Inverse of a Matrix

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Question 1 of 5

Find the inverse of the matrix:

$A=[[3,5],[2,4]]$

1. $\begin{bmatrix} -4 & 5 \\ 2 & -3 \end{bmatrix}$
2. $\begin{bmatrix} 8 & -10 \\ -4 & 6 \end{bmatrix}$
3. $\begin{bmatrix} 4 & -5 \\ -2 & 3 \end{bmatrix}$
4. $\begin{bmatrix} 2 & \frac{-5}{2} \\[0.3em] -1 & \frac{3}{2} \end{bmatrix}$

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Inverse of a $2xx2$ Matrix

$\begin{bmatrix} \color{#007DDC}{a} & \color{#9a00c7}{b} \\ \color{#9a00c7}{c} & \color{#007DDC}{d} \end{bmatrix}$ then

$A^{-1}=\frac{1}{|A|}$

$\begin{bmatrix} \color{#007DDC}{d} & -\color{#9a00c7}{b} \\ -\color{#9a00c7}{c} & \color{#007DDC}{a} \end{bmatrix}$

Determinant of a $2xx2$ Matrix

$\begin{bmatrix} \color{#007DDC}{a} & \color{#9a00c7}{b} \\ \color{#9a00c7}{c} & \color{#007DDC}{d} \end{bmatrix}$ then

$|A|=\color{#007DDC}{ad}-\color{#9a00c7}{bc}$

First, label the values of the matrix

$[[a,b],[c,d]]=[[3,5],[2,4]]$

$a=3$

$c=2$

$b=5$

$d=4$

Next, solve for the determinant

$\|A\|$	$=$	$ad$ $-$ $bc$	Determinant Formula

	$=$	$34$* $-$ $52$*	Substitute values

	$=$	$12-10$
	$=$	$2$

Finally, substitute values into the Inverse Formula

$A^(-1)$	$=$	$\frac{1}{\|A\|}$ $\begin{bmatrix} \color{#007DDC}{d} & -\color{#9a00c7}{b} \\ -\color{#9a00c7}{c} & \color{#007DDC}{a} \end{bmatrix}$	Inverse Formula

	$=$	$\frac{1}{2}$ $\begin{bmatrix} \color{#007DDC}{4} & -\color{#9a00c7}{5} \\ -\color{#9a00c7}{2} & \color{#007DDC}{3} \end{bmatrix}$	Substitute values

	$=$	$\begin{bmatrix} 2 & \frac{-5}{2} \\[0.3em] -1 & \frac{3}{2} \end{bmatrix}$

$\begin{bmatrix} 2 & \frac{-5}{2} \\[0.3em] -1 & \frac{3}{2} \end{bmatrix}$

Question 2 of 5

Find the inverse of the matrix:

$B=[[4,-5],[-3,6]]$

1. $\begin{bmatrix} \frac{2}{3} & \frac{5}{9} \\[0.3em] \frac{1}{3} & \frac{4}{9} \end{bmatrix}$
2. $\begin{bmatrix} \frac{2}{3} & -\frac{5}{9} \\[0.3em] -\frac{1}{3} & \frac{4}{9} \end{bmatrix}$
3. $\begin{bmatrix} \frac{4}{9} & \frac{5}{9} \\[0.3em] \frac{1}{3} & \frac{2}{3} \end{bmatrix}$
4. $\begin{bmatrix} \frac{4}{9} & -\frac{5}{9} \\[0.3em] -\frac{1}{3} & \frac{2}{3} \end{bmatrix}$

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Inverse of a $2xx2$ Matrix

$\begin{bmatrix} \color{#007DDC}{a} & \color{#9a00c7}{b} \\ \color{#9a00c7}{c} & \color{#007DDC}{d} \end{bmatrix}$ then

$A^{-1}=\frac{1}{|A|}$

$\begin{bmatrix} \color{#007DDC}{d} & -\color{#9a00c7}{b} \\ -\color{#9a00c7}{c} & \color{#007DDC}{a} \end{bmatrix}$

Determinant of a $2xx2$ Matrix

$\begin{bmatrix} \color{#007DDC}{a} & \color{#9a00c7}{b} \\ \color{#9a00c7}{c} & \color{#007DDC}{d} \end{bmatrix}$ then

$|A|=\color{#007DDC}{ad}-\color{#9a00c7}{bc}$

First, label the values of the matrix

$[[a,b],[c,d]]=[[4,-5],[-3,6]]$

$a=4$

$c=-3$

$b=-5$

$d=6$

Next, solve for the determinant

$\|A\|$	$=$	$ad$ $-$ $bc$	Determinant Formula

	$=$	$46$* $-$ $(-5)(-3)$*	Substitute values

	$=$	$24-15$
	$=$	$9$

Finally, substitute values into the Inverse Formula

$A^(-1)$	$=$	$\frac{1}{\|A\|}$ $\begin{bmatrix} \color{#007DDC}{d} & -\color{#9a00c7}{b} \\ -\color{#9a00c7}{c} & \color{#007DDC}{a} \end{bmatrix}$	Inverse Formula

	$=$	$\frac{1}{9}$ $\begin{bmatrix} \color{#007DDC}{6} & -\color{#9a00c7}{(-5)} \\ -\color{#9a00c7}{(-3)} & \color{#007DDC}{4} \end{bmatrix}$	Substitute values

	$=$	$\begin{bmatrix} \frac{2}{3} & \frac{5}{9} \\[0.3em] \frac{1}{3} & \frac{4}{9} \end{bmatrix}$

$\begin{bmatrix} \frac{2}{3} & \frac{5}{9} \\[0.3em] \frac{1}{3} & \frac{4}{9} \end{bmatrix}$

Question 3 of 5

Find the inverse of the matrix:

$A=[[2,0,1],[-3,1,1],[4,-1,-1]]$

1. $\begin{bmatrix} -1 & -1 & 4 \\ 1 & 1 & -3 \\ 1 & 0 & 2 \end{bmatrix}$
2. $\begin{bmatrix} 0 & 1 & 1 \\ -1 & 6 & 5 \\ 1 & -2 & -2 \end{bmatrix}$
3. $\begin{bmatrix} 0 & -1 & -1 \\ 1 & -6 & -5 \\ -1 & 2 & 2 \end{bmatrix}$
4. $\begin{bmatrix} 2 & -3 & 4 \\ 0 & 1 & -1 \\ 1 & 1 & -1 \end{bmatrix}$

