A.kaw Matrix Algebra Preliminary study Note 4. Unary Matrix Operations

"Matrix Algebra Preliminary" (Introduction to Matrix ALGEBRA) course by Prof. A.k.kaw (University of South Florida) is designed and taught.
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Summary

• Transpose
  Let $[a]$ is A $m \times n$ matrix. Then $[b]$ are the transpose of $[a]$ if $b _{ji} = a_{ij}$ for all $i $ and $j $. That was, the $i $-th row and the $j $-th column element of $[a]$ is the $j $-th row and $i $-th column element of $[b]$. Note that $[b]$ is a $n \times m$ matrix and are denoted by $[b] = [a]^{t}$. For example, $$[a] = \begin{bmatrix}1& 2& 3\\ 4& 5& 6\end{bmatrix}\rightarrow [A]^{t} = \BEGIN{BMATRIX}1&A mp 4\\ 2& 5\\ 3& 6\end{bmatrix}$$
• symmetric matrix
  A square matrix $[a]$ with real elements where $a _{ij} = a_{ji}$ for $i = 1, \cdots, n$ and $j = 1, \cdots, n$ is called A Symmetric matrix. That's, $[a]$ is A symmetric matrix if $[a] = [a]^{t}$. For example, $$[a] = \begin{bmatrix}1& 2& 3\\ 2& 4& 5\\ 3& 5& 7\end{bmatrix}$$
• skew-symmetric Matrix
  A $n \times n$ matrix is skew-symmetric if $a _{ij} =-a_{ji}$ for $i = 1, \cdots, n$ and $j = 1, \cdots, n$. That's, $[a]$ is A skew-symmetric matrix if $[a] =-[a]^{t}$. Note that the diagonal elements must is zero in a skew-symmetric matrix. For example, $$[a] = \begin{bmatrix}0& 2& 3\\ -2& 0& 5\\ -3& -5& 0\end{bmatrix}$$
• Trace of Matrix
  The trace of a $n \times n$ matrix $[a]$ is the sum of the diagonal entries of $[a]$, which is, $$\text{tr}[a] = \sum_{i=1}^ {n}a_{ii}$$ For example, $$[a] = \begin{bmatrix}1& 2& 3\\ 2& 4& 5\\ 3& 5& 7\end{bmatrix}\rightarro W \text{tr}[a] = 1 + 4 +7=12$$
• determinant
  Let $[a]$ is A $n \times n$ matrix.
  • the minor of entry $a _{ij}$ is denoted by $M _{ij}$ and is defined as the determinant of the $ (n-1) \times (n-1) $ sub-ma Trix of $[a]$, where the Sub-matrix is obtained by deleting the $i $-th row and $j $-th column of the Matrix $[a]$. The determinant is then given by $$\det (A) = \sum_{j=1}^{n} ( -1) ^{i+j}a_{ij}m_{ij},\ \text{for any}\ i=1, 2, \cdots, n$$ or $$\det (A) = \sum_{i=1}^{n} ( -1) ^{i+j}a_{ij}m_{ij},\ \text{for any}\ j=1, 2, \cdots, n$$ for example, $$[a] = \begin{bmatri x}1& 2& 3\\ 2& 4& 5\\ 3& 5& 7\end{bmatrix}$$ $$\rightarrow \det (A) = ( -1) ^{1+1}\cdot1\cdot\begin{v matrix}4& 5 \ 5& 7\end{vmatrix} + ( -1) ^{1+2}\cdot2\cdot\begin{vmatrix}2& 5 \ 3& 7\end{vmatrix} + (-1) ^{1 +3}\cdot3\cdot\begin{vmatrix}2& 4 \ 3& 5\end{vmatrix}$$ $$= (4\times7-5\times5) -2\times (2\TIMES7-3\TIMES5) + 3\times (2\TIMES5-3\TIMES4) = -1$$ Note that for a $2\times2$ matrix $[a] = \begin{bmatrix}a& b\\ c& d\end{bmatri x}$, $\det (A) = ad-bc$.
  • The number $ ( -1) ^{i+j}m_{ij}$ is called the cofactor of $a _{ij}$ and are denoted by $C _{ij}$. The formula for the determinant can then be written as $$\det (A) = \sum_{j=1}^{n}a_{ij}c_{ij},\ \text{for any}\ i=1, 2, \c Dots, n$$ or $$\det (A) = \sum_{i=1}^{n}a_{ij}c_{ij},\ \text{for any}\ j=1, 2, \cdots, n$$
  • If $[a]$ and $[b]$ is square matrices of same size, then $$\det (a\cdot B) = \det (A) \cdot\det (B) $$
  • $\det (A) = 0$ If
    • A row or a column is zero, or
    • A row (column) is proportional to another row (column).
  • If a row (column) is multiplied by $k $ to result in the matrix $[b]$, then $$\det (B) = K\cdot\det (a) $$
  • If $[b] = k\cdot[a]$, then $$\det (B) =k^{n}\det (A) $$
  • If $[a]$ is A $n \times n$ Upper or lower triangular matrix, then $$\det (A) = \prod_{i=1}^{n}a_{ii}$$
  • If $[b]$ is row-equivalent to $[a]$, then $$\begin{cases} r_i\leftrightarrow R_j: & \det (B) =-\det (A); \ \ Tr_i: & \det (b) = T\det (a), \ r_i\rightarrow R_i+tr_j: &\det (b) = \det (a). \end{cases}$$

