Inverse of a Matrix
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Question 1 of 5
1. Question
Find the inverse of the matrix:`A=[[3,5],[2,4]]`Hint
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Inverse of a `2xx2` Matrix
If \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
If \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 `=` `3*4` `-``5*2` 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
2. Question
Find the inverse of the matrix:`B=[[4,-5],[-3,6]]`Hint
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Excellent!
Incorrect
Inverse of a `2xx2` Matrix
If \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
If \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 `=` `4*6` `-``(-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
3. Question
Find the inverse of the matrix:`A=[[2,0,1],[-3,1,1],[4,-1,-1]]`Hint
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Keep Going!
Incorrect
Inverse of a `3xx3` Matrix
`A^(-1)=1/(|A|)Adj(A)`where `Adj(A)` means Adjoint `A`Determinant of a `3xx3` Matrix
If $$A=$$\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
If \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 MinorsMatrix 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)-1*4` 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)-1*4` 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)-1*4` 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)-0*4` 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}`=` `0*1-1*1` 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}`=` `2*1-1*(-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}`=` `2*1-0*(-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 CofactorsTranspose 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
4. Question
Find the inverse of the matrix:`A=[[0,2,2],[-2,3,4],[1,-1,-1]]`Hint
Help VideoCorrect
Great Work!
Incorrect
Inverse of a `3xx3` Matrix
`A^(-1)=1/(|A|)Adj(A)`where `Adj(A)` means Adjoint `A`Determinant of a `3xx3` Matrix
If $$A=$$\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
If \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 MinorsMatrix 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)-4*1` 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)-3*1` 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)-2*1` 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)-2*1` 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}`=` `2*4-2*3` 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}`=` `0*4-2*(-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}`=` `0*3-2*(-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 CofactorsTranspose 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
5. Question
Find the inverse of the matrix:`A=[[0,-1,1],[1,4,5],[1,-1,1]]`Hint
Help VideoCorrect
Exceptional!
Incorrect
Inverse of a `3xx3` Matrix
`A^(-1)=1/(|A|)Adj(A)`where `Adj(A)` means Adjoint `A`Determinant of a `3xx3` Matrix
If $$A=$$\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
If \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 MinorsMatrix 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}`=` `4*1-5*(-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}`=` `1*1-5*1` 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)-4*1` 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}`=` `-1*1-(-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}`=` `0*1-(-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}`=` `-1*5-(-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}`=` `0*5-(-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}`=` `0*4-(-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 CofactorsTranspose 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}
Quizzes
- Adding & Subtracting Matrices 1
- Adding & Subtracting Matrices 2
- Adding & Subtracting Matrices 3
- Multiplying Matrices 1
- Multiplying Matrices 2
- Multiplication Word Problems
- Determinant of a Matrix
- Inverse of a Matrix
- Solving Systems of Equations 1
- Solving Systems of Equations 2
- Gauss Jordan Elimination
- Cramer’s Rule