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Inverse of a $3xx3$ Matrix

$A^(-1)=1/(|A|)Adj(A)$

where

$Adj(A)$ means Adjoint

$A$

Determinant of a $3xx3$ Matrix

$A=$

$\begin{bmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{bmatrix}$ then

$|A|=\color{#D800AD}{a}$

$\begin{vmatrix} e & f \\ h & k \end{vmatrix}$

$-\color{#D800AD}{b}$

$\begin{vmatrix} d & f \\ g & k \end{vmatrix}$

$+\color{#D800AD}{c}$

$\begin{vmatrix} d & e \\ g & h \end{vmatrix}$

Determinant of a $2xx2$ Matrix

$\begin{bmatrix} \color{#007DDC}{a} & \color{#9a00c7}{b} \\ \color{#9a00c7}{c} & \color{#007DDC}{d} \end{bmatrix}$ then

$|A|=\color{#007DDC}{ad}-\color{#9a00c7}{bc}$

To find the inverse of a

$3xx3$ matrix, solve for its determinant and adjoint matrix. The adjoint matrix can be found by forming the Matrix of Minors, transforming it into a Matrix of Cofactors, then transposing it.

First, label the values of the matrix

$\begin{bmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{bmatrix}$

$=$

$\begin{bmatrix} \color{#D800AD}{2} & \color{#D800AD}{0} & \color{#D800AD}{1} \\ -3 & 1 & 1 \\ 4 & -1 & -1 \end{bmatrix}$

$a=2$

$d=-3$

$g=4$

$b=0$

$e=1$

$h=-1$

$c=1$

$f=1$

$k=-1$

Solve for the determinant of

$A$

$\|A\|$	$=$	$\begin{vmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{vmatrix}$	$=$	$\color{#D800AD}{a}$ $\begin{vmatrix} e & f \\ h & k \end{vmatrix}$ $-\color{#D800AD}{b}$ $\begin{vmatrix} d & f \\ g & k \end{vmatrix}$ $+\color{#D800AD}{c}$ $\begin{vmatrix} d & e \\ g & h \end{vmatrix}$	Determinant of a $3xx3$ Matrix

	$=$	$\begin{vmatrix} \color{#D800AD}{2} & \color{#D800AD}{0} & \color{#D800AD}{1} \\ -3 & 1 & 1 \\ 4 & -1 & -1 \end{vmatrix}$	$=$	$\color{#D800AD}{2}$ $\begin{vmatrix} \color{#007DDC}{1} & \color{#9a00c7}{1} \\ \color{#9a00c7}{-1} & \color{#007DDC}{-1} \end{vmatrix}$ $-\color{#D800AD}{0}$ $\begin{vmatrix} \color{#007DDC}{-3} & \color{#9a00c7}{1} \\ \color{#9a00c7}{4} & \color{#007DDC}{-1} \end{vmatrix}$ $+\color{#D800AD}{1}$ $\begin{vmatrix} \color{#007DDC}{-3} & \color{#9a00c7}{1} \\ \color{#9a00c7}{4} & \color{#007DDC}{-1} \end{vmatrix}$	Substitute values

	$=$	$[\color{#D800AD}{2}(\color{#007DDC}{1\cdot(-1)}-\color{#9a00c7}{1\cdot(-1)})]-[\color{#D800AD}{0}(\color{#007DDC}{-3\cdot(-1)}-\color{#9a00c7}{1\cdot4})]+[\color{#D800AD}{1}(\color{#007DDC}{-3\cdot(-1)}-\color{#9a00c7}{1\cdot4})]$
	$=$	$2(-1-(-1))-0(3-4)+1(3-4)$
	$=$	$2(0)-0+1(-1)$
	$=$	$0-0-1$
$\|A\|$	$=$	$-1$

Next, start solving for the Adjoint matrix of

$A$ by forming the Matrix of Minors

Matrix of Minors is a

$3xx3$ matrix. An element of it is found by getting the determinant of the

$2xx2$ matrix formed by crossing out the row and column of that element from matrix

$A$

First Element(Row

$1$ Column

$1$ ):

$A=$

$\begin{bmatrix} \color{#9E9E9E}{2} & \color{#9E9E9E}{0} & \color{#9E9E9E}{1} \\ \color{#9E9E9E}{-3} & 1 & 1 \\ \color{#9E9E9E}{4} & -1 & -1 \end{bmatrix}$

$\begin{vmatrix} 1 & 1 \\ -1 & -1 \end{vmatrix}$	$=$	$1(-1)-1(-1)$	Determinant of a $2xx2$ Matrix
	$=$	$-1-(-1)$
	$=$	$0$

Matrix of Minors:

$\begin{bmatrix} 0 & \: & \: \\ \: & \: & \: \\ \: & \: & \: \end{bmatrix}$

Second Element(Row

$1$ Column

$2$ ):

$A=$

$\begin{bmatrix} \color{#9E9E9E}{2} & \color{#9E9E9E}{0} & \color{#9E9E9E}{1} \\ -3 & \color{#9E9E9E}{1} & 1 \\ 4 & \color{#9E9E9E}{-1} & -1 \end{bmatrix}$

$\begin{vmatrix} -3 & 1 \\ 4 & -1 \end{vmatrix}$	$=$	$-3(-1)-14$	Determinant of a $2xx2$ Matrix
	$=$	$3-4$
	$=$	$-1$

Matrix of Minors:

$\begin{bmatrix} 0 & -1 & \: \\ \: & \: & \: \\ \: & \: & \: \end{bmatrix}$

Third Element(Row

$1$ Column

$3$ ):

$A=$

$\begin{bmatrix} \color{#9E9E9E}{2} & \color{#9E9E9E}{0} & \color{#9E9E9E}{1} \\ -3 & 1 & \color{#9E9E9E}{1} \\ 4 & -1 & \color{#9E9E9E}{-1} \end{bmatrix}$

