[행렬식] 행렬식의 성질과 문제풀이

Properties of Determinants

 

THEOREM 8.1: The determinant of a matrix $ A $ and its transpose $ A ^ {T} $ are equal; that is, $ |A|=|A ^ {T} | $.

Note that expanding A by column k is equivalent to expanding AT by row k.

 

THEOREM 8.2: Let $ A $ be a square matrix.

(i) If $ A $ has a row (column) of zeros, then $ |A|=0 $.

(ii) If $ A $ has two identical rows (columns), then $ |A|=0 $.

(iii) If $ A $ is triangular (i.e., $ A $ has zeros above or below the diagonal), then $ |A|= $ product of diagonal elements. Thus, in particular, $ |I|=1 $, where $ I $ is the identity matrix.

pf) (i) $ A $의 $ i $번째 행이 영행이라고 가정하자. 그러면 모든 $ j $($ 1 \leq j \leq n $)에 대하여 $ a _ {ij} =0 $임로 행렬 $ A $를 $ i $행에 대하여 행렬식을 구하면 즉,

$ |A|= \sum\limits _ {j=1} ^ {n} a _ {ij} A _ {ij} = \sum\limits _ {j=1} ^ {n} 0A _ {ij} =0 $

또는  If A has a zero row, then $det(A)=0$ follows from expanding $A$ by that row. The case where $A$ has a zero column is similar.

(ii) Proof. We prove only the row case. Switching the two rows gets back the same matrix. However, by Lemma 3, the determinant of the matrix should be multiplied by $-1$. Therefore, we get $det(A)=-det(A)$, meaning  $det(A)=0$.

(iii) Proof. We can prove the lemma by induction. First, correctness is obvious for $n = 1$. Assuming correctness for $ n \leq t-1$ (for $t \geq 2$ ), consider $n=t$ . Expanding $A$  by the first row gives:

$ det ( A)= \sum\limits _ {j=1} ^ {n} \left ( -1 \right ) ^ {1+j} a _ {1j} det ( M _ {1j} ) $ (2)

From induction we know that $$ det ( M _ {11} )= \prod _ {i=2} ^ {n} a _ {ii} $$, Furthermore, for $ j>1 $, $ det ( M _ {1j} )=0 $ because the first column of $ M _ {1j} $ contains all 0’s. It thus follows that (2) equals $$ \prod _ {i=2} ^ {n} a _ {ii} $$

 

THEOREM 8.3: Suppose $ B $ is obtained from $ A $ by an elementary row (column) operation.

(i) If two rows (columns) of $ A $ were interchanged, then $ |B|=-|A| $.

(ii) If a row (column) of $ A $ were multiplied by a scalar $ k $, then $ |B|=-k|A| $.

(iii) If a multiple of a row (column) of $ A $ were added to another row (column) of $ A $, then $ |B|=|A| $.

 

THEOREM 8.4: The determinant of a product of two matrices $ A $ and $ B $ is the product of their determinants; that is,

$$ det ( AB)=det ( A)det ( B) $$

 

THEOREM 8.5: Let $ A $ be a square matrix. Then the following are equivalent:

(i) $ A $ is invertible; that is, $ A $ has an inverse $ A ^ {-1} $.

(ii) $ AX=O $ has only the zero solution.

(iii) The determinant of $ A $ is not zero; that is, $ det ( A) \neq 0 $.

 

LEMMA 8.6: Let $ E $ be an elementary matrix. Then, for any matrix $ A $, $ |EA|=|E||A| $

 

THEOREM 8.7: Suppose $ A $ and $ B $ are similar matrices. Then $ |A|=|B| $.

