# properties of determinants linear algebra

Triangle property: If all the elements of a determinant above or below the main diagonal consist of zeros, then the determinant is equal to the product of diagonal elements. The determinant is positive or negative according to whether the linear transformation preserves or reverses the orientation of a real vector space. However notice that the second row is equal to $$2$$ times the first row. By Definition [def:twobytwodeterminant], $$\det \left(A\right) = -2$$. Pro Lite, Vedantu Both play an important role in line equations and also used to solve real-life problems in Physics, Mechanics and Optics etc. Let $$C$$ be the matrix obtained by replacing the $$j$$th row of $$B$$ by the $$i$$th row of $$B$$ multiplied by $$k$$. Invariance under row operations; if  X’ is a matrix formed by summing up the multiple of any row to another row, then det (X) = det (X’). Example $$\PageIndex{5}$$: Determinant of the Transpose. The determinant is a function which associates to a square matrix an element of the field on which it is defined (commonly the real or complex numbers). Let $$E_{ik}$$ be the elementary matrix obtained by multiplying the $$i$$th row of $$I$$ by $$k$$. You can see that this matches our answer above. True/False From Howard Anton's Linear Algebra . The assumptions state that we have $$a_{l,j}=b_{l,j}=c_{l,j}$$ for $$j\neq i$$ and for $$1\leq l\leq n$$ and $$a_{l,i}=b_{l,i}+c_{l,i}$$ for all $$1\leq l\leq n$$. Consider the matrix $$A$$ first. This implies $$\det A=0$$. This gives the next theorem. Each entry in the $$j$$th row of $$A$$ is the sum of the corresponding entries in $$j$$th rows of $$B$$ and $$C$$. The determinants will be equivalent to zero if each term of rows and columns are zero. If a determinant Δ beomes 0 while considering the value of x  = α, then (x -α) is considered as a factor of Δ. Since $$j=i+1$$, we have $(-1)^{1+j}=(-1)^{1+i+1}=-(-1)^{1+i}$ and therefore $$a_{1i}\mathrm{cof}(A)_{1i}=-b_{1j} \mathrm{cof}(B)_{1j}$$ and $$a_{1j}\mathrm{cof}(A)_{1j}=-b_{1i} \mathrm{cof}(B)_{1i}$$. First, note that $A^{T} = \left[ \begin{array}{rr} 2 & 4 \\ 5 & 3 \end{array} \right] \nonumber$. We explicitly illustrate its use with an example. Jump to navigation Jump to search. The case $$n=1$$ does not apply and thus let $$n \geq 2$$. Let $$A=\left[ \begin{array}{rr} 1 & 2 \\ 3 & 4 \end{array} \right]$$ and let $$B=\left[ \begin{array}{rr} 1 & 2 \\ 5 & 8 \end{array} \right] .$$ Find $$\det \left(B\right)$$. If $$k=0$$ then $$A=B$$ and there is nothing to prove, so we may assume $$k\neq 0$$. Let $$A$$be an $$n\times n$$ matrix and fix $$j>1$$. If $$D$$ is obtained by multiplying the $$j$$th row of $$C$$ by $$\frac 1k$$ then by (2) we have $$\det C=\frac 1k\det D$$ (recall that $$k\neq 0$$!). In Linear algebra, a determinant is a unique number that can be ascertained from a square matrix. Then, $\det \left( AB\right) =\det \left( A\right) \det \left( B\right)$. Determinant of a Identity matrix () is 1. Using Definition [def:twobytwodeterminant], the determinant is given by, $\det \left( A \right) = 1 \times 4 - 2 \times 2 = 0 \nonumber$. Fix $$j\in \{1,2, \dots ,n\}$$ such that $$j\neq i$$. The key difference between matrix and determinants are given below: The matrix is a set of numbers that are enclosed by two brackets whereas the determinants is a set of numbers that are enclosed by two bars. This is not difficult to check for $$n=2$$ (do check it!). If $$C$$ is the reduced row-echelon form of $$A$$ then we can write $$A=E_1\cdot E_2\cdot\dots\cdot E_m\cdot C$$ for some elementary matrices $$E_1,\dots, E_m$$. If two rows of a matrix are equal, its determinant is zero. Unreviewed. Notice that the second row of $$B$$ is two times the first row of $$A$$ added to the second row. Definition $$\PageIndex{1}$$: Row Operations, The row operations consist of the following. Many of the proofs in section use the Principle of Mathematical Induction. After one interchanges $$j-1$$st and $$j$$th row, we have $$i$$th row in position of $$j$$th and $$l$$th row in position of $$l-1$$st for $$i+1\leq l\leq j$$. If $$A=\left[ a_{ij} \right]$$ is an $$n\times n$$ matrix, then $$\det A$$ is defined by computing the expansion along the first row: $\label{E1} \det A=\sum_{i=1}^n a_{1,i} \mathrm{cof}(A)_{1,i}.