$\begin{vmatrix} -3 & 1 \\ 4 & -1 \end{vmatrix}$	$=$	$-3(-1)-14$	Determinant of a $2xx2$ Matrix
	$=$	$3-4$
	$=$	$-1$

Matrix of Minors:

$\begin{bmatrix} 0 & -1 & -1 \\ \: & \: & \: \\ \: & \: & \: \end{bmatrix}$

Fourth Element(Row

$2$ Column

$1$ ):

$A=$

$\begin{bmatrix} \color{#9E9E9E}{2} & 0 & 1 \\ \color{#9E9E9E}{-3} & \color{#9E9E9E}{1} & \color{#9E9E9E}{1} \\ \color{#9E9E9E}{4} & -1 & -1 \end{bmatrix}$

$\begin{vmatrix} 0 & 1 \\ -1 & -1 \end{vmatrix}$	$=$	$0(-1)-1(-1)$	Determinant of a $2xx2$ Matrix
	$=$	$0-(-1)$
	$=$	$1$

Matrix of Minors:

$\begin{bmatrix} 0 & -1 & -1 \\ 1 & \: & \: \\ \: & \: & \: \end{bmatrix}$

Fifth Element(Row

$2$ Column

$2$ ):

$A=$

$\begin{bmatrix} 2 & \color{#9E9E9E}{0} & 1 \\ \color{#9E9E9E}{-3} & \color{#9E9E9E}{1} & \color{#9E9E9E}{1} \\ 4 & \color{#9E9E9E}{-1} & -1 \end{bmatrix}$

$\begin{vmatrix} 2 & 1 \\ 4 & -1 \end{vmatrix}$	$=$	$2(-1)-14$	Determinant of a $2xx2$ Matrix
	$=$	$-2-4$
	$=$	$-6$

Matrix of Minors:

$\begin{bmatrix} 0 & -1 & -1 \\ 1 & -6 & \: \\ \: & \: & \: \end{bmatrix}$

Sixth Element(Row

$2$ Column

$3$ ):

$A=$

$\begin{bmatrix} 2 & 0 & \color{#9E9E9E}{1} \\ \color{#9E9E9E}{-3} & \color{#9E9E9E}{1} & \color{#9E9E9E}{1} \\ 4 & -1 & \color{#9E9E9E}{-1} \end{bmatrix}$

$\begin{vmatrix} 2 & 0 \\ 4 & -1 \end{vmatrix}$	$=$	$2(-1)-04$	Determinant of a $2xx2$ Matrix
	$=$	$-2-0$
	$=$	$-2$

Matrix of Minors:

$\begin{bmatrix} 0 & -1 & -1 \\ 1 & -6 & -2 \\ \: & \: & \: \end{bmatrix}$

Seventh Element(Row

$3$ Column

$1$ ):

$A=$

$\begin{bmatrix} \color{#6F6161}{2} & 0 & 1 \\ \color{#6F6161}{-3} & 1 & 1 \\ \color{#6F6161}{4} & \color{#6F6161}{-1} & \color{#6F6161}{-1} \end{bmatrix}$

$\begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix}$	$=$	$01-11$	Determinant of a $2xx2$ Matrix
	$=$	$0-1$
	$=$	$-1$

Matrix of Minors:

$\begin{bmatrix} 0 & -1 & -1 \\ 1 & -6 & -2 \\ -1 & \: & \: \end{bmatrix}$

Eighth Element(Row

$3$ Column

$2$ ):

$A=$

$\begin{bmatrix} 2 & \color{#9E9E9E}{0} & 1 \\ -3 & \color{#9E9E9E}{1} & 1 \\ \color{#9E9E9E}{4} & \color{#9E9E9E}{-1} & \color{#9E9E9E}{-1} \end{bmatrix}$

$\begin{vmatrix} 2 & 1 \\ -3 & 1 \end{vmatrix}$	$=$	$21-1(-3)$	Determinant of a $2xx2$ Matrix
	$=$	$2-(-3)$
	$=$	$5$

Matrix of Minors:

$\begin{bmatrix} 0 & -1 & -1 \\ 1 & -6 & -2 \\ -1 & 5 & \: \end{bmatrix}$

Ninth Element(Row

$3$ Column

$3$ ):

$A=$

$\begin{bmatrix} 2 & 0 & \color{#9E9E9E}{1} \\ -3 & 1 & \color{#9E9E9E}{1} \\ \color{#9E9E9E}{4} & \color{#9E9E9E}{-1} & \color{#9E9E9E}{-1} \end{bmatrix}$

$\begin{vmatrix} 2 & 0 \\ -3 & 1 \end{vmatrix}$	$=$	$21-0(-3)$	Determinant of a $2xx2$ Matrix
	$=$	$2-0$
	$=$	$2$

Matrix of Minors:

$\begin{bmatrix} 0 & -1 & -1 \\ 1 & -6 & -2 \\ -1 & 5 & 2 \end{bmatrix}$

Transform the Matrix of Minors into a Matrix of Cofactors by applying alternate signs to each element, starting with a positive sign.

	$\begin{bmatrix} 0 & -1 & -1 \\ 1 & -6 & -2 \\ -1 & 5 & 2 \end{bmatrix}$	Matrix of Minors

	$\begin{bmatrix} \color{#CC0000}{+}0 & \color{#CC0000}{-}(-1) & \color{#CC0000}{+}(-1) \\ \color{#CC0000}{-}1 & \color{#CC0000}{+}(-6) & \color{#CC0000}{-}(-2) \\ \color{#CC0000}{+}(-1) & \color{#CC0000}{-}5 & \color{#CC0000}{+}2 \end{bmatrix}$	Apply alternating signs

$=$	$\begin{bmatrix} 0 & 1 & -1 \\ -1 & -6 & 2 \\ -1 & -5 & 2 \end{bmatrix}$

This is now the Matrix of Cofactors

Transpose the Matrix of Cofactors to finally get the Adjoint matrix of

$A$

To transpose a matrix, freeze the diagonal elements and then flip it.