Selected problems

1. Let $$[a] = \begin{bmatrix}25& 3& 6\\ 7& 9& 2\end{bmatrix}$$ Find $[a]^{t}$.

Solution:

$$[a]^{t} = \begin{bmatrix}25& 7\\ 3& 9\\ 6& 2\end{bmatrix}$$

2. If $[a]$ and $[b]$ are both $n \times n$ symmetric matrices, show that $[a]+[b]$ is also symmetric.

Solution:

Let $[c]=[a]+[b]$, so we had $ $c _{ij} = A_{ij} + b_{ij} = A_{ji} + B_{ji} =c_{ji}$$ that was, $[c]=[c]^{t}$.

3. What is the trace of $$[a] = \begin{bmatrix}7& 2& 3& 4\\ -5& -5& -5& -5\\ 6& 6& 7& 9\\ -5& 2& 3& 10\end{bmatrix}$$

Solution:

$$\text{tr}[a] = 7-5+7+10=19$$

4. Find the determinant of $$[a] = \begin{bmatrix}10& -7& 0\\ -3& 2.099& 6\\ 5& -1& 5\end{bmatrix} $$

Solution:

$$\det (A) = ( -1) ^{1+1}\times10\times\begin{vmatrix}2.099& 6\\ -1& 5\end{vmatrix} + ( -1) ^{1+2}\times ( -7) \times \begin{vmatrix}-3& 6\\ 5& 5\end{vmatrix}$$ $$=10\times (2.099\TIMES5+1\TIMES6) + 7\times (-15-30) = -150.05$$

5. What is the value of a $n \times n$ matrix $\det (3[a]) $?

Solution:

$$\det (3[a]) = 3^n\det (A) $$

6. For a $5\times5$ matrix $[a]$, the first row was interchanged with the fifth row, what's the determinant of the Resulti Ng Matrix $[b]$?

Solution:

The sign would is changed if interchaged the row (column). Thus $$\det (B) =-\det (A) $$

7. What is the determinant of $$[a] = \begin{bmatrix}0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& ; 1\\ 1& 0& 0& 0\end{bmatrix}$$

Solution:

$$[A] = \begin{bmatrix}0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\\ 1& 0& 0& 0\e Nd{bmatrix}\rightarrow r_1\leftrightarrow r_4 \begin{bmatrix}1& 0& 0& 0\\ 0& 0& 1& 0\\ 0& 0&am P 0& 1\\ 0& 1& 0& 0\end{bmatrix}$$ $$\rightarrow r_2\leftrightarrow r_3 \begin{bmatrix}1& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 1& 0\\0& 1& 0& 0\end{bmatrix}$$ $$\Rightarrow R_2\leftrightarr ow R_4 \begin{bmatrix}1& 0& 0& 0\\0& 1& 0& 0\\ 0& 0& 1& 0\\0& 0& 0& 1\end{ bmatrix}=[b]$$ Thus $\det (A) = ( -1) ^{3}\det (B) =-1$.

