> R2-R1.
\[\mathbf{} \left[\begin{array} {rrr} -x & -4 & -4 \\ 0 & 2-x & 0 \\ 0 & -3 & -4-x \end{array}\right] \]
find the determinant by C1.
-x(2-x)(-4-x)
x(2-x)(4+x)
det(B)= 0
x =0, 2, -4
\[\mathbf{B} = \left[\begin{array} {rrr} 4 & -4 & -4 \\ 2 & -2 & -4 \\ 3 & -3 & -4 \end{array}\right] \] Operate C2 -> -1*C2
\[\mathbf{B} = \left[\begin{array} {rrr} 4 & 4 & -4 \\ 2 & 2 & -4 \\ 3 & 3 & -4 \end{array}\right] \] for matrix B, C1 = C2.
det(B) = 0
\[\mathbf{B} = \left[\begin{array} {rrr} 2 & -4 & -4 \\ 2 & -4 & -4 \\ 3 & -3 & -6 \end{array}\right] \]
Operate C3 -> C3/-2
\[\mathbf{B} = \left[\begin{array} {rrr} 2 & -4 & 2 \\ 2 & -4 & 2 \\ 3 & -3 & 3 \end{array}\right] \] C1 = C3.
det(B) = 0
\[\mathbf{B} = \left[\begin{array} {rrr} 8 & -4 & -4 \\ 2 & 2 & -4 \\ 3 & -3 & 0 \end{array}\right] \]
B<- matrix(c(8,2,3,-4,2,-3,-4,-4,0),3)
det(B)## [1] 0