$ If $$n=1$$ then $$\det A=a_{1,1}$$. L.H.S = $\begin{vmatrix}a & b & c\\ d & e & f\\ g & h & i\end{vmatrix}$ = $\begin{vmatrix}a & d & g\\ b & e & h\\ c & f & i\end{vmatrix}$, (Interchanging rows and columns across the diagonals), = (-1)$\begin{vmatrix}a & g & d\\ b & h & e\\ c & i & f\end{vmatrix}$ = (1)² = $\begin{vmatrix}b & h & e\\ a & g & d\\ c & i & f\end{vmatrix}$ = $\begin{vmatrix}b & h & e\\ a & g & d\\ c & i & f\end{vmatrix}$ = R.H.S, 1. To view the one-dimensional case in the same way we view higher dimensional linear transformations, we can view a as a 1×1 matrix. Theorem $$\PageIndex{5}$$: Determinant of the Transpose, Let $$A$$ be a matrix where $$A^T$$ is the transpose of $$A$$. $\begin{vmatrix}j_{1}+k_{1} & l_{1} & m_{1} \\ j_{2}+k_{2} & l_{2} & m_{2}\\ j_{3}+k_{3} & l_{3} & m_{3}\end{vmatrix}$ = $\begin{vmatrix}j_{1} & l_{1} & m_{1} \\ j_{2} & l_{2} & m_{2}\\ j_{3} & l_{3} & m_{3}\end{vmatrix}$ + $\begin{vmatrix}+k_{1} & l_{1} & m_{1} \\ +k_{2} & l_{2} & m_{2}\\ +k_{3} & l_{3} & m_{3}\end{vmatrix}$, If each term of a determinant above or below the main diagonal comprise zeroes, then the determinant is equivalent to the product of diagonal terms. The same ‘machine’ used in the previous proof will be used again. Then $$\det E_{ij}=-1$$. According to the Determinant Properties, the Value of Determinant Equals to Zero if Row is, 2. Linear Algebra Review Questions. We have seen how to compute the determinant of a matrix, and the incredible fact that we can perform expansion about any row or column to make this computation. Solution: If the entries of every row of A sum to zero, then Ax = 0 when x = (1,. . We have solved determinants using Laplace expansion but by leveraging the properties of determinants, we can solve determinants much faster. Replace a row by a multiple of another row added to itself. Using Definition [def:twobytwodeterminant] we can find the determinant as follows: $\det \left( A \right) = 3 \times 4 - 2 \times 6 = 12 - 12 = 0$ By Theorem [thm:detinverse] $$A$$ is not invertible. About "Properties of Determinants" Properties of Determinants : We can use one or more of the following properties of the determinants to simplify the evaluation of determinants. Example $$\PageIndex{3}$$: Adding a Row to Another Row. Suppose that A, B, and C are all n × n matrices and that they differ by only a row, say the k th row. Exercises on properties of determinants Problem 18.1: (5.1 #10. It behaves like a linear function of first row if all the other rows stay the same. Then, $\det\left(A^T\right) = \det \left( A \right)$. The determinant is also used in multiple variable calculus(mainly in Jacobina) and in computing the cross product of vectors. We assume $$n\geq 3$$ and (1)–(4) are true for all matrices of size $$n-1\times n-1$$. If $$A$$ is an $$n\times n$$ matrix and $$1\leq j \leq n$$, then the matrix obtained by removing $$1$$st column and $$j$$th row from $$A$$ is an $$n-1\times n-1$$ matrix (we shall denote this matrix by $$A(j)$$ below). The description of each of the 10 important properties of determinants are given below. Example $$\PageIndex{4}$$: Multiple of a Row. Let $$A$$ and $$B$$ be $$n \times n$$ matrices and $$k$$ a scalar, such that $$B = kA$$. Computing $$\det \left(A\right) \times \det \left(B\right)$$ we have $$8 \times -5 = -40$$. There are 10 important properties of determinants that are widely used. Theorem $$\PageIndex{2}$$: Multiplying a Row by a Scalar. Have questions or comments? Then matrix $$A(j)$$ used in computation