Solving Systems with Inverses

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Find the determinant of the 2x2 matrix \(\left[\begin{array}{ll}4 & -2 \\ 3 & -1\end{array}\right]\)

   បង្ហាញ​ចម្លើយ

   Write the determinant
   Subtract the products of the diagonals	\(4(-1)-3(-2)\)
   Simplify	\(\begin{array}{l}-4+6 \\ 2\end{array}\)

2. \(\left[\begin{array}{ll}6 & -2 \\ 3 & -1\end{array}\right]\)

3. \(\left[\begin{array}{ll}-4 & 8 \\ -3 & 5\end{array}\right]\)

4. For the determinant \(|\begin{array}{lllllll}4 & & & -2 & & & 3 \\ 1 & & & 0 & & & -3 \\ -2 & & & -4 & & & 2\end{array}|,\) find and then evaluate the minor of ⓐ \({a}_{1}\) ⓑ \({b}_{3}\)

   បង្ហាញ​ចម្លើយ

   ⓐ
   

   Eliminate the row and column that contains \({a}_{1}.\)
   Write the \(2\ \times \ 2\) determinant that remains.
   Evaluate.
   Simplify.

   
   
   ⓑ
   

   Eliminate the row and column that contains \({b}_{3}.\)
   Write the \(2\ \times \ 2\) determinant that remains.
   Evaluate.
   Simplify.

5. Evaluate the determinant of the 3x3 matrix by expanding by minors along the first row

   \(\left|\begin{array}{lll}2 & -3 & -1 \\ 3 & 2 & 0 \\ -1 & -1 & -2\end{array}\right|\)

   បង្ហាញ​ចម្លើយ

   Expand by minors along the first row
   Evaluate each determinant.
   Simplify.
   Simplify.
   Simplify.

6. Evaluate the determinant of the 3x3 matrix by expanding by minors along the first row. \(\left|\begin{array}{lll}-5 & -1 & -4 \\ 4 & 0 & -3 \\ 2 & -2 & 6\end{array}\right|\)

7. Given matrix A, show that \(AI=IA=A.\)

   \[A=[\begin{array}{ll}3 & 4 \\ -2 & 5\end{array}]\]
   បង្ហាញ​ចម្លើយ

   Use matrix multiplication to show that the product of \(A\) and the identity is equal to the product of the identity and A.

   \[AI=[\begin{array}{lll}3 & & 4 \\ -2 & & 5\end{array}]\begin{array}{l}\end{array}[\begin{array}{lll}1 & & 0 \\ 0 & & 1\end{array}]=[\begin{array}{llll}3⋅1+4⋅0 & & & 3⋅0+4⋅1 \\ -2⋅1+5⋅0 & & & -2⋅0+5⋅1\end{array}]=[\begin{array}{lll}3 & & 4 \\ -2 & & 5\end{array}]\]\[IA=[\begin{array}{lll}1 & & 0 \\ 0 & & 1\end{array}]\begin{array}{l}\end{array}[\begin{array}{lll}3 & & 4 \\ -2 & & 5\end{array}]=[\begin{array}{llll}1⋅3+0⋅(-2) & & & 1⋅4+0⋅5 \\ 0⋅3+1⋅(-2) & & & 0⋅4+1⋅5\end{array}]=[\begin{array}{lll}3 & & 4 \\ -2 & & 5\end{array}]\]

8. Show that the given matrices are multiplicative inverses of each other.

   \[A=[\begin{array}{lll}1 & & 5 \\ -2 & & -9\end{array}],B=[\begin{array}{lll}-9 & & -5 \\ 2 & & 1\end{array}]\]
   បង្ហាញ​ចម្លើយ

   Multiply \(AB\) and \(BA.\) If both products equal the identity, then the two matrices are inverses of each other.

   \[\begin{array}{lllllllll}AB=[\begin{array}{lll}1 & & 5 \\ -2 & & -9\end{array}]\cdot [\begin{array}{lll}-9 & & -5 \\ 2 & & 1\end{array}] \\ =[\begin{array}{lll}1(-9)+5(2) & & 1(-5)+5(1) \\ -2(-9)-9(2) & & -2(-5)-9(1)\end{array}] \\ =[\begin{array}{lll}1 & & 0 \\ 0 & & 1\end{array}]\end{array}\]\[\begin{array}{lllllllll}BA=[\begin{array}{lll}-9 & & -5 \\ 2 & & 1\end{array}]\cdot [\begin{array}{lll}1 & & 5 \\ -2 & & -9\end{array}] \\ =[\begin{array}{lll}-9(1)-5(-2) & & -9(5)-5(-9) \\ 2(1)+1(-2) & & 2(5)+1(-9)\end{array}] \\ =[\begin{array}{lll}1 & & 0 \\ 0 & & 1\end{array}]\end{array}\]

   \(A\) and \(B\) are inverses of each other.