 

1. Compute the determinant of each of the following matrices:

(a) $ A= \left[ \matrix {2 & 3 & 4\\5 & 6 & 7\\8 & 9 & 1} \right] $, (b) $ B= \left[ \matrix {4 & 6 & 8 & 9\\0 & 2 & 7 & 3\\0 & 0 & 5 & 6\\0 & 0 & 0 & 3} \right] $, (c) $ C= \left[ \matrix { \frac {1} {2} & -1 & - \frac {1} {4} \\ \frac {3} {4} & \frac {1} {2} & -1\\1 & -4 & 1} \right] $ 

(a) One can simplify the entries by first subtracting twice the first row from the second row—that is, by applying the row operation "Replace $ R _ {2} $ by $ -2R _ {1} +R _ {2} $". Then

$ |A|= \left| \matrix {2 & 3 & 4\\5 & 6 & 7\\8 & 9 & 1} \right| = \left| \matrix {2 & 3 & 4\\1 & 0 & -1\\8 & 9 & 1} \right| =0-24+36-0+18-3=27 $

(b) $ B $ is triangular, so $ |B|= $product of the diagonal entries $ =120 $.

(c) The arithmetic is simpler if fractions are first eliminated. Hence, multiply the first row $ R _ {1} $ by $ 6 $ and the second row $ R _ {2} $ by $ 4 $. Then

$ |24C|= \left| \matrix {3 & -6 & -2\\3 & 2 & -4\\1 & -4 & 1} \right| =6+24+24+4-48+18=28 $, $ |C|= \frac {28} {24} = \frac {7} {6} $

 

2. Compute the determinant of each of the following matrices:

(a) $ A= \left[ \matrix {2 & 5 & -3 & -2\\-2 & -3 & 2 & -5\\1 & 3 & -2 & 2\\-1 & -6 & 4 & 3} \right] $

(b) $ B= \left[ \matrix {6 & 2 & 1 & 0 & 5\\2 & 1 & 1 & -2 & 1\\1 & 1 & 2 & -2 & 3\\3 & 0 & 2 & 3 & -1\\3 & 0 & 2 & 3 & 2} \right] $

 

sol) (a) Use $ a _ {31} =1 $ as a pivot to put 0’s in the first column, by applying the row operations "Replace $ R _ {1} $ by $ -2R _ {3} +R _ {1} $", "Replace $ R _ {2} $ by $ 2R _ {3} +R _ {1} $", and "Replace $ R _ {4} $ by $ R _ {3} +R _ {4} $". Then

$ |A|= \left| \matrix {2 & 5 & -3 & -2\\-2 & -3 & 2 & -5\\1 & 3 & -2 & 2\\-1 & -6 & 4 & 3} \right| = \left| \matrix {0 & -1 & 1 & -6\\0 & 3 & -2 & -1\\1 & 3 & -2 & 2\\0 & -3 & 2 & 5} \right| $$ = \left| \matrix {-1 & 1 & -6\\3 & -2 & -1\\-3 & 2 & 5} \right| $ $ =-4 $

(b) First reduce $ |B| $ to a determinant of order $ 4 $, and then to a determinant of order $ 3 $, for which we can use Fig. 8-1. First use $ c _ {22} =1 $ as a pivot to put 0’s in the second column, by applying the row operations

"Replace $ R _ {1} $ by $ -2R _ {2} +R _ {1} $", "Replace $ R _ {3} $ by $ -R _ {2} +R _ {3} $", and "Replace $ R _ {5} $ by $ R _ {2} +R _ {5} $". Then

$ |B|= \left| \matrix {2 & 0 & -1 & 4 & 3\\2 & 1 & 1 & -2 & 1\\-1 & 0 & 1 & 0 & 2\\3 & 0 & 2 & 3 & -1\\3 & 0 & 2 & 3 & 2} \right| = \left| \matrix {2 & -1 & 4 & 3\\-1 & 1 & 0 & 2\\3 & 2 & 3 & -1\\1 & -2 & 2 & 3} \right| = \left| \matrix {1 & 1 & 4 & 5\\0 & 1 & 0 & 0\\5 & 2 & 3 & -5\\-1 & -2 & 2 & 7} \right| $ $ = \left| \matrix {1 & 4 & 5\\5 & 3 & -5\\-1 & 2 & 7} \right| =-34 $

 

Cofactors, Classical Adjoints, Minors, Principal Minors

4. Let $ A= \left[ \matrix {2 & 1 & -3 & 4\\5 & -4 & 7 & -2\\4 & 0 & 6 & -3\\3 & -2 & 5 & 2} \right] $

(a) Find $ A _ {23} $, the cofactor (signed minor) of $ 7 $ in $ A $.

