Question

Find AB, BA , A+B , A-B , ${\mathbf{A}}^{\mathbf{2}}$, ${\mathbf{B}}^{\mathbf{2}}$,5.A,3,B . Observe that $\mathbf{AB}\mathbf{\ne }\mathbf{BA}$. Show that $\mathbf{(}\mathbf{A}\mathbf{-}\mathbf{B}\mathbf{)}\mathbf{(}\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{)}\mathbf{\ne }\mathbf{(}\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{)}\mathbf{(}\mathbf{A}\mathbf{-}\mathbf{B}\mathbf{)}\mathbf{\ne }{\mathbf{A}}^{\mathbf{2}}\mathbf{-}{\mathbf{B}}^{\mathbf{2}}$. Show that $\mathbf{det}\mathbf{\left(}\mathbf{AB}\mathbf{\right)}\mathbf{=}\mathbf{det}\mathbf{\left(}\mathbf{BA}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\mathbf{detA}\mathbf{\right)}\mathbf{\left(}\mathbf{detB}\mathbf{\right)}$, but that $\mathbf{det}\mathbf{(}\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{)}\mathbf{\ne }\mathbf{detA}\mathbf{+}\mathbf{detB}$. Show that $\mathbf{det}\mathbf{\left(}\mathbf{5}\mathbf{A}\mathbf{\right)}\mathbf{\ne }\mathbf{5}\mathbf{detA}$ and find n so that $\mathbf{det}\mathbf{\left(}\mathbf{5}\mathbf{A}\mathbf{\right)}\mathbf{=}{\mathbf{5}}^{\mathbf{n}}\mathbf{detA}$. Find similar results for $\mathbf{det}\mathbf{\left(}\mathbf{3}\mathbf{B}\mathbf{\right)}$. Remember that the point of doing these simple problems by hand is to learn how to manipulate determinants and matrices correctly. Check your answers by computer.

role="math" localid="1658986967380" $\mathbf{A}\mathbf{=}\mathbf{\left(}\begin{array}{ccc}\mathbf{1}& \mathbf{0}& \mathbf{2}\\ \mathbf{3}& \mathbf{-}\mathbf{1}& \mathbf{0}\\ \mathbf{0}& \mathbf{5}& \mathbf{1}\end{array}\mathbf{\right)}\mathbf{,}\mathbf{B}\mathbf{=}\mathbf{\left(}\begin{array}{ccc}\mathbf{1}& \mathbf{1}& \mathbf{0}\\ \mathbf{0}& \mathbf{2}& \mathbf{1}\\ \mathbf{3}& \mathbf{-}\mathbf{1}& \mathbf{0}\end{array}\mathbf{\right)}$

Short answer

By finding the product , sum, and subtraction of required matrices AB ,BA ,A+B,A-B ,${\mathrm{A}}^2$ ,${\mathrm{B}}^2$,5.A, 3.B it can be proved that $(\mathrm{A}-\mathrm{B})(\mathrm{A}+\mathrm{B})\ne (\mathrm{A}+\mathrm{B})(\mathrm{A}-\mathrm{B})\ne {\mathrm{A}}^2-{\mathrm{B}}^2$,$\det \left(\mathrm{AB}\right)=\det \left(\mathrm{BA}\right)=\left(\mathrm{detA}\right)\left(\mathrm{detB}\right)$ , and $\det \left(5\mathrm{A}\right)\ne 5\mathrm{detA}$ .

Step-by-step solution

Definition of Matrix multiplication:

Matrix multiplication is a binary operation that creates a matrix by multiplying two matrices together. For matrix multiplication to work, the number of columns in the first matrix must equal the number of rows in the second matrix.

For example, if matrix A and B is defined as $\left(3\times 2\right)$and $\left(3\times 2\right)$then, the product of the matrices A and B is meaningless as the columns in the first matrix (here, 2) is not equals to the number of rows in the second matrix (here, 3).

Given parameters:

The given matrices are

$\mathrm{A}=\left(\begin{array}{ccc}1& 0& 2\\ 3& -1& 0\\ 0& 5& 1\end{array}\right),\mathrm{B}=\left(\begin{array}{ccc}1& 1& 0\\ 0& 2& 1\\ 3& -1& 0\end{array}\right)$

The products, sum and subtraction of two of meaningful matrices is to be find.