9. Show that the following two matrices are inverses of each other.

   \[A=[\begin{array}{lll}1 & & 4 \\ -1 & & -3\end{array}],B=[\begin{array}{lll}-3 & & -4 \\ 1 & & 1\end{array}]\]
   បង្ហាញ​ចម្លើយ

   \[\begin{array}{lllllllllllllllll}AB=[\begin{array}{lll}1 & & 4 \\ -1 & & -3\end{array}]\begin{array}{l}\end{array}[\begin{array}{lll}-3 & & -4 \\ 1 & & 1\end{array}]=[\begin{array}{lll}1(-3)+4(1) & & 1(-4)+4(1) \\ -1(-3)+-3(1) & & -1(-4)+-3(1)\end{array}]=[\begin{array}{lll}1 & & 0 \\ 0 & & 1\end{array}] \\ BA=[\begin{array}{lll}-3 & & -4 \\ 1 & & 1\end{array}]\begin{array}{l}\end{array}[\begin{array}{lll}1 & & 4 \\ -1 & & -3\end{array}]=[\begin{array}{lll}-3(1)+-4(-1) & & -3(4)+-4(-3) \\ 1(1)+1(-1) & & 1(4)+1(-3)\end{array}]=[\begin{array}{lll}1 & & 0 \\ 0 & & 1\end{array}]\end{array}\]

10. Use matrix multiplication to find the inverse of the given matrix.

    \[A=[\begin{array}{lll}1 & & -2 \\ 2 & & -3\end{array}]\]
    បង្ហាញ​ចម្លើយ

    For this method, we multiply \(A\) by a matrix containing unknown constants and set it equal to the identity.

    \[[\begin{array}{ll}1 & -2 \\ 2 & -3\end{array}]\ [\begin{array}{ll}a & b \\ c & d\end{array}]=[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}]\]

    Find the product of the two matrices on the left side of the equal sign.

    \[[\begin{array}{ll}1 & -2 \\ 2 & -3\end{array}]\ [\begin{array}{ll}a & b \\ c & d\end{array}]=[\begin{array}{ll}1a-2c & 1b-2d \\ 2a-3c & 2b-3d\end{array}]\]

    Next, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, 1. Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is 0.

    \[\begin{array}{l}1a-2c=1\ {R}_{1} \\ 2a-3c=0\ {R}_{2}\end{array}\]

    Using row operations, multiply and add as follows: \((-2){R}_{1}+{R}_{2}\to {R}_{2}.\) Add the equations, and solve for \(c.\)

    \[\begin{array}{l}1a-2c=1\ \\ 0+1c=-2 \\ c=-2\end{array}\]

    Back-substitute to solve for \(a.\)

    \[\begin{array}{l}a-2(-2)=1\ \\ a+4=1\ \\ a=-3\end{array}\]

    Write another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, 0. Set the entry in row 2, column 2 equal to the corresponding entry of the identity.

    \[\begin{array}{ll}1b-2d=0 & {R}_{1} \\ 2b-3d=1 & {R}_{2}\end{array}\]

    Using row operations, multiply and add as follows: \((-2){R}_{1}+{R}_{2}={R}_{2}.\) Add the two equations and solve for \(d.\)

    \[\begin{array}{l}1b-2d=0 \\ \frac{0+1d=1}{\ d=1} \\ \end{array}\]

    Once more, back-substitute and solve for \(b.\)

    \[\begin{array}{l}b-2(1)=0 \\ b-2=0 \\ b=2\end{array}\]\[{A}^{-1}=[\begin{array}{lll}-3 & & 2 \\ -2 & & 1\end{array}]\]

11. Use the formula to find the multiplicative inverse of

    \[A=[\begin{array}{ll}1 & -2 \\ 2 & -3\end{array}]\]
    បង្ហាញ​ចម្លើយ

    Using the formula, we have

    \[\begin{array}{lll}{A}^{-1}=\frac{1}{(1)(-3)-(-2)(2)}[\begin{array}{ll}-3 & 2 \\ -2 & 1\end{array}] \\ =\frac{1}{-3+4}[\begin{array}{ll}-3 & 2 \\ -2 & 1\end{array}] \\ =[\begin{array}{ll}-3 & 2 \\ -2 & 1\end{array}]\end{array}\]

