Answer to Question #190381 in Linear Algebra for Regomoditswe Dibob

Question #190381

Consider the given matrix B =   2 2 0 1 0 1 0 1 1   . Find detB and use it to determine whether or not B is invertible, and if so, find B −1 . (Hint: Use the matrix equation BX = I)


1
Expert's answer
2021-05-18T06:32:01-0400
B=(220101011)B=\begin{pmatrix} 2 & 2 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix}

detB=220101011\text{det}B=\begin{vmatrix} 2 & 2 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{vmatrix}

=(1)1+120111+(1)2+112011=(-1)^{1+1}\cdot 2\cdot\begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix}+(-1)^{2+1}\cdot 1\cdot\begin{vmatrix} 2 & 0 \\ 1 & 1 \end{vmatrix}

+(1)3+100111=+(-1)^{3+1}\cdot 0\cdot\begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix}=

=2(01)(20)=40=2(0-1)-(2-0)=-4\not=0

Therefore the matrix BB is invertible and B1B^{-1} exists.


Augment the matrix with the identity matrix:


[220100101010011001]\begin{bmatrix} 2 & 2 & 0 & & 1 & 0 & 0 \\ 1 & 0 & 1 & & 0 & 1 & 0 \\ 0 & 1 & 1 & & 0 & 0 & 1 \\ \end{bmatrix}



R1=R12R_1=\dfrac{R_1}{2}


[1101/200101010011001]\begin{bmatrix} 1 & 1 & 0 & & 1/2 & 0 & 0 \\ 1 & 0 & 1 & & 0 & 1 & 0 \\ 0 & 1 & 1 & & 0 & 0 & 1 \\ \end{bmatrix}


R2=R2R1R_2=R_2-R_1


[1101/2000111/210011001]\begin{bmatrix} 1 & 1 & 0 & & 1/2 & 0 & 0 \\ 0 & -1 & 1 & & -1/2 & 1 & 0 \\ 0 & 1 & 1 & & 0 & 0 & 1 \\ \end{bmatrix}


R2=R2R_2=-R_2


[1101/2000111/210011001]\begin{bmatrix} 1 & 1 & 0 & & 1/2 & 0 & 0 \\ 0 & 1 & -1 & & 1/2 & -1 & 0 \\ 0 & 1 & 1 & & 0 & 0 & 1 \\ \end{bmatrix}

R1=R1R2R_1=R_1-R_2


[1010100111/210011001]\begin{bmatrix} 1 & 0 & 1 & & 0 & 1 & 0 \\ 0 & 1 & -1 & & 1/2 & -1 & 0 \\ 0 & 1 & 1 & & 0 & 0 & 1 \\ \end{bmatrix}


R3=R3R2R_3=R_3-R_2


[1010100111/2100021/211]\begin{bmatrix} 1 & 0 & 1 & & 0 & 1 & 0 \\ 0 & 1 & -1 & & 1/2 & -1 & 0 \\ 0 & 0 & 2 & & -1/2 & 1 & 1 \\ \end{bmatrix}


R3=R3/2R_3=R_3/2


[1010100111/2100011/41/21/2]\begin{bmatrix} 1 & 0 & 1 & & 0 & 1 & 0 \\ 0 & 1 & -1 & & 1/2 & -1 & 0 \\ 0 & 0 & 1 & & -1/4 & 1/2 & 1/2 \\ \end{bmatrix}


R1=R1R3R_1=R_1-R_3


[1001/41/21/20111/2100011/41/21/2]\begin{bmatrix} 1 & 0 & 0 & & 1/4 & 1/2 & -1/2 \\ 0 & 1 & -1 & & 1/2 & -1 & 0 \\ 0 & 0 & 1 & & -1/4 & 1/2 & 1/2 \\ \end{bmatrix}


R2=R2+R3R_2=R_2+R_3


[1001/41/21/20101/41/21/20011/41/21/2]\begin{bmatrix} 1 & 0 & 0 & & 1/4 & 1/2 & -1/2 \\ 0 & 1 & 0 & & 1/4 & -1/2 & 1/2 \\ 0 & 0 & 1 & & -1/4 & 1/2 & 1/2 \\ \end{bmatrix}

On the left is the identity matrix. On the right is the inverse matrix.



B1=(1/41/21/21/41/21/21/41/21/2)B^{-1}=\begin{pmatrix} 1/4 & 1/2 & -1/2 \\ 1/4 & -1/2 & 1/2 \\ -1/4 & 1/2 & 1/2 \end{pmatrix}


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