	$\begin{bmatrix} \color{#CC0000}{0} & 1 & -1 \\ -1 & \color{#CC0000}{-6} & 2 \\ -1 & -5 & \color{#CC0000}{2} \end{bmatrix}$	Matrix of Cofactor

	$\begin{bmatrix} \color{#CC0000}{0} & -1 & -1 \\ 1 & \color{#CC0000}{-6} & -5 \\ -1 & 2 & \color{#CC0000}{2} \end{bmatrix}$	Transpose the matrix

This is the Adjoint matrix of

$A$

Finally, substitute the components into the Inverse Formula

$|A|=-1$

$A^(-1)$	$=$	$1/(\|A\|)Adj(A)$	Inverse Formula

	$=$	$1/(-1)[[0,-1,-1],[1,-6,-5],[-1,2,2]]$	Substitute components

	$=$	$-[[0,-1,-1],[1,-6,-5],[-1,2,2]]$

	$=$	$[[0,1,1],[-1,6,5],[1,-2,-2]]$

$[[0,1,1],[-1,6,5],[1,-2,-2]]$

Question 4 of 5

Find the inverse of the matrix:

$A=[[0,2,2],[-2,3,4],[1,-1,-1]]$

1. $\begin{bmatrix} \frac{1}{2} & 0 & 1 \\[0.3em] 1 & -1 & -2 \\[0.3em] -\frac{1}{2} & 1 & 2 \end{bmatrix}$
2. $\begin{bmatrix} -\frac{1}{2} & 1 & 0 \\[0.3em] 2 & -1 & -2 \\[0.3em] \frac{1}{2} & 1 & 2 \end{bmatrix}$
3. $\begin{bmatrix} \frac{1}{3} & 0 & 1 \\[0.3em] 1 & 0 & -2 \\[0.3em] -\frac{1}{3} & 1 & 2 \end{bmatrix}$
4. $\begin{bmatrix} 1 & 0 & \frac{1}{2} \\[0.3em] -1 & 1 & -2 \\[0.3em] -\frac{1}{2} & 1 & 0 \end{bmatrix}$

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Inverse of a $3xx3$ Matrix

$A^(-1)=1/(|A|)Adj(A)$

where

$Adj(A)$ means Adjoint

$A$

Determinant of a $3xx3$ Matrix

$A=$

$\begin{bmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{bmatrix}$ then

$|A|=\color{#D800AD}{a}$

$\begin{vmatrix} e & f \\ h & k \end{vmatrix}$

$-\color{#D800AD}{b}$

$\begin{vmatrix} d & f \\ g & k \end{vmatrix}$

$+\color{#D800AD}{c}$

$\begin{vmatrix} d & e \\ g & h \end{vmatrix}$

Determinant of a $2xx2$ Matrix

$\begin{bmatrix} \color{#007DDC}{a} & \color{#9a00c7}{b} \\ \color{#9a00c7}{c} & \color{#007DDC}{d} \end{bmatrix}$ then

$|A|=\color{#007DDC}{ad}-\color{#9a00c7}{bc}$

To find the inverse of a

$3xx3$ matrix, solve for its determinant and adjoint matrix. The adjoint matrix can be found by forming the Matrix of Minors, transforming it into a Matrix of Cofactors, then transposing it.

First, solve for the determinant of

$A$

$\|A\|$	$=$	$\begin{vmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{vmatrix}$	$=$	$\color{#D800AD}{a}$ $\begin{vmatrix} e & f \\ h & k \end{vmatrix}$ $-\color{#D800AD}{b}$ $\begin{vmatrix} d & f \\ g & k \end{vmatrix}$ $+\color{#D800AD}{c}$ $\begin{vmatrix} d & e \\ g & h \end{vmatrix}$	Determinant of a $3xx3$ Matrix

	$=$	$\begin{vmatrix} \color{#D800AD}{0} & \color{#D800AD}{2} & \color{#D800AD}{2} \\ -2 & 3 & 4 \\ 1 & -1 & -1 \end{vmatrix}$	$=$	$\color{#D800AD}{0}$ $\begin{vmatrix} \color{#007DDC}{3} & \color{#9a00c7}{4} \\ \color{#9a00c7}{-1} & \color{#007DDC}{-1} \end{vmatrix}$ $-\color{#D800AD}{2}$ $\begin{vmatrix} \color{#007DDC}{-2} & \color{#9a00c7}{4} \\ \color{#9a00c7}{1} & \color{#007DDC}{-1} \end{vmatrix}$ $+\color{#D800AD}{2}$ $\begin{vmatrix} \color{#007DDC}{-2} & \color{#9a00c7}{3} \\ \color{#9a00c7}{1} & \color{#007DDC}{-1} \end{vmatrix}$	Substitute values

	$=$	$[\color{#D800AD}{0}(\color{#007DDC}{3\cdot(-1)}-\color{#9a00c7}{4\cdot(-1)})]-[\color{#D800AD}{2}(\color{#007DDC}{-2\cdot(-1)}-\color{#9a00c7}{4\cdot1})]+[\color{#D800AD}{2}(\color{#007DDC}{-2\cdot(-1)}-\color{#9a00c7}{3\cdot1})]$
	$=$	$0-2(2-4)+2(2-3)$
	$=$	$0-2(-2)+2(-1)$
	$=$	$0+4-2$
$\|A\|$	$=$	$2$

Next, start solving for the Adjoint matrix of

$A$ by forming the Matrix of Minors

Matrix of Minors is a

$3xx3$ matrix. An element of it is found by getting the determinant of the

$2xx2$ matrix formed by crossing out the row and column of that element from matrix

$A$

First Element(Row

$1$ Column

$1$ ):

$A=$

$\begin{bmatrix} \color{#9E9E9E}{0} & \color{#9E9E9E}{2} & \color{#9E9E9E}{2} \\ \color{#9E9E9E}{-2} & 3 & 4 \\ \color{#9E9E9E}{1} & -1 & -1 \end{bmatrix}$

$\begin{vmatrix} 3 & 4 \\ -1 & -1 \end{vmatrix}$	$=$	$3(-1)-4(-1)$	Determinant of a $2xx2$ Matrix
	$=$	$-3-(-4)$
	$=$	$1$