Finding product of the matrices:

Find the product of the matrix AB.

$$\begin{gathered} \mathrm{AB}=\left(\begin{array}{ccc}1& 0& 2\\ 3& -1& 0\\ 0& 5& 1\end{array}\right)\left(\begin{array}{ccc}1& 1& 0\\ 0& 2& 1\\ 3& -1& 0\end{array}\right) \\ =\left(\begin{array}{ccc}\left(1\times 1\right)+\left(0\times 0\right)+\left(2\times 3\right)& \left(1\times 1\right)+\left(0\times 2\right)+\left(2\times \left(-1\right)\right)& \left(1\times 0\right)+\left(0\times 1\right)+\left(2\times 0\right)\\ \left(3\times 1\right)+\left(\left(-1\right)\times 0\right)+\left(0\times 3\right)& \left(3\times 1\right)+\left(\left(-1\right)\times 2\right)+\left(0\times \left(-1\right)\right)& \left(3\times 0\right)+\left(\left(-1\right)\times 1\right)+\left(0\times 0\right)\\ \left(0\times 1\right)+\left(5\times 0\right)+\left(1\times 3\right)& \left(0\times 1\right)+\left(5\times 2\right)\left(1\times \left(-1\right)\right)& \left(0\times 0\right)+\left(5\times 1\right)+\left(1\times 0\right)\end{array}\right) \\ =\left(\begin{array}{ccc}7& -1& 0\\ 3& 1& -1\\ 3& 9& 5\end{array}\right) \end{gathered}$$

Find the product of the matrix BA.

$$\begin{gathered} \mathrm{BA}=\left(\begin{array}{ccc}1& 1& 0\\ 0& 2& 1\\ 3& -1& 0\end{array}\right)\left(\begin{array}{ccc}1& 0& 2\\ 3& -1& 0\\ 0& 5& 1\end{array}\right) \\ =\left(\begin{array}{ccc}4& -1& 2\\ 6& 3& 1\\ 0& 1& 6\end{array}\right) \end{gathered}$$

Find the addition and subtraction of the matrices

Find the addition of matrix A+B.

$$\begin{gathered} \mathrm{A}+\mathrm{B}=\left(\begin{array}{ccc}1& 0& 2\\ 3& -1& 0\\ 0& 5& 1\end{array}\right)+\left(\begin{array}{ccc}1& 1& 0\\ 0& 2& 1\\ 3& -1& 0\end{array}\right) \\ =\left(\begin{array}{ccc}2& 1& 2\\ 3& 1& 1\\ 3& 4& 1\end{array}\right) \end{gathered}$$

Find the addition of matrix A-B.

$$\begin{gathered} \mathrm{A}-\mathrm{B}=\left(\begin{array}{ccc}1& 0& 2\\ 3& -1& 0\\ 0& 5& 1\end{array}\right)-\left(\begin{array}{ccc}1& 1& 0\\ 0& 2& 1\\ 3& -1& 0\end{array}\right) \\ =\left(\begin{array}{ccc}0& -1& 2\\ 3& -3& -1\\ -3& 6& 1\end{array}\right) \end{gathered}$$

Find the squares and scalar multiplication of matrices of the matrices:

Find the square of matrix A.

$$\begin{gathered} {\mathrm{A}}^2=\left(\begin{array}{ccc}1& 0& 2\\ 3& -1& 0\\ 0& 5& 1\end{array}\right)-\left(\begin{array}{ccc}1& 0& 2\\ 3& -1& 0\\ 0& -5& 1\end{array}\right) \\ =\left(\begin{array}{ccc}1& 10& 4\\ 0& 1& 6\\ 15& 0& 1\end{array}\right) \end{gathered}$$

Find the Square of matrix B.

$$\begin{gathered} {\mathrm{B}}^2=\left(\begin{array}{ccc}1& 1& 2\\ 0& 2& 1\\ 3& -1& 0\end{array}\right)\left(\begin{array}{ccc}1& 1& 0\\ 0& 2& 1\\ 3& -1& 0\end{array}\right) \\ =\left(\begin{array}{ccc}1& 3& 1\\ 3& 3& 2\\ 3& 1& -1\end{array}\right) \end{gathered}$$

Find the scalar multiplication 5A.