of $$\mathrm{cof}(A)_{1,j}$$ has a row consisting of zeros, and by our inductive assumption $$\mathrm{cof}(A)_{1,j}=0$$. Does this mean that det A = 1? Then we can write $$A= E_1\cdot E_2\cdot \dots\cdot E_m C$$. Make sure to just spend 10 mins watching this video to get general idea on what determinant is. Property 1 : The determinant of a matrix remains unaltered if its rows are changed into columns and columns into rows. Pro Lite, Vedantu Finally, consider the next theorem for the last row operation, that of adding a multiple of a row to another row. A square matrix is a matrix that has equal number of rows and columns. Therefore by our inductive assumption we have $$\mathrm{cof}(A)_{1j}=\mathrm{cof}(B)_{1j}+\mathrm{cof}(C)_{1j}$$ for $$j\neq i$$. In future sections, we will see that using the following properties can greatly assist in finding determinants. We have that $$a_{1,i}=b_{1,j}$$ and also that $$A(i)=B(j)$$. The determinants of a matrix will be equivalent to 0 under the following situations: A row or column is a constant multiple of other row or columns. This concept is discussed in Appendix A.2 and is reviewed here for convenience. The determinant is a linear function. This section includes some important proofs on determinants and cofactors. What is the Key Difference Between Matrices and Determinants? That is, | A| = | A T | . Therefore $$\det A=\det A^T$$. The determinant of a matrix A is denoted det(A), det A, or |A|. Using Properties of Determinant, Prove That, $\begin{vmatrix}x & y & z\\ y & z & x\\ z & x & y\end{vmatrix}$ = (x + y + z)(xy + yz + zx - x² - y² - z²). Also, $$\det C$$ is either 0 or 1 (depending on whether $$C=I$$ or not) and in either case $$\det C=\det C^T$$. Δ = $\begin{vmatrix}x & y & z\\ y & z & x\\ z & x & y\end{vmatrix}$ = $\begin{vmatrix}x+y+z & y & z\\ y+z+x & z & x\\ z+x+y & z & y\end{vmatrix}$ [Operating C$_{1}$ ⟶ C$_{1}$ + C$_{2}$ + C$_{3}$], =  (x + y + z) $\begin{vmatrix}1 & y & 0\\ 1 & z & x\\ 1 & x & y\end{vmatrix}$, = (x + y + z) $\begin{vmatrix}1 & y & z\\ 0 & z-y & x-z\\ 1 & x-y & y-z\end{vmatrix}$[Operating (R$_{2}$ ⟶ R$_{2}$ - R$_{1}$ and (R$_{3}$ ⟶ R$_{3}$ - R$_{1}$)], = ( x + y + z) [(z - y)(y - z)- (x - y) (x - z ), = ( x + y + z) (xy + yz + zx - x² - y² - Z²), 2. The above discussions allow us to now prove Theorem [thm:welldefineddeterminant]. 2. Then $$\det E_{ik}=k$$. Then $$\det \left( A\right) =\det \left( B \right)$$. Let $$C$$ be the reduced row-echelon form of $$A$$. Since these matrices are used in computation of cofactors $$\mathrm{cof}(A)_{1,i}$$, for $$1\leq i\neq n$$, the inductive assumption applies to these matrices. If each term of rows or columns is similar to the column of some other row (or column) then the determinant is equivalent to zero. Then $$A^T=C^T\cdot E_m^T\cdot \dots \cdot E_2^T\cdot E_1$$. Properties of Determinants : 1. Therefore $$\mathrm{cof}(A)_{1l}=k\mathrm{cof}(B)_{1l}$$ for $$l\neq i$$, and for all $$l$$ we have $$a_{1l} \mathrm{cof}(A)_{1l}=k b_{1l}\mathrm{cof}(B)_{1l}$$. The determinant of a square matrix with one row or one column of zeros is equal to zero. Show that $$\det \left( A \right) = 0$$. For each matrix, determine if it is invertible. In the above determinants of the cofactor matrix,Cij denotes the cofactor of the elements aij in Δ. Assume that (2) is true for all $$n-1\times n-1$$ matrices. We will not begin by stating such a formula. Spanning Set of Null Space of Matrix. First we check that the assertion is true for $$n=2$$ (the case $$n=1$$ is either completely trivial or meaningless). Finally, since $$\det A=\det A^T$$ by Theorem [thm:T.T], we conclude that the cofactor expansion along row $$1$$ of $$A$$ is equal to the cofactor expansion along row $$1$$ of $$A^T$$, which is equal to the cofactor expansion along column $$1$$ of $$A$$. To see this, suppose the first row of $$A$$ is equal to $$-1$$ times the second row. Properties of Determinants