12. Use the formula to find the inverse of matrix \(A.\) Verify your answer by augmenting with the identity matrix.

    \[A=[\begin{array}{ll}1 & -1 \\ 2 & \ 3\end{array}]\]
    បង្ហាញ​ចម្លើយ

    \({A}^{-1}=[\begin{array}{ll}\frac{3}{5} & \frac{1}{5} \\ -\frac{2}{5} & \frac{1}{5}\end{array}]\)

13. Find the inverse, if it exists, of the given matrix.

    \[A=[\begin{array}{ll}3 & 6 \\ 1 & 2\end{array}]\]
    បង្ហាញ​ចម្លើយ

    We will use the method of augmenting with the identity.

    \[[\begin{array}{ll}3 & 6 \\ 1 & 2\end{array}|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}]\]

    1. Switch row 1 and row 2. \[[\begin{array}{ll}1 & 3 \\ 3 & 2\end{array}|\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}]\]
    2. Multiply row 1 by −3 and add it to row 2. \[[\begin{array}{ll}1 & 2 \\ 0 & 0\end{array}|\begin{array}{ll}1 & 0 \\ -3 & 1\end{array}]\]
    3. There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse.

14. Given the \(3\times 3\) matrix \(A,\) find the inverse.

    \[A=[\begin{array}{lll}2 & 3 & 1 \\ 3 & 3 & 1 \\ 2 & 4 & 1\end{array}]\]
    បង្ហាញ​ចម្លើយ

    Augment \(A\) with the identity matrix, and then begin row operations until the identity matrix replaces \(A.\) The matrix on the right will be the inverse of \(A.\)

    \[[\begin{array}{lll}2 & 3 & 1 \\ 3 & 3 & 1 \\ 2 & 4 & 1\end{array}\ |\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}]\overset{\text{Interchange }{R}_{2}\ \text{and }{R}_{1}}{\to }[\begin{array}{lll}3 & 3 & 1 \\ 2 & 3 & 1 \\ 2 & 4 & 1\end{array}\ |\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}]\]\[-{R}_{2}+{R}_{1}={R}_{1}\to [\begin{array}{lll}1 & 0 & 0 \\ 2 & 3 & 1 \\ 2 & 4 & 1\end{array}\ |\ \begin{array}{lll}-1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}]\]\[-{R}_{2}+{R}_{3}={R}_{3}\to [\begin{array}{lll}1 & 0 & 0 \\ 2 & 3 & 1 \\ 0 & 1 & 0\end{array}\ |\ \begin{array}{lll}-1 & 1 & 0 \\ 1 & 0 & 0 \\ -1 & 0 & 1\end{array}]\]\[{R}_{3}\ ↔{R}_{2}\to [\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 3 & 1\end{array}\ |\ \begin{array}{lll}-1 & 1 & 0 \\ -1 & 0 & 1 \\ 1 & 0 & 0\end{array}]\]\[-2{R}_{1}+{R}_{3}={R}_{3}\to [\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 3 & 1\end{array}\ |\ \begin{array}{lll}-1 & 1 & 0 \\ -1 & 0 & 1 \\ 3 & -2 & 0\end{array}]\]\[-3{R}_{2}+{R}_{3}={R}_{3}\to [\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\ |\ \begin{array}{lll}-1 & 1 & 0 \\ -1 & 0 & 1 \\ 6 & -2 & -3\end{array}]\]

    Thus,

    \[{A}^{-1}=B=[\begin{array}{lll}-1 & \ 1 & \ 0 \\ -1 & \ 0 & \ 1 \\ 6 & -2 & -3\end{array}\ ]\]

15. Find the inverse of the \(3\times 3\) matrix.

    \[A=[\begin{array}{lll}\ 2 & -17 & 11 \\ -1 & \ 11 & -7 \\ 0 & \ 3 & -2\end{array}]\]
    បង្ហាញ​ចម្លើយ

    \({A}^{-1}=[\begin{array}{lll}1 & 1 & \ 2 \\ 2 & 4 & -3 \\ 3 & 6 & -5\end{array}]\)

16. Solve the given system of equations using the inverse of a matrix.

    \[\begin{array}{l}3x+8y=5 \\ 4x+11y=7\end{array}\]
    បង្ហាញ​ចម្លើយ

    Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.