$$\begin{gathered} 5A=5\times \left(\begin{array}{ccc}1& 0& 2\\ 3& -1& 0\\ 0& 5& 1\end{array}\right) \\ =\left(\begin{array}{ccc}5& 0& 10\\ 15& -5& 0\\ 0& 25& 5\end{array}\right) \end{gathered}$$

Find the scalar multiplication 3B

$$\begin{gathered} 3B=3\times \left(\begin{array}{ccc}1& 1& 0\\ 0& 2& 1\\ 3& -1& 0\end{array}\right) \\ =\left(\begin{array}{ccc}3& 3& 0\\ 0& 6& 1\\ 9& -3& 0\end{array}\right) \end{gathered}$$

Show that the multiplication of addition and subtraction of matrices is not equal.

Find the product of (A+B)(A-B) .

$$\begin{gathered} \left(A+B\right)\left(A-B\right)=\left(\begin{array}{ccc}2& 1& 2\\ 3& 1& 1\\ 3& 4& 1\end{array}\right)\left(\begin{array}{ccc}0& -1& 2\\ 3& -3& -1\\ -3& 6& 1\end{array}\right) \\ =\left(\begin{array}{ccc}-3& 6& 7\\ 0& 0& 6\\ 9& -9& 3\end{array}\right) \end{gathered}$$

Find the product of (A-B)(A+B) .

$$\begin{gathered} \left(A-B\right)\left(A+B\right)=\left(\begin{array}{ccc}0& -1& 2\\ 3& -1& -1\\ -3& 6& 1\end{array}\right)\left(\begin{array}{ccc}2& 1& 2\\ 3& 1& 1\\ 3& 4& 1\end{array}\right) \\ =\left(\begin{array}{ccc}3& 7& 1\\ -6& -4& 2\\ 15& 7& 1\end{array}\right) \end{gathered}$$

Find the subtraction of squares of matrices ${\mathrm{A}}^2-{\mathrm{B}}^2$.

$$\begin{gathered} {\mathrm{A}}^2-{\mathrm{B}}^2=\left(\begin{array}{ccc}1& 10& 4\\ 0& 1& 6\\ 15& 0& 1\end{array}\right)-\left(\begin{array}{ccc}1& 3& 1\\ 3& 3& 2\\ 3& 1& -1\end{array}\right) \\ =\left(\begin{array}{ccc}0& 7& 3\\ -3& -2& 4\\ 12& -1& 0\end{array}\right) \end{gathered}$$

Therefore, it is showed that,

$(\mathrm{A}+\mathrm{B})(\mathrm{A}-\mathrm{B})\ne (\mathrm{A}-\mathrm{B})(\mathrm{A}+\mathrm{B})\ne {\mathrm{A}}^2-{\mathrm{B}}^2$

Show that the determinant of the product of matrices and determinant of matrices product is equal:

Find the determinant of AB.

$$\begin{gathered} det\left(AB\right)=\left|\begin{array}{ccc}7& -1& 0\\ 3& 1& -1\\ 3& 9& 5\end{array}\right| \\ =\left|\begin{array}{cc}1& -1\\ 9& 5\end{array}\right|+1\left|\begin{array}{cc}3& -1\\ 3& 5\end{array}\right| \\ =7\left(5+9\right)+\left(15+3\right) \\ =116 \end{gathered}$$

Find the determinant of BA.

role="math" localid="1658990125245" $$\begin{gathered} \det \left(\mathrm{BA}\right)=\left|\begin{array}{ccc}4& -1& 2\\ 6& 3& 1\\ 0& 1& 6\end{array}\right| \\ =4\left|\begin{array}{cc}3& 1\\ 1& 6\end{array}\right|-6\left|\begin{array}{cc}-1& 2\\ 1& 6\end{array}\right| \\ =4\left(18-1\right)-6\left(-6-2\right) \\ =116 \end{gathered}$$

Find the determinant of matrix A.

$$\begin{gathered} det\left(A\right)=\left|\begin{array}{ccc}1& 0& 2\\ 3& -1& 0\\ 0& 5& 1\end{array}\right| \\ =1\left|\begin{array}{cc}-1& 0\\ 5& 1\end{array}\right|+2\left|\begin{array}{cc}3& -1\\ 0& 5\end{array}\right| \\ =-1+2\left(15\right) \\ =-29 \end{gathered}$$