of Matrices: Determinant evaluated across any row or column is same. Similarly, the square matrix of 3x3 order has three rows and three columns. Let $$A$$ be an $$n\times n$$ matrix and let $$B$$ be a matrix which results from multiplying some row of $$A$$ by a scalar $$k$$. In order to prove the general case, one needs the following fact. 4. In particular $$a_{1i}=kb_{1i}$$, and for $$l\neq i$$ matrix $$A(l)$$ is obtained from $$B(l)$$ by multiplying one of its rows by $$k$$. Assume first that $$C=I$$. Again by Definition [def:twobytwodeterminant] we have $\det \left( B \right) = 2 \times 1 - 5 \times 3 = 2 - 15 = -13$ By Theorem [thm:detinverse] $$B$$ is invertible and the determinant of the inverse is given by \begin{aligned} \det \left( A^{-1} \right) &=& \frac{1}{\det(A)} \\ &=& \frac{1}{-13} \\ &=& -\frac{1}{13}\end{aligned}. Three simple properties completely describe the determinant. 1. Knowing that $$\det \left( A \right) =-2$$, find $$\det \left( B \right)$$. The determinant is considered an important function … Linear Algebra/Properties of Determinants. Linear Algebra/Determinant. Instead, we … Notice that the first row of $$B$$ is $$5$$ times the first row of $$A$$, while the second row of $$B$$ is equal to the second row of $$A$$. If $$A$$ is obtained by interchanging $$i$$th and $$j$$th rows of $$B$$ (with $$i\neq j$$), then $$\det A=-\det B$$. Vedantu academic counsellor will be calling you shortly for your Online Counselling session. If so, find the determinant of the inverse. If rows and columns are interchanged, the determinant remains unaltered. Notice that this theorem is true when we multiply one row of the matrix by $$k$$. Note that if a matrix $$A$$ contains a row which is a multiple of another row, $$\det \left(A\right)$$ will equal $$0$$. Examples Problems on Properties of Determinants. Problem 8. The determinant is considered an important function as it satisfies some additional properties of determinants that are derived from the following conditions. Suppose we were to multiply all $$n$$ rows of $$A$$ by $$k$$ to obtain the matrix $$B$$, so that $$B = kA$$. Jump to navigation Jump to search. If $$A$$ is obtained by multiplying $$i$$th row of $$B$$ by $$k$$ and adding it to $$j$$th row of $$B$$ ($$i\neq j$$) then $$\det A=\det B$$. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. First we recall the definition of a determinant. \end{aligned}\]. The determinant is required to hold these properties: It is linear on the rows of the … (3) This is a consequence of (1). The case $$n=2$$ is easily checked directly (and it is strongly suggested that you do check it). If two rows of $$A$$ are identical then $$\det A=0$$. That is, $\begin{vmatrix}x_{1} & x_{2} & x_{3} \\ 0 & y_{2} & y_{3}\\ 0 & 0 & z_{3}\end{vmatrix}$ = $\begin{vmatrix}x_{1} & 0 & 0\\ x_{2} & y_{2} & 0 \\ x_{3} & y_{3} & z_{3}\end{vmatrix}$ = X$_{1}$Y$_{2}$Z$_{3}$, Δ = $\begin{vmatrix}x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23}\\ x_{31} & x_{32} & x_{33}\end{vmatrix}$ then Δ$_{1}$ = $\begin{vmatrix}z_{11} & z_{12} & z_{13} \\ z_{21} & z_{22} & z_{23}\\ z_{31} & z_{32} & z_{33}\end{vmatrix}$ = Δ$^{2}$. 1. Now assume $$C\neq I$$. |K| or det K. The determinants and its properties are useful as they enable us to obtain the same outcomes with distinct and simpler configurations of elements. This is because of property 2, the exchange rule. Next, we assume that the assertion is true for $$n-1$$ (where $$n\geq 3$$) and prove it for $$n$$. Browse other questions tagged linear-algebra determinant or ask your own question. We have that $$a_{ji}=k b_{ji}$$ for $$1\leq j\leq n$$. The determinant is a linear function. There are several other major properties of determinants which do not involve row (or column) operations. Then $\det \left( AB\right) =\det \left( A\right) \det \left( B\right)$. Watch the recordings here on Youtube! By [E1] we have (using all