    \[A=[\begin{array}{ll}3 & 8 \\ 4 & 11\end{array}],X=[\begin{array}{l}x \\ y\end{array}],B=[\begin{array}{l}5 \\ 7\end{array}]\]

    Then

    \[[\begin{array}{ll}3 & 8 \\ 4 & 11\end{array}]\ [\begin{array}{l}x \\ y\end{array}]=[\begin{array}{l}5 \\ 7\end{array}]\]

    First, we need to calculate \({A}^{-1}.\) Using the formula to calculate the inverse of a 2 by 2 matrix, we have:

    \[\begin{array}{lll}{A}^{-1}=\frac{1}{ad-bc}[\begin{array}{ll}d & -b \\ -c & a\end{array}] \\ \ =\frac{1}{3(11)-8(4)}[\begin{array}{ll}11 & -8 \\ -4 & 3\end{array}] \\ \ =\frac{1}{1}[\begin{array}{ll}11 & -8 \\ -4 & 3\end{array}]\end{array}\]

    So,

    \[{A}^{-1}=[\begin{array}{ll}11 & -8 \\ -4 & \text{}\text{}\ 3\end{array}]\]

    Now we are ready to solve. Multiply both sides of the equation by \({A}^{-1}.\)

    \[\begin{array}{lllllll}\ ({A}^{-1})AX=({A}^{-1})B \\ [\begin{array}{ll}11 & -8 \\ -4 & 3\end{array}]\ [\begin{array}{ll}3 & 8 \\ 4 & 11\end{array}]\ [\begin{array}{l}x \\ y\end{array}]=[\begin{array}{ll}11 & -8 \\ -4 & 3\end{array}]\ [\begin{array}{l}5 \\ 7\end{array}] \\ [\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}]\ [\begin{array}{l}x \\ y\end{array}]=[\begin{array}{l}11(5)+(-8)7 \\ -4(5)+3(7)\end{array}] \\ [\begin{array}{l}x \\ y\end{array}]=[\begin{array}{l}-1 \\ 1\end{array}]\end{array}\]

    The solution is \((-1,1).\)

17. Solve the following system using the inverse of a matrix.

    \[\begin{array}{l}5x+15y+56z=35\ \\ -4x-11y-41z=-26 \\ -x-3y-11z=-7\ \end{array}\]
    បង្ហាញ​ចម្លើយ

    Write the equation \(AX=B.\)

    \[[\begin{array}{lll}5 & 15 & 56 \\ -4 & -11 & -41 \\ -1 & -3 & -11\end{array}]\ [\begin{array}{l}x \\ y \\ z\end{array}]=[\begin{array}{l}35 \\ -26 \\ -7\end{array}]\]

    First, we will find the inverse of \(A\) by augmenting with the identity.

    \[[\begin{array}{lll}5 & 15 & 56 \\ -4 & -11 & -41 \\ -1 & -3 & -11\end{array}\ |\ \begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}]\]

    Multiply row 1 by \(\frac{1}{5}.\)

    \[[\begin{array}{lll}1 & 3 & \frac{56}{5} \\ -4 & -11 & -41 \\ -1 & -3 & -11\end{array}\ |\ \begin{array}{lll}\frac{1}{5} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}]\]

    Multiply row 1 by 4 and add to row 2.

    \[[\begin{array}{lll}1 & 3 & \frac{56}{5} \\ 0 & 1 & \frac{19}{5} \\ -1 & -3 & -11\end{array}\ |\ \begin{array}{lll}\frac{1}{5} & 0 & 0 \\ \frac{4}{5} & 1 & 0 \\ 0 & 0 & 1\end{array}]\]

    Add row 1 to row 3.

    \[[\begin{array}{lll}1 & 3 & \frac{56}{5} \\ 0 & 1 & \frac{19}{5} \\ 0 & 0 & \frac{1}{5}\end{array}\ |\ \begin{array}{lll}\frac{1}{5} & 0 & 0 \\ \frac{4}{5} & 1 & 0 \\ \frac{1}{5} & 0 & 1\end{array}]\]

    Multiply row 2 by −3 and add to row 1.

    \[[\begin{array}{lll}1 & 0 & -\frac{1}{5} \\ 0 & 1 & \frac{19}{5} \\ 0 & 0 & \frac{1}{5}\end{array}\ |\ \begin{array}{lll}-\frac{11}{5} & -3 & 0 \\ \frac{4}{5} & 1 & 0 \\ \frac{1}{5} & 0 & 1\end{array}]\]