8. Find the determinant of $$[a]=\begin{bmatrix}0& 0& 0\\ 2& 3& 5\\ 6& 9& 2\end{bmatrix}$$

Solution:

$\det (A) =0$ since the first row is zero.

9. Find the determinant of $$[a]=\begin{bmatrix}0& 0& 2& 3\\ 0& 2& 3& 5\\ 6& 7& 2& 3\\ 6.6& 7.7& 2.2& 3.3\end{bmatrix}$$

Solution:

Since $R _4 = 1.1r_3$, so $\det (A) = 0$.

Find the determinant of $$[a]=\begin{bmatrix}5& 0& 0& 0\\ 0& 3& 0& 0\\ 2& 5& 6& 0\ \ 1& 2& 3& 9\end{bmatrix}$$

Solution:

This is a lower triangular matrix and hence $$\det (a) = 5\times3\times6\times9=810$$

Given The Matrix $$[a]=\begin{bmatrix}125& 25& 5& 1\\ 512& 64& 8& 1\\ 1157& 89& 13& Amp 1\\ 8& 4& 2& 1\end{bmatrix}$$ and $\det (A) = -32400$. Find the determinant of $$[a_1]=\begin{bmatrix}125& 25& 5& 1\\ 512& 64& 8& 1\\ 1141& 81& 9 & -1\\ 8& 4& 2& 1\end{bmatrix};$$ $$[a_2]=\begin{bmatrix}125& 25& 1& 5\\ 512& 64& 1&am P 8\\ 1157& 89& 1& 13\\ 8& 4& 1& 2\end{bmatrix};$$ $$[A_3] = \begin{bmatrix} 125& 25& 5& 1\\ 1157& 89& 13& 1\\ 512& 64& 8& 1\\8& 4& 2& 1\end{bmatrix};$$ $$[A_4] = \begin{bmat rix} 125& 25& 5& 1\\ 1157& 89& 13& 1\\ 8& 4& 2& 1\\ 512& 64& 8& 1\end{bmat rix};$$ $$[A_5] = \begin{bmatrix} 125& 25& 5& 1\\ 512& 64& 8& 1\\ 1157& 89& 13& 1\\ 16 & 8& 4& 2 \end{bmatrix}.$$

solution:

$$[a]=\begin{bmatrix}125& 25& 5& 1\\ 512& 64& 8& 1\\ 1157& 89& 13& 1\\ 8& 4& 2 & 1\end{bmatrix}\rightarrow r_3-2r_4 \begin{bmatrix}125& 25& 5& 1\\ 512& 64& 8& 1\\ 1141& 81& 9& -1\\ 8& 4& 2& 1\end{bmatrix}=[a_1]$$ Thus $\det (a_1) = \det (A) =-32400$. $$[a]=\begin{bmatrix}125& 25& 5& 1\\ 512& 64& 8& 1\\ 1157& 89& 13& 1\\ 8& 4& 2 & 1\end{bmatrix}\rightarrow c_3\leftrightarrow c_4 \begin{bmatrix}125& 25& 1& 5\\ 512& 64& 1 & 8\\ 1157& 89& 1& 13\\ 8& 4& 1& 2\end{bmatrix} = [a_2]$$ Thus $\det (a_2) =-\det (A) =32400$. $$[a]=\begin{bmatrix}125& 25& 5& 1\\ 512& 64& 8& 1\\ 1157& 89& 13& 1\\ 8& 4& 2 & 1\end{bmatrix}\rightarrow R_2\leftrightarrow R_3\begin{bmatrix} 125& 25& 5& 1\\ 1157& 89& 13& Amp 1\\ 512& 64& 8& 1\\8& 4& 2& 1\end{bmatrix}= [a_3]$$ Thus $\det (a_3) =-\det (A) = 32400$. $$[a]=\begin{bmatrix}125& 25& 5& 1\\ 512& 64& 8& 1\\ 1157& 89& 13& 1\\ 8& 4& 2 & 1\end{bmatrix}\rightarrow \begin{cases} r_2\leftrightarrow r_3\\ R ' _3\leftrightarrow R_4\end{cases} \begin{ bmatrix} 125& 25& 5& 1\\ 1157& 89& 13& 1\\ 8& 4& 2& 1\\ 512& 64& 8& 1\end{ Bmatrix} = [a_4]$$ Thus $\det (a_4) = ( -1) ^2\det (A) = -32400$. $$[a]=\begin{bmatrix}125& 25& 5& 1\\ 512& 64& 8& 1\\ 1157& 89& 13& 1\\ 8& 4& 2 & 1\end{bmatrix} \Rightarrow 2R_4\begin{bmatrix} 125& 25& 5& 1\\ 512& 64& 8& 1\\ 1157& 89& Amp 13& 1\\ 16& 8& 4& 2 \end{bmatrix} = [a_5]$$ Thus $\det (a_5) = 2\det (A) = -64800$.