Matrix of Minors:

$\begin{bmatrix} 1 & \: & \: \\ \: & \: & \: \\ \: & \: & \: \end{bmatrix}$

Second Element(Row

$1$ Column

$2$ ):

$A=$

$\begin{bmatrix} \color{#9E9E9E}{0} & \color{#9E9E9E}{2} & \color{#9E9E9E}{2} \\ -2 & \color{#9E9E9E}{3} & 4 \\ 1 & \color{#9E9E9E}{-1} & -1 \end{bmatrix}$

$\begin{vmatrix} -2 & 4 \\ 1 & -1 \end{vmatrix}$	$=$	$-2(-1)-41$	Determinant of a $2xx2$ Matrix
	$=$	$2-4$
	$=$	$-2$

Matrix of Minors:

$\begin{bmatrix} 1 & -2 & \: \\ \: & \: & \: \\ \: & \: & \: \end{bmatrix}$

Third Element(Row

$1$ Column

$3$ ):

$A=$

$\begin{bmatrix} \color{#9E9E9E}{0} & \color{#9E9E9E}{2} & \color{#9E9E9E}{2} \\ -2 & 3 & \color{#9E9E9E}{4} \\ 1 & -1 & \color{#9E9E9E}{-1} \end{bmatrix}$

$\begin{vmatrix} -2 & 3 \\ 1 & -1 \end{vmatrix}$	$=$	$-2(-1)-31$	Determinant of a $2xx2$ Matrix
	$=$	$2-3$
	$=$	$-1$

Matrix of Minors:

$\begin{bmatrix} 1 & -2 & -1 \\ \: & \: & \: \\ \: & \: & \: \end{bmatrix}$

Fourth Element(Row

$2$ Column

$1$ ):

$A=$

$\begin{bmatrix} \color{#9E9E9E}{0} & 2 & 2 \\ \color{#9E9E9E}{-2} & \color{#9E9E9E}{3} & \color{#9E9E9E}{4} \\ \color{#9E9E9E}{1} & -1 & -1 \end{bmatrix}$

$\begin{vmatrix} 2 & 2 \\ -1 & -1 \end{vmatrix}$	$=$	$2(-1)-2(-1)$	Determinant of a $2xx2$ Matrix
	$=$	$-2-(-2)$
	$=$	$0$

Matrix of Minors:

$\begin{bmatrix} 1 & -2 & -1 \\ 0 & \: & \: \\ \: & \: & \: \end{bmatrix}$

Fifth Element(Row

$2$ Column

$2$ ):

$A=$

$\begin{bmatrix} 0 & \color{#9E9E9E}{2} & 2 \\ \color{#9E9E9E}{-2} & \color{#9E9E9E}{3} & \color{#9E9E9E}{4} \\ 1 & \color{#9E9E9E}{-1} & -1 \end{bmatrix}$

$\begin{vmatrix} 0 & 2 \\ 1 & -1 \end{vmatrix}$	$=$	$0(-1)-21$	Determinant of a $2xx2$ Matrix
	$=$	$0-2$
	$=$	$-2$

Matrix of Minors:

$\begin{bmatrix} 1 & -2 & -1 \\ 0 & -2 & \: \\ \: & \: & \: \end{bmatrix}$

Sixth Element(Row

$2$ Column

$3$ ):

$A=$

$\begin{bmatrix} 0 & 2 & \color{#9E9E9E}{2} \\ \color{#9E9E9E}{-2} & \color{#9E9E9E}{3} & \color{#9E9E9E}{4} \\ 1 & -1 & \color{#9E9E9E}{-1} \end{bmatrix}$

$\begin{vmatrix} 0 & 2 \\ 1 & -1 \end{vmatrix}$	$=$	$0(-1)-21$	Determinant of a $2xx2$ Matrix
	$=$	$0-2$
	$=$	$-2$

Matrix of Minors:

$\begin{bmatrix} 1 & -2 & -1 \\ 0 & -2 & -2 \\ \: & \: & \: \end{bmatrix}$

Seventh Element(Row

$3$ Column

$1$ ):

$A=$

$\begin{bmatrix} \color{#9E9E9E}{0} & 2 & 2 \\ \color{#9E9E9E}{-2} & 3 & 4 \\ \color{#9E9E9E}{1} & \color{#9E9E9E}{-1} & \color{#9E9E9E}{-1} \end{bmatrix}$

$\begin{vmatrix} 2 & 2 \\ 3 & 4 \end{vmatrix}$	$=$	$24-23$	Determinant of a $2xx2$ Matrix
	$=$	$8-6$
	$=$	$2$

Matrix of Minors:

$\begin{bmatrix} 1 & -2 & -1 \\ 0 & -2 & -2 \\ 2 & \: & \: \end{bmatrix}$

Eighth Element(Row

$3$ Column

$2$ ):

$A=$

$\begin{bmatrix} 0 & \color{#9E9E9E}{2} & 2 \\ -2 & \color{#9E9E9E}{3} & 4 \\ \color{#9E9E9E}{1} & \color{#9E9E9E}{-1} & \color{#9E9E9E}{-1} \end{bmatrix}$

$\begin{vmatrix} 0 & 2 \\ -2 & 4 \end{vmatrix}$	$=$	$04-2(-2)$	Determinant of a $2xx2$ Matrix
	$=$	$0-(-4)$
	$=$	$4$

Matrix of Minors:

$\begin{bmatrix} 1 & -2 & -1 \\ 0 & -2 & -2 \\ 2 & 4 & \: \end{bmatrix}$

Ninth Element(Row

$3$ Column

$3$ ):

$A=$

$\begin{bmatrix} 0 & 2 & \color{#9E9E9E}{2} \\ -2 & 3 & \color{#9E9E9E}{4} \\ \color{#9E9E9E}{1} & \color{#9E9E9E}{-1} & \color{#9E9E9E}{-1} \end{bmatrix}$

$\begin{vmatrix} 0 & 2 \\ -2 & 3 \end{vmatrix}$	$=$	$03-2(-2)$	Determinant of a $2xx2$ Matrix
	$=$	$0-(-4)$
	$=$	$4$

Matrix of Minors:

$\begin{bmatrix} 1 & -2 & -1 \\ 0 & -2 & -2 \\ 2 & 4 & 4 \end{bmatrix}$

Transform the Matrix of Minors into a Matrix of Cofactors by applying alternate signs to each element, starting with a positive sign.

