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 $\mathbf{det}\mathbf{(}\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{)}\mathbf{\ne }\mathbf{detA}\mathbf{+}\mathbf{detB}\mathbf{.}$ Show that $\mathbf{det}\mathbf{\left(}\mathbf{5}\mathbf{A}\mathbf{\right)}\mathbf{\ne }\mathbf{5}\mathbf{detA}$ and find n so that localid="1658983435079" $\mathbf{det}\mathbf{\left(}\mathbf{5}\mathbf{A}\mathbf{\right)}\mathbf{=}{\mathbf{5}}^{\mathbf{n}}\mathbf{detA}\mathbf{.}$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.

localid="1658983077106" $\mathbf{A}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\mathbf{2}& \mathbf{5}\\ \mathbf{-}\mathbf{1}& \mathbf{3}\end{array}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{B}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\mathbf{-}\mathbf{1}& \mathbf{4}\\ \mathbf{0}& \mathbf{2}\end{array}\mathbf{\right)}$

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).

The given matrices are

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

Find the product of the matrix .

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

Find the product of the matrix .

$$\begin{gathered} \mathrm{A}=\left(\begin{array}{cc}-1& 4\\ 0& 2\end{array}\right)\left(\begin{array}{cc}2& -5\\ -1& 3\end{array}\right) \\ =\left(\begin{array}{cc}\left(-1\right)\left(2\right)+\left(4\right)\left(-1\right)& \left(-1\right)\left(-5\right)+\left(4\right)\left(3\right)\\ \left(0\right)\left(2\right)+\left(2\right)\left(-1\right)& \left(0\right)\left(-5\right)+\left(2\right)\left(3\right)\end{array}\right) \\ =\left(\begin{array}{cc}-6& 17\\ -2& 6\end{array}\right) \end{gathered}$$

Find the addition of matrix A+B .

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

Find the addition of matrix A-B .

$$\begin{gathered} \mathrm{A}-\mathrm{B}=\left(\begin{array}{cc}2& -5\\ -1& 3\end{array}\right)-\left(\begin{array}{cc}-1& 4\\ 0& 2\end{array}\right) \\ =\left(\begin{array}{cc}2-\left(-1\right)& -5-4\\ -1+0& 3-2\end{array}\right) \\ =\left(\begin{array}{cc}3& -9\\ -1& 1\end{array}\right) \end{gathered}$$

Find the square of matrix A .

$$\begin{gathered} A^2=\left(\begin{array}{cc}2& -5\\ -1& 3\end{array}\right)\left(\begin{array}{cc}2& -5\\ -1& 3\end{array}\right) \\ =\left(\begin{array}{cc}\left(2\right)\left(2\right)+\left(-5\right)\left(-1\right)& \left(2\right)\left(-5\right)+\left(-5\right)\left(3\right)\\ \left(-1\right)\left(2\right)+\left(3\right)\left(-1\right)& \left(-1\right)\left(-5\right)+\left(3\right)\left(3\right)\end{array}\right) \\ =\left(\begin{array}{cc}9& -25\\ -5& 14\end{array}\right) \end{gathered}$$

Find the Square of matrix B .

$$\begin{gathered} B^2=\left(\begin{array}{cc}-1& -4\\ 0& 2\end{array}\right)\left(\begin{array}{cc}-1& 4\\ 0& 2\end{array}\right) \\ =\left(\begin{array}{cc}\left(-1\right)\left(-1\right)+\left(4\right)\left(0\right)& \left(-1\right)\left(4\right)+\left(4\right)\left(2\right)\\ \left(0\right)\left(-1\right)+\left(2\right)\left(0\right)& \left(0\right)\left(4\right)+\left(2\right)\left(2\right)\end{array}\right) \\ =\left(\begin{array}{cc}1& 4\\ 0& 4\end{array}\right) \end{gathered}$$

Find the scalar multiplication 5A .

$$\begin{gathered} 5\mathrm{A}=5\times \left(\begin{array}{cc}2& -5\\ -16& 3\end{array}\right) \\ =\left(\begin{array}{cc}10& -25\\ -5& 15\end{array}\right) \end{gathered}$$

Find the scalar multiplication 3B.