Find the determinant of $$[a] = \begin{bmatrix}25& 5& 1\\ 64& 8& 1\\ 144& 12& 5\end{bmatrix}$$

Solution:

$$\det (A) = ( -1) ^{1+3}a_{13}m_{13}+ ( -1) ^{2+3}a_{23}m_{23} + ( -1) ^{3+3}a_{33}m_{33}$$ $$ = \begin{vmatrix}64& 8\\ 144& 12\end{vmatrix}-\begin{vmatrix}25& 5\\ 144& 12\end{vmatrix} + 5\times \begin{vmatrix}25& 5\\ 64&am P 8\end{vmatrix} = -564$$

Show that if $[a][b]=[i]$, where $[a]$, $[b]$ and $[i]$ are matrices of $n \times n$ size and $[i]$ was an identity matr IX, then $\det (A) \neq0$ and $\det (B) \neq0$.

Solution: $$\det (a) \det (b) =\det (AB) =\det (I) = 1$$ $$\rightarrow \det (a) \neq0,\ \det (b) \neq0.$$

If the determinant of a $4\times4$ matrix $[a]$ is given as a, then what is the determinant of $5[a]$?

Solution:

$$\det (K[a]) =k^n\det (a) $$ $$\rightarrow \det (5[a]) = 5^4\det (a) = 625\times20=12500$$

If The matrix product $[a][b][b]$ is defined, which is $ ([a][b][c]) ^{t}$?

Solution:

$$ ([a][b]) ^{t}=[b]^{t}[a]^{t}$$$$\rightarrow ([A][b][c]) ^{t}=[c]^{t} ([a][b]) ^{t}=[c]^{t}[b]^{t}[a]^{t}$$

The determinant of the matrix $$[a] = \begin{bmatrix}25& 5& 1\\ 0& 3& 8\\ 0& 9& A\end{bmatrix} $$ is 50. What is the value of $a $?

Solution:

$$\det (A) = 25\times\begin{vmatrix}3& 8\\ 9& A\end{vmatrix} = 25\times (3a-72) =50$$ $$\Rightarrow a={74\over3}$$

$[a]$ is a $5\times 5$ matrix and a matrix $[b]$ are obtained by the row operations of replacing ROW1 with ROW3, and th En Row3 is replaced by a linear combination of $2\times$row3$+4\times$row2. If $\det (A) =17$, then who is the value of $\det (B) $?

Solution:

The process is $$[a]\rightarrow r_1\leftrightarrow r_3 \rightarrow 2r_3\rightarrow r_3+4r_2\rightarrow [B]$$ Thus $$\det ( B) = ( -1) \times2\cdot\det (A) = -34$$

A.kaw Matrix Algebra Preliminary study Note 4. Unary Matrix Operations