	$\begin{bmatrix} 1 & -2 & -1 \\ 0 & -2 & -2 \\ 2 & 4 & 4 \end{bmatrix}$	Matrix of Minors

	$\begin{bmatrix} \color{#CC0000}{+}1 & \color{#CC0000}{-}(-2) & \color{#CC0000}{+}(-1) \\ \color{#CC0000}{-}0 & \color{#CC0000}{+}(-2) & \color{#CC0000}{-}(-2) \\ \color{#CC0000}{+}2 & \color{#CC0000}{-}4 & \color{#CC0000}{+}4 \end{bmatrix}$	Apply alternating signs

$=$	$\begin{bmatrix} 1 & 2 & -1 \\ 0 & -2 & 2 \\ 2 & -4 & 4 \end{bmatrix}$

This is now the Matrix of Cofactors

Transpose the Matrix of Cofactors to finally get the Adjoint matrix of

$A$

To transpose a matrix, freeze the diagonal elements and then flip it.

	$\begin{bmatrix} \color{#CC0000}{1} & 2 & -1 \\ 0 & \color{#CC0000}{-2} & 2 \\ 2 & -4 & \color{#CC0000}{4} \end{bmatrix}$	Matrix of Cofactor

	$\begin{bmatrix} \color{#CC0000}{1} & 0 & 2 \\ 2 & \color{#CC0000}{-2} & -4 \\ -1 & 2 & \color{#CC0000}{4} \end{bmatrix}$	Transpose the matrix

This is the Adjoint matrix of

$A$

Finally, substitute the components into the Inverse Formula

$|A|=2$

$A^(-1)$	$=$	$1/(\|A\|)Adj(A)$	Inverse Formula

	$=$	$1/2[[1,0,2],[2,-2,-4],[-1,2,4]]$	Substitute components

	$=$	$\begin{bmatrix} \frac{1}{2} & 0 & 1 \\[0.3em] 1 & -1 & -2 \\[0.3em] -\frac{1}{2} & 1 & 2 \end{bmatrix}$

$\begin{bmatrix} \frac{1}{2} & 0 & 1 \\[0.3em] 1 & -1 & -2 \\[0.3em] -\frac{1}{2} & 1 & 2 \end{bmatrix}$

Question 5 of 5

Find the inverse of the matrix:

$A=[[0,-1,1],[1,4,5],[1,-1,1]]$

1. $\begin{bmatrix} 9 & 2 & -1 \\ 4 & 1 & -1 \\ -5 & -1 & 1 \end{bmatrix}$
2. $\begin{bmatrix} 9 & 4 & -5 \\ -1 & 1 & -1 \\ -1 & 2 & 1 \end{bmatrix}$
3. $\begin{bmatrix} 5 & 2 & 1 \\ 4 & -1 & -1 \\ -9 & -1 & 1 \end{bmatrix}$
4. $\begin{bmatrix} -9 & -2 & 1 \\ -2 & 1 & -1 \\ 3 & 1 & -2 \end{bmatrix}$

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Inverse of a $3xx3$ Matrix

$A^(-1)=1/(|A|)Adj(A)$

where

$Adj(A)$ means Adjoint

$A$

Determinant of a $3xx3$ Matrix

$A=$

$\begin{bmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{bmatrix}$ then

$|A|=\color{#D800AD}{a}$

$\begin{vmatrix} e & f \\ h & k \end{vmatrix}$

$-\color{#D800AD}{b}$

$\begin{vmatrix} d & f \\ g & k \end{vmatrix}$

$+\color{#D800AD}{c}$

$\begin{vmatrix} d & e \\ g & h \end{vmatrix}$

Determinant of a $2xx2$ Matrix

$\begin{bmatrix} \color{#007DDC}{a} & \color{#9a00c7}{b} \\ \color{#9a00c7}{c} & \color{#007DDC}{d} \end{bmatrix}$ then

$|A|=\color{#007DDC}{ad}-\color{#9a00c7}{bc}$

To find the inverse of a

$3xx3$ matrix, solve for its determinant and adjoint matrix. The adjoint matrix can be found by forming the Matrix of Minors, transforming it into a Matrix of Cofactors, then transposing it.

First, solve for the determinant of

$A$

$\|A\|$	$=$	$\begin{vmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{vmatrix}$	$=$	$\color{#D800AD}{a}$ $\begin{vmatrix} e & f \\ h & k \end{vmatrix}$ $-\color{#D800AD}{b}$ $\begin{vmatrix} d & f \\ g & k \end{vmatrix}$ $+\color{#D800AD}{c}$ $\begin{vmatrix} d & e \\ g & h \end{vmatrix}$	Determinant of a $3xx3$ Matrix

	$=$	$\begin{vmatrix} \color{#D800AD}{0} & \color{#D800AD}{-1} & \color{#D800AD}{-1} \\ 1 & 4 & 5 \\ 1 & -1 & 1 \end{vmatrix}$	$=$	$\color{#D800AD}{0}$ $\begin{vmatrix} \color{#007DDC}{4} & \color{#9a00c7}{5} \\ \color{#9a00c7}{-1} & \color{#007DDC}{1} \end{vmatrix}$ $-\color{#D800AD}{-1}$ $\begin{vmatrix} \color{#007DDC}{1} & \color{#9a00c7}{5} \\ \color{#9a00c7}{1} & \color{#007DDC}{1} \end{vmatrix}$ $+\color{#D800AD}{-1}$ $\begin{vmatrix} \color{#007DDC}{1} & \color{#9a00c7}{4} \\ \color{#9a00c7}{1} & \color{#007DDC}{-1} \end{vmatrix}$	Substitute values

