In the previous chapter, we have studied about matrices and algebra of matrices. We have also learnt that a system of algebraic equations can be expressed in the form of matrices. This means, a system of linear equations like \(a_{1}x+b_{1}y=c_{1}\) \(a_{2}x+b_{2}y=c_{2}\) can be represented as \([\begin{smallmatrix}a_{1}&b_{1}\\ a_{2}&b_{2}\end{smallmatrix}][\begin{smallmatrix}x\\ y\end{smallmatrix}]=[\begin{smallmatrix}c_{1}\\ c_{2}\end{smallmatrix}]\). Now, this system of equations has a unique solution or not, is determined by the number \(a_{1}b_{2}-a_{2}b_{1}.\) (Recall that if \(\frac{a_{1}}{a_{2}}\ne\frac{b_{1}}{b_{2}}\) or, \(a_{1}b_{2}-a_{2}b_{1}\ne0\) then the system of linear equations has a unique solution). The number \(a_{1}b_{2}-a_{2}b_{1}\) which determines uniqueness of solution is associated with the matrix \(A=[\begin{smallmatrix}a_{1}&b_{1}\\ a_{2}&b_{2}\end{smallmatrix}]\) and is called the determinant of A or det A. Determinants have wide applications in Engineering, Science, Economics, Social Science, etc.

In this chapter, we shall study determinants up to order three only with real entries. Also, we will study various properties of determinants, minors, cofactors and applications of determinants in finding the area of a triangle, adjoint and inverse of a square matrix, consistency and inconsistency of system of linear equations and solution of linear equations in two or three variables using inverse of a matrix.

To every square matrix \(A=[a_{ij}]\) of order n, we can associate a number (real or complex) called determinant of the square matrix A, where \(a_{ij}=(i,j)^{ih}\) element of A. Reprint 2025-26 This may be thought of as a function which associates each square matrix with a unique number (real or complex). If M is the set of square matrices, K is the set of numbers (real or complex) and \(f:M\rightarrow K\) is defined by \(f(A)=k\) where \(A\in M\) and \(k\in K\), then \(f(A)\) is called the determinant of A. It is also denoted by | \(|A|\) or det A or A.

If \(A=[\begin{smallmatrix}a&b\\ c&d\end{smallmatrix}],\) then determinant of A is written as \(|A|=|\begin{smallmatrix}a&b\\ c&d\end{smallmatrix}|=det(A)\)

1. For matrix A, \(|A|\) is read as determinant of A and not modulus of A.
2. Only square matrices have determinants.

4.2.1 Determinant of a matrix of order one

Let \(A=[a]\) be the matrix of order 1, then determinant of A is defined to be equal to a

4.2.2 Determinant of a matrix of order two

Let \(A=[\begin{smallmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{smallmatrix}]\) be a matrix of order \(2\times2\) of order 2 then the determinant of A is defined as: \(det(A)=|A|=\Delta=|\begin{smallmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{smallmatrix}|=a_{11}a_{22}-a_{21}a_{12}\)

Determinant of a matrix of order three can be determined by expressing it in terms of second order determinants. This is known as expansion of a determinant along a row (or a column). There are six ways of expanding a determinant of order 3 corresponding to each of three rows \((R_{1}\), \(R_{2}\) and \(R_{3}\)) and three columns \((C_{1}\), \(C_{2}\) and \(C_{3})\) giving the same value as shown below.

Consider the determinant of square matrix \(A=[a_{ij}]_{3\times3}\) i.e., \(|A|=|\begin{smallmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{smallmatrix}|\)

Expansion along first Row (R1)

Step 1 Multiply first element \(a_{11}\) of \(R_{1}\) by \((-1)^{(1+1)}[(-1)^{\text{sum of suffixes in } a_{11}}]\) and with the second order determinant obtained by deleting the first row and first column \((C_{1})\) of \(|A|\) as \(a_{11}\) lies in \(R_{1}\) and \(C_{1}\), i.e., \((-1)^{1+1}a_{11}|\begin{smallmatrix}a_{22}&a_{23}\\ a_{32}&a_{33}\end{smallmatrix}|\)