$3B=3\times \left(\begin{array}{cc}-1& 4\\ 0& 6\end{array}\right)$

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

$$\begin{gathered} \left(A+B\right)\left(A-B\right)=\left(\begin{array}{cc}1& -1\\ -1& 5\end{array}\right)\left(\begin{array}{cc}3& -9\\ -1& 1\end{array}\right) \\ =\left(\begin{array}{cc}4& -10\\ -8& 14\end{array}\right) \end{gathered}$$

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

$$\begin{gathered} \left(\mathrm{A}-\mathrm{B}\right)\left(\mathrm{A}+\mathrm{B}\right)=\left(\begin{array}{cc}3& -9\\ -1& 1\end{array}\right)\left(\begin{array}{cc}1& -1\\ -1& 5\end{array}\right) \\ =\left(\begin{array}{cc}12& -48\\ -2& 6\end{array}\right) \end{gathered}$$

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

$$\begin{gathered} A^2-B^2=\left(\begin{array}{cc}9& -25\\ -5& 14\end{array}\right)-\left(\begin{array}{cc}1& 4\\ 0& 4\end{array}\right) \\ =\left(\begin{array}{cc}8& -29\\ -5& 10\end{array}\right) \end{gathered}$$

Therefore, it is showed that $(\mathrm{A}+\mathrm{B})(\mathrm{A}-\mathrm{B})\ne (\mathrm{A}-\mathrm{B})(\mathrm{A}+\mathrm{B})\ne {\mathrm{A}}^2-{\mathrm{B}}^2$.

Find the determinant of AB.

$$\begin{gathered} \det \left(\mathrm{AB}\right)=\left|\begin{array}{cc}-2& -2\\ 1& 2\end{array}\right| \\ =-4+2 \\ =-2 \end{gathered}$$

Find the determinant of BA.

$$\begin{gathered} \det \left(\mathrm{AB}\right)=\left|\begin{array}{cc}-6& 17\\ -2& 6\end{array}\right| \\ =-36+34 \\ =-2 \end{gathered}$$

Find the determinant of matrix A .

role="math" localid="1658984830439" $$\begin{gathered} \det \left(A\right)=\left|\begin{array}{cc}2& -5\\ -1& 3\end{array}\right| \\ =6-5 \\ =1 \end{gathered}$$

Find the determinant of matrix B .

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

Find the product of determinant of matrix A and B .

$$\begin{gathered} \det \left(\mathrm{A}\right)\det \left(\mathrm{B}\right)=1\times \left(-2\right) \\ =-2 \end{gathered}$$

Therefore,

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

Find the determinant of A+B .

$$\begin{gathered} det\left(A+B\right)=\left|\begin{array}{cc}1& -1\\ -1& 5\end{array}\right| \\ =5-1 \\ =4 \end{gathered}$$

Find the sum of determinant of matrix A and B

$$\begin{gathered} \det \left(\mathrm{A}\right)+\det \left(\mathrm{B}\right)=1-2 \\ =-1 \end{gathered}$$

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

Find the determinant of 5A .

$$\begin{gathered} \det \left(5\mathrm{A}\right)=\left|\begin{array}{cc}10& -25\\ -5& 15\end{array}\right| \\ =150-125 \\ =25 \\ =5^2\times 1 \end{gathered}$$

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

role="math" localid="1658985741127" $$\begin{gathered} 5\left(\det \left(\mathrm{A}\right)\right)=5\times 1 \\ =5 \end{gathered}$$

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

Find the determinant of 3B.

$$\begin{gathered} det\left(3B\right)=\left|\begin{array}{cc}-3& 12\\ 0& 6\end{array}\right| \\ =-18-0 \\ =-18 \\ =3^2\times \left(-2\right) \end{gathered}$$

Find the role="math" localid="1658985726400" $3\left(\det \right(\mathrm{B}\left)\right)$.

role="math" localid="1658985733523" $$\begin{gathered} 3\left(\det \left(\mathrm{A}\right)\right)=3\times \left(-2\right) \\ =-6 \end{gathered}$$

Therefore,

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

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 two dimensional square matrices,$\det (\mathrm{k}\cdot \mathrm{A})={\mathrm{k}}^2\left(\det \right(\mathrm{A}\left)\right)$and therefore n=2.