	$=$	$[\color{#D800AD}{0}(\color{#007DDC}{4\cdot1}-\color{#9a00c7}{5\cdot(-1)})]-[\color{#D800AD}{(-1)}(\color{#007DDC}{1\cdot1}-\color{#9a00c7}{5\cdot1})]+[\color{#D800AD}{(-1)}(\color{#007DDC}{1\cdot(-1)}-\color{#9a00c7}{4\cdot1})]$
	$=$	$0-(-1)(1-5)+(-1)((-1)-4)$
	$=$	$0+1(-4)-1(-5)$
	$=$	$0-4+5$
$\|A\|$	$=$	$1$

Next, start solving for the Adjoint matrix of

$A$ by forming the Matrix of Minors

Matrix of Minors is a

$3xx3$ matrix. An element of it is found by getting the determinant of the

$2xx2$ matrix formed by crossing out the row and column of that element from matrix

$A$

First Element(Row

$1$ Column

$1$ ):

$A=$

$\begin{bmatrix} \color{#9E9E9E}{0} & \color{#9E9E9E}{-1} & \color{#9E9E9E}{-1} \\ \color{#9E9E9E}{1} & 4 & 5 \\ \color{#9E9E9E}{1} & -1 & 1 \end{bmatrix}$

$\begin{vmatrix} 4 & 5 \\ -1 & 1 \end{vmatrix}$	$=$	$41-5(-1)$	Determinant of a $2xx2$ Matrix
	$=$	$4-(-5)$
	$=$	$9$

Matrix of Minors:

$\begin{bmatrix} 9 & \: & \: \\ \: & \: & \: \\ \: & \: & \: \end{bmatrix}$

Second Element(Row

$1$ Column

$2$ ):

$A=$

$\begin{bmatrix} \color{#9E9E9E}{0} & \color{#9E9E9E}{-1} & \color{#9E9E9E}{-1} \\ 1 & \color{#9E9E9E}{4} & 5 \\ 1 & \color{#9E9E9E}{-1} & 1 \end{bmatrix}$

$\begin{vmatrix} 1 & 5 \\ 1 & 1 \end{vmatrix}$	$=$	$11-51$	Determinant of a $2xx2$ Matrix
	$=$	$1-5$
	$=$	$-4$

Matrix of Minors:

$\begin{bmatrix} 9 & -4 & \: \\ \: & \: & \: \\ \: & \: & \: \end{bmatrix}$

Third Element(Row

$1$ Column

$3$ ):

$A=$

$\begin{bmatrix} \color{#9E9E9E}{0} & \color{#9E9E9E}{-1} & \color{#9E9E9E}{-1} \\ 1 & 4 & \color{#9E9E9E}{5} \\ 1 & -1 & \color{#9E9E9E}{1} \end{bmatrix}$

$\begin{vmatrix} 1 & 4 \\ 1 & -1 \end{vmatrix}$	$=$	$1(-1)-41$	Determinant of a $2xx2$ Matrix
	$=$	$-1-4$
	$=$	$-5$

Matrix of Minors:

$\begin{bmatrix} 9 & -4 & -5 \\ \: & \: & \: \\ \: & \: & \: \end{bmatrix}$

Fourth Element(Row

$2$ Column

$1$ ):

$A=$

$\begin{bmatrix} \color{#9E9E9E}{0} & -1 & -1 \\ \color{#9E9E9E}{1} & \color{#9E9E9E}{4} & \color{#9E9E9E}{5} \\ \color{#9E9E9E}{1} & -1 & 1 \end{bmatrix}$

$\begin{vmatrix} -1 & -1 \\ -1 & 1 \end{vmatrix}$	$=$	$-11-(-1)(-1)$	Determinant of a $2xx2$ Matrix
	$=$	$-1-1$
	$=$	$-2$

Matrix of Minors:

$\begin{bmatrix} 9 & -4 & -5 \\ -2 & \: & \: \\ \: & \: & \: \end{bmatrix}$

Fifth Element(Row

$2$ Column

$2$ ):

$A=$

$\begin{bmatrix} 0 & \color{#9E9E9E}{-1} & -1 \\ \color{#9E9E9E}{1} & \color{#9E9E9E}{4} & \color{#9E9E9E}{5} \\ 1 & \color{#9E9E9E}{-1} & 1 \end{bmatrix}$

$\begin{vmatrix} 0 & -1 \\ 1 & 1 \end{vmatrix}$	$=$	$01-(-1)1$	Determinant of a $2xx2$ Matrix
	$=$	$0-(-1)$
	$=$	$1$

Matrix of Minors:

$\begin{bmatrix} 9 & -4 & -5 \\ -2 & 1 & \: \\ \: & \: & \: \end{bmatrix}$

Sixth Element(Row

$2$ Column

$3$ ):

$A=$

$\begin{bmatrix} 0 & -1 & \color{#9E9E9E}{-1} \\ \color{#9E9E9E}{1} & \color{#9E9E9E}{4} & \color{#9E9E9E}{5} \\ 1 & -1 & \color{#9E9E9E}{1} \end{bmatrix}$

$\begin{vmatrix} 0 & -1 \\ 1 & -1 \end{vmatrix}$	$=$	$0(-1)-(-1)1$	Determinant of a $2xx2$ Matrix
	$=$	$0-(-1)$
	$=$	$1$

Matrix of Minors:

$\begin{bmatrix} 9 & -4 & -5 \\ -2 & 1 & 1 \\ \: & \: & \: \end{bmatrix}$

Seventh Element(Row

$3$ Column

$1$ ):

$A=$

$\begin{bmatrix} \color{#9E9E9E}{0} & -1 & -1 \\ \color{#9E9E9E}{1} & 4 & 5 \\ \color{#9E9E9E}{1} & \color{#9E9E9E}{-1} & \color{#9E9E9E}{1} \end{bmatrix}$

$\begin{vmatrix} -1 & -1 \\ 4 & 5 \end{vmatrix}$	$=$	$-15-(-1)4$	Determinant of a $2xx2$ Matrix
	$=$	$-5-(-4)$
	$=$	$-1$

Matrix of Minors:

$\begin{bmatrix} 9 & -4 & -5 \\ -2 & 1 & 1 \\ -1 & \: & \: \end{bmatrix}$

Eighth Element(Row

$3$ Column

$2$ ):