Step 2 Multiply 2nd element \(a_{12}\) of \(R_{1}\) by \((-1)^{1+2}[(-1)^{\text{sum of suffixes in } a_{12}}]\) and the second order determinant obtained by deleting the first row \((R_{1})\) and second column \((C_{2})\) of \(|A|\) as \(a_{12}\) lies in \(R_{1}\) and \(C_{2},\) i.e., \((-1)^{1+2}a_{12}|\begin{smallmatrix}a_{21}&a_{23}\\ a_{31}&a_{33}\end{smallmatrix}|\)

Step 3 Multiply 3rd element of \(R_{1}\) by \((-1)^{1+3}[(-1)^{\text{sum of suffixes in } a_{13}}]\) and the second order determinant obtained by deleting the first row \((R_{1})\) and third column \((C_{3})\) of \(|A|\) as \(a_{13}\) lies in \(R_{1}\) and \(C_{3}\). i.e., \((-1)^{1+3}a_{13}|\begin{smallmatrix}a_{21}&a_{22}\\ a_{31}&a_{32}\end{smallmatrix}|\)

Step 4 Finally, the expansion of the determinant of A, that is, \(|A|\) written as the sum of all three terms obtained in steps 1, 2 and 3 above is given by \(|A|=(-1)^{1+1}a_{11}|\begin{smallmatrix}a_{22}&a_{23}\\ a_{32}&a_{33}\end{smallmatrix}|+(-1)^{1+2}a_{12}|\begin{smallmatrix}a_{21}&a_{23}\\ a_{31}&a_{33}\end{smallmatrix}| +(-1)^{1+3}a_{13}|\begin{smallmatrix}a_{21}&a_{22}\\ a_{31}&a_{32}\end{smallmatrix}|\) or \(|A|=a_{11}(a_{22}a_{33}-a_{32}a_{23})-a_{12}(a_{21}a_{33}-a_{31}a_{23}) +a_{13}(a_{21}a_{32}-a_{31}a_{22})\) \(=a_{11}a_{22}a_{33}-a_{11}a_{32}a_{23}-a_{12}a_{21}a_{33}+a_{12}a_{31}a_{23}+a_{13}a_{21}a_{32}\) \(-a_{13} a_{31} a_{22}\)

Expanding along \(R_{2}\) we get \(|A|=|\begin{smallmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{smallmatrix}|\) \(|A|=(-1)^{2+1}a_{21}|\begin{smallmatrix}a_{12}&a_{13}\\ a_{32}&a_{33}\end{smallmatrix}|+(-1)^{2+2}a_{22}|\begin{smallmatrix}a_{11}&a_{13}\\ a_{31}&a_{33}\end{smallmatrix}| +(-1)^{2+3}a_{23}|\begin{smallmatrix}a_{11}&a_{12}\\ a_{31}&a_{32}\end{smallmatrix}|\) \(=-a_{21}(a_{12}a_{33}-a_{32}a_{13})+a_{22}(a_{11}a_{33}-a_{31}a_{13})\) \(-a_{23}(a_{11}a_{32}-a_{31}a_{12})\) \(|A|=-a_{21}a_{12}a_{33}+a_{21}a_{32}a_{13}+a_{22}a_{11}a_{33}-a_{22}a_{31}a_{13}-a_{23}a_{11}a_{32}\) + A23 A31 A12 \(=a_{11}a_{22}a_{33}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}\) \(-a_{13}a_{31}a_{22}\)