    Multiply row 3 by 5.

    \[[\begin{array}{lll}1 & 0 & -\frac{1}{5} \\ 0 & 1 & \frac{19}{5} \\ 0 & 0 & 1\end{array}\ |\ \begin{array}{lll}-\frac{11}{5} & -3 & 0 \\ \frac{4}{5} & 1 & 0 \\ 1 & 0 & 5\end{array}]\]

    Multiply row 3 by \(\frac{1}{5}\) and add to row 1.

    \[[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & \frac{19}{5} \\ 0 & 0 & 1\end{array}\ |\ \begin{array}{lll}-2 & -3 & 1 \\ \frac{4}{5} & 1 & 0 \\ 1 & 0 & 5\end{array}]\]

    Multiply row 3 by \(-\frac{19}{5}\) and add to row 2.

    \[[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\ |\ \begin{array}{lll}-2 & -3 & 1 \\ -3 & 1 & -19 \\ 1 & 0 & 5\end{array}]\]

    So,

    \[{A}^{-1}=[\begin{array}{lll}-2 & -3 & 1 \\ -3 & 1 & -19 \\ 1 & 0 & 5\end{array}]\]

    Multiply both sides of the equation by \({A}^{-1}.\) We want \({A}^{-1}AX={A}^{-1}B:\)

    \[[\begin{array}{lll}-2 & -3 & 1 \\ -3 & 1 & -19 \\ 1 & 0 & 5\end{array}]\ [\begin{array}{lll}5 & 15 & 56 \\ -4 & -11 & -41 \\ -1 & -3 & -11\end{array}]\ [\begin{array}{l}x \\ y \\ z\end{array}]=[\begin{array}{lll}-2 & -3 & 1 \\ -3 & 1 & -19 \\ 1 & 0 & 5\end{array}]\ [\begin{array}{l}35 \\ -26 \\ -7\end{array}]\]

    Thus,

    \[{A}^{-1}B=[\begin{array}{l}-70+78-7 \\ -105-26+133 \\ 35+0-35\end{array}]=[\begin{array}{l}1 \\ 2 \\ 0\end{array}]\]

    The solution is \((1,2,0).\)

18. Solve the system using the inverse of the coefficient matrix.

    \[\begin{array}{l}\ 2x-17y+11z=0 \\ \ -x+11y-7z=8 \\ \ 3y-2z=-2\end{array}\]
    បង្ហាញ​ចម្លើយ

    \(X=[\begin{array}{l}4 \\ 38 \\ 58\end{array}]\)

19. Solve the system of equations with matrix inverses using a calculator

    \[\begin{array}{l}2x+3y+z=32 \\ 3x+3y+z=-27 \\ 2x+4y+z=-2\end{array}\]
    បង្ហាញ​ចម្លើយ

    On the matrix page of the calculator, enter the coefficient matrix as the matrix variable \([A],\) and enter the constant matrix as the matrix variable \([B].\)

    \[[A]=[\begin{array}{lll}2 & 3 & 1 \\ 3 & 3 & 1 \\ 2 & 4 & 1\end{array}],\ [B]=[\begin{array}{l}32 \\ -27 \\ -2\end{array}]\]

    On the home screen of the calculator, type in the multiplication to solve for \(X,\) calling up each matrix variable as needed.

    \[{[A]}^{-1}\times [B]\]

    Evaluate the expression.

    \[[\begin{array}{l}-59 \\ -34 \\ 252\end{array}]\]

20. In a previous section, we showed that matrix multiplication is not commutative, that is, \(AB\ne BA\) in most cases. Can you explain why matrix multiplication is commutative for matrix inverses, that is, \({A}^{-1}A=A{A}^{-1}?\)

    បង្ហាញ​ចម្លើយ

    If \({A}^{-1}\) is the inverse of \(A,\) then \(A{A}^{-1}=I,\) the identity matrix. Since \(A\) is also the inverse of \({A}^{-1},{A}^{-1}A=I.\) You can also check by proving this for a \(2\times 2\) matrix.

21. Does every \(2\times 2\) matrix have an inverse? Explain why or why not. Explain what condition is necessary for an inverse to exist.

22. Can you explain whether a \(2\times 2\) matrix with an entire row of zeros can have an inverse?

    បង្ហាញ​ចម្លើយ

    No, because \(ad\) and \(bc\) are both 0, so \(ad-bc=0,\) which requires us to divide by 0 in the formula.