Find the determinant of matrix B

$$\begin{gathered} det\left(B\right)=\left|\begin{array}{ccc}1& 1& 0\\ 0& 2& 1\\ 3& -1& 0\end{array}\right| \\ =1\left|\begin{array}{cc}2& 1\\ -1& 0\end{array}\right|-1\left|\begin{array}{cc}0& 1\\ 3& 0\end{array}\right| \\ =1\left(1\right)-1\left(-3\right) \\ =4 \end{gathered}$$

Find the product of determinant of matrix A and B

$$\begin{gathered} \det \left(\mathrm{A}\right)\det \left(\mathrm{B}\right)=29\times 4 \\ =116 \end{gathered}$$

Therefore,

$$\begin{gathered} \det \left(\mathrm{AB}\right)=\det \left(\mathrm{BA}\right) \\ =\det \left(\mathrm{A}\right)\det \left(\mathrm{B}\right) \end{gathered}$$

Show that the determinant of the sum of matrices and determinant of matrices sum is not equal:

Find the determinant of A+B.

$$\begin{gathered} det\left(A+B\right)=\left|\begin{array}{ccc}2& 1& 2\\ 3& 1& 1\\ 3& 4& 1\end{array}\right| \\ =2\left|\begin{array}{cc}1& 1\\ 4& 1\end{array}\right|-1\left|\begin{array}{cc}3& 1\\ 3& 1\end{array}\right|+2\left|\begin{array}{cc}3& 1\\ 3& 4\end{array}\right| \\ =2\left(1-4\right)-1\left(3-3\right)+2\left(12-3\right) \\ =12 \end{gathered}$$

Find the sum of determinant of matrix A and B

$$\begin{gathered} \det \left(\mathrm{A}\right)+\det \left(\mathrm{B}\right)=29+4 \\ =33 \end{gathered}$$

Therefore,

$\det (\mathrm{A}+\mathrm{B})\ne \det \left(\mathrm{A}\right)+\det \left(\mathrm{B}\right)$

Show that the determinant of scalar multiple matrices and product of scalar with determinant of matrices is not equal:

Find the determinant of 5A

$$\begin{gathered} det\left(5A\right)=\left|\begin{array}{ccc}5& 0& 10\\ 15& -5& 0\\ 0& 25& 5\end{array}\right| \\ =5\left|\begin{array}{cc}-5& 0\\ 25& 5\end{array}\right|+10\left|\begin{array}{cc}15& -5\\ 0& 25\end{array}\right| \\ =5\left(-25\right)+10\left(375\right) \\ =3625 \end{gathered}$$

Find the $5\left(\det \right(\mathrm{A}\left)\right)$

$$\begin{gathered} 5\left(\det \right(\mathrm{A}\left)\right)=5\times 129 \\ =145 \end{gathered}$$

Therefore,$\det \left(5\mathrm{A}\right)\ne 5\left(\det \right(\mathrm{A}\left)\right)$

Find the determinant of 3B.

$\det \left(3\mathrm{B}\right)=\left|\begin{array}{ccc}3& 3& 0\\ 0& 6& 1\\ 9& -3& 0\end{array}\right|$

role="math" localid="1658991181154" $$\begin{gathered} \det \left(3\mathrm{B}\right)=3\left|\begin{array}{cc}6& 1\\ -3& 0\end{array}\right|-3\left|\begin{array}{cc}0& 1\\ 9& 0\end{array}\right| \\ =3\left(-3\right)-3\left(-9\right) \\ =18 \end{gathered}$$

Find the $3\left(\det \right(\mathrm{B}\left)\right)$.

$$\begin{gathered} 3\left(\det \left(\mathrm{B}\right)\right)=3\times 4 \\ =12 \end{gathered}$$

Therefore,

$\det \left(3\mathrm{B}\right)\ne 3\left(\det \right(\mathrm{B}\left)\right)$

Find the value of   to show determinant of scalar multiple matrices and product of scalar with determinant of matrices is equal:

Find the value of n so that $\det \left(5\mathrm{A}\right)=5\left(\det \right(\mathrm{A}\left)\right)$and $\det \left(3\mathrm{B}\right)=3\left(\det \right(\mathrm{B}\left)\right)$.

Since, the matrices A and B are three dimensional square matrices,$\det (\mathrm{k}\cdot \mathrm{A})={\mathrm{k}}^3\left(\det \right(\mathrm{A}\left)\right)$therefore,n=3.