By expanding along \(C_{1}\) we get \(|A|=|\begin{smallmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{smallmatrix}|\) \(|A|=a_{11}(-1)^{1+1}|\begin{smallmatrix}a_{22}&a_{23}\\ a_{32}&a_{33}\end{smallmatrix}|+a_{21}(-1)^{2+1}|\begin{smallmatrix}a_{12}&a_{13}\\ a_{32}&a_{33}\end{smallmatrix}| +a_{31}(-1)^{3+1}|\begin{smallmatrix}a_{12}&a_{13}\\ a_{22}&a_{23}\end{smallmatrix}|\) \(=a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{21}(a_{12}a_{33}-a_{13}a_{32})+a_{31}(a_{12}a_{23}-a_{13}a_{22})\) \(|A|=a_{11}a_{22}a_{33}-a_{11}a_{23}a_{32}-a_{21}a_{12}a_{33}+a_{21}a_{13}a_{32}+a_{31}a_{12}a_{23}\) \(-a_{13}a_{31}a_{22}\) \(-a_{31}a_{13}a_{22}\) \(=a_{11}a_{22}a_{33}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}\) (3)

Clearly, values of A in (1), (2) and (3) are equal. It is left as an exercise to the reader to verify that the values of |A| by expanding along \(R_{3}\), \(C_{2}\) and \(C_{3}\) are equal to the value of \(|A|\) obtained in (1), (2) or (3).

Hence, expanding a determinant along any row or column gives same value.

1. For easier calculations, we shall expand the determinant along that row or column which contains maximum number of zeros.
2. While expanding, instead of multiplying by \((-1)^{i+j}\), we can multiply by +1 or-1 according as \((i+j)\) is even or odd.
3. Let \(A=[\begin{smallmatrix}2&2\\ 4&0\end{smallmatrix}]\) and \(B=[\begin{smallmatrix}1&1\\ 2&0\end{smallmatrix}]\) Then, it is easy to verify that \(A=2B\) Also \(|A|=0-8=-8\) and \(|B|=0-2=-2.\) Observe that, \(|A|=4(-2)=2^{2}|B|\) or \(|A|=2^{n}|B|\) , where \(n=2\) is the order of square matrices A and B.

In general, if \(A=kB\) where A and B are square matrices of order n, then | \(A|=k^{n}\) | B , where \(n=1,2,3\)

In earlier classes, we have studied that the area of a triangle whose vertices are \((x_{1},y_{1})\), \((x_2, y_2)\) and \((x_{3},y_{3})\), is given by the expression \(\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+ x_{3}(y_{1}-y_{2})]\) Now this expression can be written in the form of a determinant as \(A=\frac{1}{2}|\begin{smallmatrix}x_{1}&y_{1}&1\\ x_{2}&y_{2}&1\\ x_{3}&y_{3}&1\end{smallmatrix}|\)

Remarks

1. Since area is a positive quantity, we always take the absolute value of the determinant in (1).
2. If area is given, use both positive and negative values of the determinant for calculation.
3. The area of the triangle formed by three collinear points is zero.

In this section, we will learn to write the expansion of a determinant in compact form using minors and cofactors.

Definition 1 Minor of an element \(a_{ij}\) of a determinant is the determinant obtained by deleting its ith row and jth column in which element \(a_{ij}\) lies. Minor of an element \(a_{ij}\) is denoted by \(M_{ij}\)

Remark Minor of an element of a determinant of order \(n(n\ge2)\) is a determinant of order \(n-1\).

Cofactor of an element \(a_{ij}\), denoted by \(A_{ij}\) is defined by \(A_{ij}=(-1)^{i+j}M_{ij}\) where \(M_{ij}\) is minor of \(a_{ij}.\)

If elements of a row (or column) are multiplied with cofactors of any other row (or column), then their sum is zero. For example, \(\Delta = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}\) But \(a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = a_{11}(-1)^{2+1}|\begin{smallmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{smallmatrix}| + a_{12}(-1)^{2+2}|\begin{smallmatrix} a_{11} & a_{13} \\ a_{31} & a_{33} \end{smallmatrix}| + a_{13}(-1)^{2+3}|\begin{smallmatrix} a_{11} & a_{12} \\ a_{31} & a_{32} \end{smallmatrix}|\) \(= -a_{11}(a_{12}a_{33} - a_{13}a_{32}) + a_{12}(a_{11}a_{33} - a_{13}a_{31}) - a_{13}(a_{11}a_{32} - a_{12}a_{31})\) \(= -a_{11}a_{12}a_{33} + a_{11}a_{13}a_{32} + a_{12}a_{11}a_{33} - a_{12}a_{13}a_{31} - a_{13}a_{11}a_{32} + a_{13}a_{12}a_{31} = 0\) Similarly, we can try for other rows and columns.