 

sol) (a) Take the determinant of the submatrix of A obtained by deleting row $ 2 $ and column $ 3 $ (those which contain the $ 7 $), and multiply the determinant by $ ( -1) ^ {2+3} $:

$$ A _ {23} =- \left| \matrix {2 & 1 & 4\\4 & 0 & -3\\3 & -2 & 2} \right| =- ( -61)=61 $$

The exponent $ 2+3 $ comes from the subscripts of $ A _ {23} $—that is, from the fact that $ 7 $ appears in row $ 2 $ and column $ 3 $.

 

5. Let $ B= \left[ \matrix {1 & 1 & 1\\2 & 3 & 4\\5 & 8 & 9} \right] $.

Find : (a) $ |B| $,

(b) $ adjB $,

(c) $ B ^ {-1} $ using $ adjB $

sol) 

(a) $ |B|=-2 $              

(b) $ adjB= \left[ \matrix {-5 & -1 & 1\\2 & 4 & -2\\1 & -3 & 1} \right] $

(c) Because $ |B| \neq 0 $, $ B ^ {-1} = \frac {1} {|B|} \left ( adjB \right ) $$ = \frac {1} {-2} \left[ \matrix {-5 & -1 & 1\\2 & 4 & -2\\1 & -3 & 1} \right] = \left[ \matrix { \frac {5} {2} & \frac{1} {2} & - \frac {1} {2} \\-1 & -2 & 1\\- \frac{1}  {2} & \frac{3}  {2} & -\frac {1}  {2} } \right] $

 

Determinants and Systems of Linear Equations

 

6. Use determinants to solve the system $ { \begin {cases} 3y+2x=z+1\\3x+2z=8-5y\\3z-1=x-2y\end {cases} } $

sol) First arrange the equation in standard form, then compute the determinant $ D $ of the matrix of coefficients:

$ \left\{ \begin{split} 2x+3y-z=1\\3x+5y+2z=8\\x-2y-3z=-1 \end{split} \right. $ and $ D= \left| \matrix {2 & 3 & -1\\3 & 5 & 2\\1 & -2 & -3} \right| =22 $

Because $ D \neq 0 $, the system has a unique solution. To compute $ N _ {x} ,~N _ {y} ,~N _ {z} $, we replace, respectively, the coefficients of $ x,~y,~z $ in the matrix of coefficients by the constant terms. Then

$ N _ {x} = \left| \matrix {1 & 3 & -1\\8 & 5 & 2\\-1 & -2 & -3} \right| =66 $, $ N _ {y} = \left| \matrix {2 & 1 & -1\\3 & 8 & 2\\1 & -1 & -3} \right| =-22 $, $ N _ {z} = \left| \matrix {2 & 3 & 1\\3 & 5 & 8\\1 & -2 & -1} \right| =44 $

Thus,

$ x= \frac {N _ {x} } {D} = \frac {66} {22} =3 $, $ y= \frac {N _ {y} } {D} = \frac {-22} {22} =-1 $, $ z= \frac {N _ {z} } {D} = \frac {44} {22} =3 $

 

7. Consider the system $ { \begin {cases} kx+y+z=1\\x+ky+z=1\\x+y+kz=1\end {cases} } $

Use determinants to find those values of k for which the system has

(a) a unique solution, (b) more than one solution, (c) no solution.

 

(sol) (a) The system has a unique solution when $ D \neq 0 $, where D is the determinant of the matrix of coefficients.

Compute

$$ D= \left| \matrix {k & 1 & 1\\1 & k & 1\\1 & 1 & k} \right| = \left ( k-1 \right ) ^ {2} \left ( k+2 \right ) $$

Thus, the system has a unique solution when

$ \left ( k-1 \right ) ^ {2} \left ( k+2 \right ) \neq 0 $, when $ k \neq 1 $ and $ k \neq 2 $

(b and c) Gaussian elimination shows that the system has more than one solution when $ k=1 $, and the system has no solution when $ k=-2 $