$A=$

$\begin{bmatrix} 0 & \color{#9E9E9E}{-1} & -1 \\ 1 & \color{#9E9E9E}{4} & 5 \\ \color{#9E9E9E}{1} & \color{#9E9E9E}{-1} & \color{#9E9E9E}{1} \end{bmatrix}$

$\begin{vmatrix} 0 & -1 \\ 1 & 5 \end{vmatrix}$	$=$	$05-(-1)1$	Determinant of a $2xx2$ Matrix
	$=$	$0-(-1)$
	$=$	$1$

Matrix of Minors:

$\begin{bmatrix} 9 & -4 & -5 \\ -2 & 1 & 1 \\ -1 & 1 & \: \end{bmatrix}$

Ninth Element(Row

$3$ Column

$3$ ):

$A=$

$\begin{bmatrix} 0 & -1 & \color{#9E9E9E}{-1} \\ 1 & 4 & \color{#9E9E9E}{5} \\ \color{#9E9E9E}{1} & \color{#9E9E9E}{-1} & \color{#9E9E9E}{1} \end{bmatrix}$

$\begin{vmatrix} 0 & -1 \\ 1 & 4 \end{vmatrix}$	$=$	$04-(-1)1$	Determinant of a $2xx2$ Matrix
	$=$	$0-(-1)$
	$=$	$1$

Matrix of Minors:

$\begin{bmatrix} 9 & -4 & -5 \\ -2 & 1 & 1 \\ -1 & 1 & 1 \end{bmatrix}$

Transform the Matrix of Minors into a Matrix of Cofactors by applying alternate signs to each element, starting with a positive sign.

	$\begin{bmatrix} 9 & -4 & -5 \\ -2 & 1 & 1 \\ -1 & 1 & 1 \end{bmatrix}$	Matrix of Minors

	$\begin{bmatrix} \color{#CC0000}{+}9 & \color{#CC0000}{-}(-4) & \color{#CC0000}{+}(-5) \\ \color{#CC0000}{-}(-2) & \color{#CC0000}{+}1 & \color{#CC0000}{-}1 \\ \color{#CC0000}{+}(-1) & \color{#CC0000}{-}1 & \color{#CC0000}{+}1 \end{bmatrix}$	Apply alternating signs

$=$	$\begin{bmatrix} 9 & 4 & -5 \\ 2 & 1 & -1 \\ -1 & -1 & 1 \end{bmatrix}$

This is now the Matrix of Cofactors

Transpose the Matrix of Cofactors to finally get the Adjoint matrix of

$A$

To transpose a matrix, freeze the diagonal elements and then flip it.

	$\begin{bmatrix} \color{#CC0000}{9} & 4 & -5 \\ 2 & \color{#CC0000}{1} & -1 \\ -1 & -1 & \color{#CC0000}{1} \end{bmatrix}$	Matrix of Cofactor

	$\begin{bmatrix} \color{#CC0000}{9} & 2 & -1 \\ 4 & \color{#CC0000}{1} & -1 \\ -5 & -1 & \color{#CC0000}{1} \end{bmatrix}$	Transpose the matrix

This is the Adjoint matrix of

$A$

Finally, substitute the components into the Inverse Formula

$|A|=1$

$A^(-1)$	$=$	$1/(\|A\|)Adj(A)$	Inverse Formula

	$=$	$1/1[[9,2,-1],[4,1,-1],[-5,-1,1]]$	Substitute components

	$=$	$\begin{bmatrix} 9 & 2 & -1 \\ 4 & 1 & -1 \\ -5 & -1 & 1 \end{bmatrix}$

$\begin{bmatrix} 9 & 2 & -1 \\ 4 & 1 & -1 \\ -5 & -1 & 1 \end{bmatrix}$

Inverse of a Matrix

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Quiz summary

Information

1. Question

Hint

Well Done!

Inverse of a $2xx2$ Matrix

Determinant of a $2xx2$ Matrix

2. Question

Hint

Excellent!

Inverse of a $2xx2$ Matrix

Determinant of a $2xx2$ Matrix

3. Question

Hint

Keep Going!

Inverse of a $3xx3$ Matrix

Determinant of a $3xx3$ Matrix

Determinant of a $2xx2$ Matrix

4. Question

Hint

Great Work!

Inverse of a $3xx3$ Matrix

Determinant of a $3xx3$ Matrix

Determinant of a $2xx2$ Matrix

5. Question

Hint

Exceptional!

Inverse of a $3xx3$ Matrix

Determinant of a $3xx3$ Matrix

Determinant of a $2xx2$ Matrix

Quizzes

Inverse of a Matrix

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Quiz summary

Information

1. Question

Hint

Well Done!

Inverse of a 2×22xx2 Matrix

Determinant of a 2×22xx2 Matrix

2. Question

Hint

Excellent!

Inverse of a 2×22xx2 Matrix

Determinant of a 2×22xx2 Matrix

3. Question

Hint

Keep Going!

Inverse of a 3×33xx3 Matrix

Determinant of a 3×33xx3 Matrix

Determinant of a 2×22xx2 Matrix

4. Question

Hint

Great Work!

Inverse of a 3×33xx3 Matrix

Determinant of a 3×33xx3 Matrix

Determinant of a 2×22xx2 Matrix

5. Question

Hint

Exceptional!

Inverse of a 3×33xx3 Matrix

Determinant of a 3×33xx3 Matrix

Determinant of a 2×22xx2 Matrix

Quizzes

Inverse of a $2xx2$ Matrix

Determinant of a $2xx2$ Matrix

Inverse of a $2xx2$ Matrix

Determinant of a $2xx2$ Matrix

Inverse of a $3xx3$ Matrix

Determinant of a $3xx3$ Matrix

Determinant of a $2xx2$ Matrix

Inverse of a $3xx3$ Matrix

Determinant of a $3xx3$ Matrix

Determinant of a $2xx2$ Matrix

Inverse of a $3xx3$ Matrix

Determinant of a $3xx3$ Matrix

Determinant of a $2xx2$ Matrix