In the previous chapter, we have studied inverse of a matrix. In this section, we shall discuss the condition for existence of inverse of a matrix. To find inverse of a matrix, we shall first define adjoint of a matrix.

4.5.1 Adjoint of a matrix

Definition 3 The adjoint of a square matrix \(A = [a_{ij}]_{n \times n}\) is defined as the transpose of the matrix \([A_{ij}]_{n \times n}\), where \(A_{ij}\) is the cofactor of the element \(a_{ij}\). Adjoint of the matrix A is denoted by adj A. Let \(A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}\) Then \(adj A = Transpose of \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix} = \begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix}\)

For a square matrix of order 2, given by \(A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}\). The adj A can also be obtained by interchanging \(a_{11}\) and \(a_{22}\) and by changing signs of \(a_{12}\) and \(a_{21}\), i.e., We state the following theorem without proof.

If A be any given square matrix, then \(A(adj A) = (adj A)A = |A|I\), where I is the identity matrix of the same order A.

Verification Let \(A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}\), then \(adj A = \begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix}\) Since sum of product of elements of a row with their corresponding cofactors is equal to \(|A|\) and otherwise zero, we have \(A(adj A) = \begin{bmatrix} |A| & 0 & 0 \\ 0 & |A| & 0 \\ 0 & 0 & |A| \end{bmatrix} = |A|\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = |A|I\) Similarly, we can verify \((adj A) A = |A| I\) Hence A (adj A) = (adj A) A = |A| I

Definition 4 A square matrix A is said to be singular if \(|A| = 0\). Definition 5 A square matrix A is said to be non-singular if \(|A| \neq 0\).

We state the following theorems without proof. Theorem 2 If A and B are nonsingular matrices of the same order, then AB and BA are also nonsingular matrices of the same order. Theorem 3 The determinant of the product of matrices is equal to product of their respective determinants, that is, \(|AB| = |A| |B|\), where A and B are square matrices of the same order.

We know that \((adj A)A = |A|I = \begin{bmatrix} |A| & 0 & 0 \\ 0 & |A| & 0 \\ 0 & 0 & |A| \end{bmatrix}\) Taking determinant of both sides, we get \(|(adj A)A| = \begin{vmatrix} |A| & 0 & 0 \\ 0 & |A| & 0 \\ 0 & 0 & |A| \end{vmatrix}\) \(|adj A| |A| = |A|^3 \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = |A|^3(1) = |A|^3\) In general, if A is a square matrix of order n, then \(|adj(A)| = |A|^{n-1}\).

A square matrix A is invertible if and only if A is nonsingular matrix. Proof Let A be an invertible matrix of order n and I be the identity matrix of order n. Then there exists a square matrix B of order n such that \(AB = BA = I\). Now \(AB = I\). So \(|AB| = |I|\) or \(|A| |B| = 1\) This gives \(|A| \neq 0\). Hence A is nonsingular. Conversely, let A be nonsingular. Then \(|A| \neq 0\). Now \(A(adj A) = (adj A)A = |A| I\) (Theorem 1) or \(A(\frac{1}{|A|}adj A) = (\frac{1}{|A|}adj A)A = I\) or \(AB = BA = I\), where \(B = \frac{1}{|A|}adj A\) Thus A is invertible and \(A^{-1} = \frac{1}{|A|}adj A\)

In this section, we shall discuss application of determinants and matrices for solving the system of linear equations in two or three variables and for checking the consistency of the system of linear equations.

Consistent system: A system of equations is said to be consistent if its solution (one or more) exists. Inconsistent system: A system of equations is said to be inconsistent if its solution does not exist.