equalities established above) \begin{aligned} \det A&=\sum_{l=1}^n a_{1,l} \mathrm{cof}(A)_{1,l}\\ &=\sum_{l\neq i} a_{1,l}(\mathrm{cof}(B)_{1,l}+\mathrm{cof}(C)_{1,l})+ (b_{1,i}+c_{1,i})\mathrm{cof}(A)_{1,i}\\ &= \det B+\det C\end{aligned} This proves that the assertion is true for all $$n$$ and completes the proof. Now, let’s compute $$\det \left(B\right)$$ using Theorem [thm:multiplyingrowbyscalar] and see if we obtain the same answer. But $$i$$th and $$j$$th rows of $$C$$ are proportional. If those entries add to one, show that det(A − I) = 0. Approach 2 (axiomatic): we formulate properties that the determinant should have. All of the properties of determinant listed so far have been multiplicative. There are many important properties of determinants. Describe the solution set of a homogeneous linear system if the determinant of the matrix of coefficients is nonzero. The situation for matrix addition and determinants is less elegant: $$\det (A + B)$$ has no pleasant identity. Then In each term, the factor if i > j i. Show that this determinant is zero. Problem 7. In linear algebra, we can compute the determinants of square matrices. However, this row operation will result in a row of zeros. Thus the proof is complete. Since it is in reduced row-echelon form, its last row consists of zeros and by (4) of Example [exa:EX1] the last row of $$CB$$ consists of zeros. Example $$\PageIndex{4}$$: The Determinant of a Product, Compare $$\det \left( AB\right)$$ and $$\det \left( A\right) \det \left( B\right)$$ for $A=\left[ \begin{array}{rr} 1 & 2 \\ -3 & 2 \end{array} \right] ,B=\left[ \begin{array}{rr} 3 & 2 \\ 4 & 1 \end{array} \right]$, $AB=\left[ \begin{array}{rr} 1 & 2 \\ -3 & 2 \end{array} \right] \left[ \begin{array}{rr} 3 & 2 \\ 4 & 1 \end{array} \right] = \left[ \begin{array}{rr} 11 & 4 \\ -1 & -4 \end{array} \right] \nonumber$, and so by Definition [def:twobytwodeterminant], $\det \left( AB\right) =\det \left[ \begin{array}{rr} 11 & 4 \\ -1 & -4 \end{array} \right] = -40 \nonumber$, $\det \left( A\right) =\det \left[ \begin{array}{rr} 1 & 2 \\ -3 & 2 \end{array} \right] = 8 \nonumber$, $\det \left( B\right) =\det \left[ \begin{array}{rr} 3 & 2 \\ 4 & 1 \end{array} \right] = -5 \nonumber$. A matrix is invertible, nonsingular if and only if the value of determinant is not equal to zero, So if the determinant is zero, the matrix is singular and does not have an inverse. This is the same answer as above and you can see that $$\det \left( A\right) \det \left( B\right) =8\times \left( -5\right) =-40 = \det \left(AB\right)$$. If $$A$$ is obtained by multiplying $$i$$th row of $$B$$ by $$k$$ then $$\det A=k\det B$$. Hence, $$\det \left(A\right) = \det \left(A^T\right)$$. Then $$A=E_1\cdot E_2\cdot \dots\cdot E_m$$ and $$AB= E_1\cdot E_2\cdot \dots\cdot E_m B$$. Where do these determinants come from for line intersection? Notice that the rows of $$B$$ are the rows of $$A$$ but switched. The first is the determinant of a product of matrices. These properties are true for determinants of any order. Maybe I can try to say it in words. Theorem $$\PageIndex{3}$$: Adding a Multiple of a Row to Another Row. Maybe I can try to say it in words. 2. If $$A$$ is an $$n\times n$$ matrix such that one of its rows consists of zeros, then $$\det A=0$$. Legal. We have therefore proved the case of (1) when $$j=i+1$$. Some basic properties of determinants are given below: If In is the identity matrix of the order m ×m, then det(I) is equal to1, If the matrix XT is the transpose of matrix X, then det (XT) = det (X), If matrix X-1 is the inverse of matrix X, then det (X-1) = 1/det (x) = det(X)-1, If two square matrices x and y are of equal  size, then det (XY) = det (X) det (Y), If matrix X retains size a × a and C is a constant, then det (CX) = Ca det (X), If A, B, and C are three positive semidefinite matrices of equal size, then the following equation holds along with the corollary det (A+B) ≥ det(A) + det (B) for A,B, C ≥ 0 det (A+B+C) + det C ≥ det (A+B) + det (B+C). Theorem $$\PageIndex{4}$$: Determinant of a Product, Let $$A$$ and $$B$$ be two $$n\times n$$ matrices. $\det A=\sum_{l=1}^n a_{1l}\mathrm{cof}(A)_{1l} =-\sum_{l=1}^n b_{1l} B_{1l} =\det B.