23. Can a matrix with an entire column of zeros have an inverse? Explain why or why not.

24. Can a matrix with zeros on the diagonal have an inverse? If so, find an example. If not, prove why not. For simplicity, assume a \(2\times 2\) matrix.

    បង្ហាញ​ចម្លើយ

    Yes. Consider the matrix \([\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}].\) The inverse is found with the following calculation: \({A}^{-1}=\frac{1}{0(0)-1(1)}[\begin{array}{ll}0 & -1 \\ -1 & 0\end{array}]=[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}].\)

25. \(A=[\begin{array}{ll}1 & 0 \\ -1 & 1\end{array}],\ B=[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}]\)

26. \(A=[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}],\ B=[\begin{array}{ll}-2 & 1 \\ \frac{3}{2} & -\frac{1}{2}\end{array}]\)

    បង្ហាញ​ចម្លើយ

    \(AB=BA=[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}]=I\)

27. \(A=[\begin{array}{ll}4 & 5 \\ 7 & 0\end{array}],\ B=[\begin{array}{ll}0 & \frac{1}{7} \\ \frac{1}{5} & -\frac{4}{35}\end{array}]\)

28. \(A=[\begin{array}{ll}-2 & \frac{1}{2} \\ 3 & -1\end{array}],\ B=[\begin{array}{ll}-2 & -1 \\ -6 & -4\end{array}]\)

    បង្ហាញ​ចម្លើយ

    \(AB=BA=[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}]=I\)

29. \(A=[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 1 & 1\end{array}],\ B=\frac{1}{2}[\begin{array}{lll}2 & 1 & -1 \\ 0 & 1 & 1 \\ 0 & -1 & 1\end{array}]\)

30. \(A=[\begin{array}{lll}1 & 2 & 3 \\ 4 & 0 & 2 \\ 1 & 6 & 9\end{array}],\ B=\frac{1}{4}[\begin{array}{lll}6 & 0 & -2 \\ 17 & -3 & -5 \\ -12 & 2 & 4\end{array}]\)

    បង្ហាញ​ចម្លើយ

    \(AB=BA=[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}]=I\)

31. \(A=[\begin{array}{lll}3 & 8 & 2 \\ 1 & 1 & 1 \\ 5 & 6 & 12\end{array}],\ B=\frac{1}{36}[\begin{array}{lll}-6 & 84 & -6 \\ 7 & -26 & 1 \\ -1 & -22 & 5\end{array}]\)

32. \([\begin{array}{ll}3 & -2 \\ 1 & 9\end{array}]\)

    បង្ហាញ​ចម្លើយ

    \(\frac{1}{29}[\begin{array}{ll}9 & 2 \\ -1 & 3\end{array}]\)

33. \([\begin{array}{ll}-2 & 2 \\ 3 & 1\end{array}]\)

34. \([\begin{array}{ll}-3 & 7 \\ 9 & 2\end{array}]\)

    បង្ហាញ​ចម្លើយ

    \(\frac{1}{69}[\begin{array}{ll}-2 & 7 \\ 9 & 3\end{array}]\)

35. \([\begin{array}{ll}-4 & -3 \\ -5 & 8\end{array}]\)

36. \([\begin{array}{ll}1 & 1 \\ 2 & 2\end{array}]\)

    បង្ហាញ​ចម្លើយ

    There is no inverse

37. \([\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}]\)

38. \([\begin{array}{ll}0.5 & 1.5 \\ 1 & -0.5\end{array}]\)

    បង្ហាញ​ចម្លើយ

    \(\frac{4}{7}[\begin{array}{ll}0.5 & 1.5 \\ 1 & -0.5\end{array}]\)

39. \([\begin{array}{lll}1 & 0 & 6 \\ -2 & 1 & 7 \\ 3 & 0 & 2\end{array}]\)

40. \([\begin{array}{lll}0 & 1 & -3 \\ 4 & 1 & 0 \\ 1 & 0 & 5\end{array}]\)

    បង្ហាញ​ចម្លើយ

    \(\frac{1}{17}[\begin{array}{lll}-5 & 5 & -3 \\ 20 & -3 & 12 \\ 1 & -1 & 4\end{array}]\)

ព្យាយាម​របស់​អ្នក​ផ្ទាល់

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

បន្ថែម​ទៀត​ក្នុង Linear Algebra

DeterminantsMatrix inverseEigenvalues and eigenvectorsMatrix multiplicationRow reductionVector spaces, span and linear independenceOrthogonality, projections and least squaresDiagonalisation and matrix powersLinear transformations and change of basis