Solution of system of linear equations using inverse of a matrix

Let us express the system of linear equations as a matrix equation and solve it using inverse of the coefficient matrix. Consider the system of equations \(a_1x + b_1y + c_1z = d_1\) \(a_2x + b_2y + c_2z = d_2\) \(a_3x + b_3y + c_3z = d_3\) Let \(A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}\), \(X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\) and \(B = \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix}\). Then, the given system of equations can be written as, \(AX = B\).

Case I: If A is a nonsingular matrix, then its inverse exists. Now \(AX = B\) \(A^{-1}(AX) = A^{-1}B\) (premultiplying by \(A^{-1}\)) \((A^{-1}A)X = A^{-1}B\) (by associative property) \(IX = A^{-1}B\) \(X = A^{-1}B\) This matrix equation provides unique solution for the given system of equations as inverse of a matrix is unique. This method of solving system of equations is known as Matrix Method.

Case II: If A is a singular matrix, then \(|A| = 0\). In this case, we calculate \((adj A)B\). If \((adj A)B \neq O\), (O being zero matrix), then solution does not exist and the system of equations is called inconsistent. If \((adj A)B = O\), then system may be either consistent or inconsistent according as the system have either infinitely many solutions or no solution.

• Determinant of a matrix A = [a₁₁] of order 1 is given by |A| = a₁₁.
• Determinant of a matrix A = \(\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}\) of order 2 is given by \(|A| = a_{11}a_{22} - a_{12}a_{21}\).
• Determinant of a matrix A = \(\begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}\) of order 3 is given by \(|A| = a_1(b_2c_3 - b_3c_2) - b_1(a_2c_3 - a_3c_2) + c_1(a_2b_3 - a_3b_2)\).
• For any square matrix A, the area of a triangle with vertices \((x_1, y_1), (x_2, y_2)\) and \((x_3, y_3)\) is given by \(\frac{1}{2}|\begin{smallmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{smallmatrix}|\).
• Minor of an element \(a_{ij}\) of the determinant of matrix A is the determinant obtained by deleting the \(i^{th}\) row and \(j^{th}\) column.
• Cofactor of an element \(a_{ij}\) of the determinant of matrix A is \(A_{ij} = (-1)^{i+j} M_{ij}\).
• Value of determinant of a matrix A is obtained by the sum of the product of elements of a row (or a column) with their corresponding cofactors.
• If elements of a row (or column) are multiplied with cofactors of any other row (or column), then their sum is zero.
• Adjoint of A, denoted by adj A, is the transpose of the cofactor matrix.
• \(A(adj A) = (adj A)A = |A|I\).
• A square matrix A is said to be singular or non-singular according as \(|A|=0\) or \(|A| \neq 0\).
• If AB = BA = I, where B is the inverse of A, then \(A^{-1}=B\).
• \(A^{-1} = \frac{1}{|A|}(adj A)\).
• A system of linear equations is consistent if it has one or more solutions, and inconsistent if its solution does not exist.
• For a system of equations \(AX=B\):
  • If \(|A| \neq 0\), there is a unique solution given by \(X=A^{-1}B\).
  • If \(|A| = 0\) and \((adj A)B \neq 0\), then there is no solution.
  • If \(|A| = 0\) and \((adj A)B = 0\), then the system has either no solution or infinitely many solutions.

Evaluate the determinants in Exercises 1 and 2. \(|\begin{smallmatrix}2&4\\ -5&-1\end{smallmatrix}|\)

Evaluate the determinants (i) \(|\begin{smallmatrix}cos~\theta&-sin~\theta\\ sin~\theta&cos~\theta\end{smallmatrix}|\) (ii) \(|\begin{smallmatrix}x^{2}-x+1&x-1\\ x+1&x+1\end{smallmatrix}|\) *

Evaluate the determinants If \(A=[\begin{smallmatrix}1&2\\ 4&2\end{smallmatrix}]\) then show that \(|2A|=4|A|\)

Evaluate the determinants If \(A=[\begin{smallmatrix}1&0&1\\ 0&1&2\\ 0&0&4\end{smallmatrix}],\) then show that \(|3~A|=27|A|\)

