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

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"Matrix Algebra Preliminary" (Introduction to Matrix ALGEBRA) course by Prof. A.k.kaw (University of South Florida) is designed and taught.
PDF format Learning note Download (academia.edu)
4th. Download the course handout (PDF)

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

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