$. This exercise is recommended for all readers. By Definition [def:twobytwodeterminant], $$\det \left( A\right) =-2.$$ We can also compute $$\det \left(B\right)$$ using Definition [def:twobytwodeterminant], and we see that $$\det \left(B\right) = -10$$. When we switch two rows of a matrix, the determinant is multiplied by $$-1$$. Laplace’s Formula and the Adjugate Matrix. 0. Let $$E_{ijk}$$ be the elementary matrix obtained by multiplying $$i$$th row of $$I$$ by $$k$$ and adding it to its $$j$$th row. Featured on Meta Feature Preview: New Review Suspensions Mod UX The next theorem demonstrates the effect on the determinant of a matrix when we multiply a row by a scalar. If this is true, it follows that $\det(A^{-1}) = \frac{1}{\det(A)} \nonumber$, Example $$\PageIndex{6}$$: Determinant of an Invertible Matrix. By Theorem $$\PageIndex{1}$$ since two rows of $$A$$ have been switched, $$\det \left(B\right) = - \det \left(A\right) = - \left(-2\right) = 2$$. The reflection property of determinants defines that determinants do no change if rows are transformed into columns and columns are transformed into rows. (2) This is like (1)… but much easier. The following provides an essential property of the determinant, as well as a useful way to determine if a matrix is invertible. Then $$A(l)$$ is obtained from $$B(l)$$ by interchanging two of its rows (draw a picture) and by our assumption $\label{E2} \mathrm{cof}(A)_{1,l}=-\mathrm{cof}(B)_{1,l}.$, Now consider $$a_{1,i} \mathrm{cof}(A)_{1,l}$$. It follows that $$\det \left(A\right) = 2 \times 3 - 4 \times 5 = -14$$ and $$\det \left(A^T\right) = 2 \times 3 - 5 \times 4 = -14$$. When Can We Get the Determinant of a Matrix Equivalent to Zero? We need to prove that $\det A=\sum_{i=1}^n a_{j,i} \mathrm{cof}(A)_{j,i}.$ Let us prove the case when $$j=2$$. The case when $$j>2$$ is very similar; we still have $$minor(B)_{1,i}=minor (A)_{j,i}$$ but checking that $$\det B=-\sum_{i=1}^n a_{j,i} \mathrm{cof}(A)_{j,i}$$ is slightly more involved. Let $$l\in \{1, \dots, n\}\setminus \{i,j\}$$. If all the elements of a row (or column) are zeros, then the value of the determinant is zero. Putting this together with [E2] into [E1] we see that if in the formula for $$\det A$$ we change the sign of each of the summands we obtain the formula for $$\det B$$. On the other hand, if $$j=i$$ then $$a_{1,j}=0$$. The determinant of a matrix is zero if each element of the matrix is equal to zero. Solution: Interchanging the rows and columns across the diagonals by making use of reflection property and then using the switching property of determination we can get the desired outcome. Then we have $$a_{ij}=0$$ for $$1\leq j\leq n$$. Once this is accomplished, by the Principle of Mathematical Induction we can conclude that the statement is true for all $$n\times n$$ matrices for every $$n\geq 2$$. Now the cofactor expansion along column $$j$$ of $$A$$ is equal to the cofactor expansion along row $$j$$ of $$A^T$$, which is by the above result just proved equal to the cofactor expansion along row 1 of $$A^T$$, which is equal to the cofactor expansion along column $$1$$ of $$A$$. The following example is straightforward and strongly recommended as a means for getting used to definitions. Answer. Expanding an $$n\times n$$ matrix along any row or column always gives the same result, which is the determinant. Let $$A=\left[ \begin{array}{rr} 1 & 2 \\ 3 & 4 \end{array} \right] ,\ B=\left[ \begin{array}{rr} 5 & 10 \\ 3 & 4 \end{array} \right] .$$ Knowing that $$\det \left( A \right) =-2$$, find $$\det \left( B \right)$$. 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