Evaluate the determinants

1. \(|\begin{smallmatrix}3&-1&-2\\ 0&0&-1\\ 3&-5&0\end{smallmatrix}|\)

2. \(|\begin{smallmatrix}3&-4&5\\ 1&1&-2\\ 2&3&1\end{smallmatrix}|\)

3. \(|\begin{smallmatrix}0&1&2\\ -1&0&-3\\ -2&3&0\end{smallmatrix}|\)

4. \(|\begin{smallmatrix}2&-1&-2\\ 0&2&-1\\ 3&-5&0\end{smallmatrix}|\)

If \(A=[\begin{smallmatrix}1&1&-2\\ 2&1&-3\\ 5&4&-9\end{smallmatrix}],\) find |A|

Find values of x, if (i) \(|\begin{smallmatrix}2&4\\ 5&1\end{smallmatrix}|=|\begin{smallmatrix}2x&4\\ 6&x\end{smallmatrix}|\) (ii) \(|\begin{smallmatrix}2&3\\ 4&5\end{smallmatrix}|=|\begin{smallmatrix}x&3\\ 2x&5\end{smallmatrix}|\)

If \(|\begin{smallmatrix}x&2\\ 18&x\end{smallmatrix}|=|\begin{smallmatrix}6&2\\ 18&6\end{smallmatrix}|\) then x is equal to (A) 6
(B) ± 6 (C) -6

Show that points A(\(a, b+c\)), B(\(b, c+a\)), and C(\(c, a+b\)) are collinear.

Find the values of \(k\) if the area of the triangle is 4 sq. units and the vertices are: (i) (\(k\), 0), (4, 0), (0, 2) (ii) (-2, 0), (0, 4), (0, \(k\))

If the area of a triangle is 35 sq units with vertices (2, -6), (5, 4), and (\(k\), 4), then \(k\) is: (A) 12 (B) -2 (C) -12, -2 (D) 12, -2

Write the Minors and Cofactors of the elements of the following determinants: (i) \[\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}\] (ii) \[\begin{vmatrix} a & c \\ b & d \end{vmatrix}\]

Write the Minors and Cofactors of the elements of the following determinants: (i) \[\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}\] (ii) \[\begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}\]

Using Cofactors of elements of the second row, evaluate: \[\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}\]

Using Cofactors of elements of the third column, evaluate: \[\Delta = \begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}\]

If \(\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}\) and \(A_{ij}\) is the Cofactor of \(a_{ij}\), then the value of \(\Delta\) is given by: (A) \(a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33}\) (B) \(a_{11}A_{11} + a_{12}A_{21} + a_{13}A_{31}\) (C) \(a_{21}A_{11} + a_{22}A_{12} + a_{23}A_{13}\) (D) \(a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}\)

Find the adjoint of the matrix: \[\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\]

Find the adjoint of the matrix: \[\begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix}\]

Verify \(A(\text{adj } A) = (\text{adj } A)A = |A|I\) for: \[A = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}\]

Verify \(A(\text{adj } A) = (\text{adj } A)A = |A|I\) for: \[A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}\]

Find the inverse of the matrix (if it exists): \[\begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}\]

Find the inverse of the matrix (if it exists): \[\begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}\]

Find the inverse of the matrix (if it exists): \[\begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix}\]

Find the inverse of the matrix (if it exists): \[\begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}\]

Find the inverse of the matrix (if it exists): \[\begin{bmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{bmatrix}\]

Find the inverse of the matrix (if it exists): \[\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}\]

Find the inverse of the matrix (if it exists): \[\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}\]

Let \(A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\) and \(B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}\). Verify that \((AB)^{-1} = B^{-1}A^{-1}\).

If \(A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\), show that \(A^2 - 5A + 7I = O\). Hence find \(A^{-1}\).

For the matrix \(A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}\), find the numbers \(a\) and \(b\) such that \(A^2 + aA + bI = O\).

For the matrix \(A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}\), show that \(A^3 - 6A^2 + 5A + 11I = O\). Hence, find \(A^{-1}\).

If \(A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}\), verify that \(A^3 - 6A^2 + 9A - 4I = O\) and hence find \(A^{-1}\).

Let A be a nonsingular square matrix of order \(3 \times 3\). Then \(|\text{adj } A|\) is equal to: (A) \(|A|\) (B) \(|A|^2\) (C) \(|A|^3\) (D) \(3|A|\)

If A is an invertible matrix of order 2, then \(\det(A^{-1})\) is equal to: (A) \(\det(A)\) (B) \(\frac{1}{\det(A)}\) (C) 1 (D) 0

Examine the consistency of the system of equations: \(x + 2y = 2\) \(2x + 3y = 3\)

Examine the consistency of the system of equations: \(2x - y = 5\) \(x + y = 4\)

Examine the consistency of the system of equations: \(x + 3y = 5\) \(2x + 6y = 8\)

Examine the consistency of the system of equations: \(x + y + z = 1\) \(2x + 3y + 2z = 2\) \(ax + ay + 2az = 4\)

Examine the consistency of the system of equations: \(3x - y - 2z = 2\) \(2y - z = -1\) \(3x - 5y = 3\)

Examine the consistency of the system of equations: \(5x - y + 4z = 5\) \(2x + 3y + 5z = 2\) \(5x - 2y + 6z = -1\)

Solve the system of linear equations using the matrix method: \(5x + 2y = 4\) \(7x + 3y = 5\)

Solve the system of linear equations using the matrix method: \(2x - y = -2\) \(3x + 4y = 3\)

Solve the system of linear equations using the matrix method: \(4x - 3y = 3\) \(3x - 5y = 7\)

Solve the system of linear equations using the matrix method: \(5x + 2y = 3\) \(3x + 2y = 5\)

Solve the system of linear equations using the matrix method: \(2x + y + z = 1\) \(x - 2y - z = \frac{3}{2}\) \(3y - 5z = 9\)

Solve the system of linear equations using the matrix method: \(x - y + z = 4\) \(2x + y - 3z = 0\) \(x + y + z = 2\)

Solve the system of linear equations using the matrix method: \(2x + 3y + 3z = 5\) \(x - 2y + z = -4\) \(3x - y - 2z = 3\)

Solve the system of linear equations using the matrix method: \(x - y + 2z = 7\) \(3x + 4y - 5z = -5\) \(2x - y + 3z = 12\)

If \(A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}\), find \(A^{-1}\). Using \(A^{-1}\), solve the system of equations: \(2x - 3y + 5z = 11\) \(3x + 2y - 4z = -5\) \(x + y - 2z = -3\)

Prove that the determinant is independent of \(\theta\): \[\begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}\]

Evaluate: \[\begin{vmatrix} \cos\alpha \cos\beta & \cos\alpha \sin\beta & -\sin\alpha \\ -\sin\beta & \cos\beta & 0 \\ \sin\alpha \cos\beta & \sin\alpha \sin\beta & \cos\alpha \end{vmatrix}\]

If \(A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}\), find \((AB)^{-1}\).

Let \(A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}\). Verify that: (i) \([\text{adj } A]^{-1} = \text{adj}(A^{-1})\) (ii) \((A^{-1})^{-1} = A\)

Evaluate: \[\begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{vmatrix}\]

Evaluate: \[\begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix}\]

Solve the system of equations: \[\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4\] \[\frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1\] \[\frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2\]

If x, y, z are nonzero real numbers, then the inverse of matrix \(A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}\) is: (A) \(\begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix}\) (B) \(xyz\begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix}\) (C) \(\frac{1}{xyz}\begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}\) (D) \(\frac{1}{xyz}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)

Let \(A = \begin{bmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{bmatrix}\), where \(0 \le \theta \le 2\pi\). Then: (A) \(\det(A) = 0\) (B) \(\det(A) \in (2, \infty)\) (C) \(\det(A) \in (2, 4)\) (D) \(\det(A) \in [2, 4]\)