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DETERMINANTS 

Class 12 PCM . Mathematics . Mathematics Part I . Determinants

Chapter 4: DETERMINANTS 

Ch 4

Mathematics

Class 12 PCM

Lecture 1

Introduction

Reference video · Lecture 1Introduction

Meaning and Use of Determinants

Determinant of a Two-Equation Coefficient Matrix
$$\det(A)=a_1b_2-a_2b_1$$

For a two-variable linear system, a non-zero coefficient determinant guarantees a unique solution.

Test for a Unique Solution

  1. Write the coefficient matrix of the system.
  2. Calculate its determinant.
    $$\det(A)=a_1b_2-a_2b_1$$
  3. If the determinant is non-zero, the system has a unique solution.
    $$\det(A)\ne 0$$
Worked example

Does the system \(2x+4y=6\), \(x+3y=5\) have a unique solution?

  1. Find the coefficient determinant for the system.
    $$\begin{vmatrix}2&4\\1&3\end{vmatrix}=2(3)-4(1)=2$$
  2. Since it is non-zero, the system has a unique solution.
    $$2\ne 0$$
Answer: Yes. Its coefficient determinant is \(2\), which is non-zero.

Determinant of Order One

Order One
$$\det([a])=a$$
Worked example

Evaluate \(\det([7])\).

  1. The determinant equals the only entry.
    $$\det([7])=7$$
Answer: \(7\)

Determinant of Order Two

Order Two
$$\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}=a_{11}a_{22}-a_{12}a_{21}$$
Worked example

Evaluate \(\begin{vmatrix}2&4\\-1&2\end{vmatrix}\).

  1. Apply the order-two formula.
    $$\begin{vmatrix}2&4\\-1&2\end{vmatrix}=2(2)-4(-1)$$
  2. Simplify.
    $$4+4=8$$
Answer: \(8\)
⚠ Common mistake

Do not add the diagonal products: the order-two rule is main-diagonal product minus off-diagonal product.

Determinant of Order Three

Expansion rule

For an entry in row \(i\) and column \(j\), use sign \((-1)^{i+j}\), delete its row and column to form the order-two determinant, and add the resulting terms. Expansion along any row or column gives the same value.

$$\det(A)=a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31})$$
Order-three Determinant: Six-term Form
$$a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}-a_{12}a_{21}a_{33}-a_{11}a_{23}a_{32}$$

This equivalent form is useful for checking an expansion.

Expansion Properties and Examples

  • Sign pattern: Along the first row the signs are \(+,-,+\); generally, the sign at \((i,j)\) is \((-1)^{i+j}\).
  • Efficient choice: Expand along a row or column containing the most zeros.
  • Same value: Expansions along all three rows and all three columns give the same determinant.
  • Zero-column example: \(\begin{vmatrix}1&2&4\\-1&3&0\\4&1&0\end{vmatrix}=-52\).
  • Trigonometric example: \(\begin{vmatrix}0&\sin\alpha&-\cos\alpha\\-\sin\alpha&0&\sin\beta\\\cos\alpha&-\sin\beta&0\end{vmatrix}=0\); expansion along the first row gives \(\sin\alpha\sin\beta\cos\alpha-\cos\alpha\sin\alpha\sin\beta=0\).
Scaling a Determinant
$$|A|=k^n|B|$$

If \(A=kB\) and both matrices have order \(n\), scaling every entry by \(k\) scales the determinant by \(k^n\).

Expand an Order-three Determinant

  1. Choose a row or column, preferably one with the most zeros.
  2. For each entry, delete its row and column to form an order-two determinant.
  3. Multiply each entry by its minor and sign \((-1)^{i+j}\), then add the terms.
Worked example

Evaluate \(\begin{vmatrix}1&2&4\\-1&3&0\\4&1&0\end{vmatrix}\).

  1. Expand down the third column, which contains two zeros.
    $$\begin{vmatrix}1&2&4\\-1&3&0\\4&1&0\end{vmatrix}=4\begin{vmatrix}-1&3\\4&1\end{vmatrix}$$
  2. Evaluate the remaining order-two determinant.
    $$4((-1)(1)-3(4))=4(-13)=-52$$
Answer: \(-52\)
Practice

Evaluate \(\begin{vmatrix}x&x+1\\x-1&x\end{vmatrix}\).

Hint: Use the order-two rule and simplify the difference of squares.
Key step: \(x^2-(x+1)(x-1)\)
Solution: \(x^2-(x^2-1)=1\).
Practice

If \(A=3B\), where \(B\) is an order-three matrix and \(|B|=-2\), find \(|A|\).

Hint: Use the scaling rule \(|A|=k^n|B|\).
Key step: \(|A|=3^3|B|\)
Solution: \(|A|=27(-2)=-54\).
Practice

Find \(x\) if \(\begin{vmatrix}3&x\\x&1\end{vmatrix}=\begin{vmatrix}3&2\\4&1\end{vmatrix}\).

Hint: Evaluate both order-two determinants, then solve the resulting equation.
Key step: \(3-x^2=3-8\)
Solution: \(x^2=8\), so \(x=\pm2\sqrt{2}\).

Quick recap

  • A determinant is a number assigned only to a square matrix; its notations include \(|A|\), \(\det(A)\), and \(\Delta\).
  • For two linear equations, a non-zero coefficient determinant guarantees a unique solution.
  • An order-one determinant is its entry; an order-two determinant is the main-diagonal product minus the off-diagonal product.
  • An order-three determinant can be expanded along any row or column using signed order-two determinants.
  • The expansion sign at position \((i,j)\) is \((-1)^{i+j}\); choosing a line with zeros simplifies calculations.
  • If \(A=kB\) for matrices of order \(n\), then \(|A|=k^n|B|\).
Lecture 2

Area Of A Triangle

Reference video · Lecture 2Area Of A Triangle

Area of a Triangle

Definition

Area of a triangle using determinants — For vertices \((x_1,y_1)\), \((x_2,y_2)\) and \((x_3,y_3)\), the triangle’s area is half the absolute value of their coordinate determinant.

Key properties

  • Positive area: Take the absolute value of the determinant.
  • Given area: Use both positive and negative determinant values when solving.
  • Collinear points: Three collinear points form a triangle of area \(0\).
  • Line test: A point \((x,y)\) is on the line through two fixed points if the determinant for the three points is \(0\).
Triangle area
$$\Delta=\frac{1}{2}\left|\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}\right|=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$$

Use the determinant of the three coordinate rows; the expanded form is equivalent.

Finding a triangle’s area

  1. Write the three vertices as rows with a final coordinate of \(1\).
    $$\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}$$
  2. Evaluate the determinant, take its absolute value, and divide by \(2\).
    $$\Delta=\frac{1}{2}|D|$$
Worked example

Find the area of the triangle with vertices \((3,8)\), \((-4,2)\) and \((5,1)\).

  1. Find the area for vertices \((3,8)\), \((-4,2)\) and \((5,1)\).
    $$\Delta=\frac{1}{2}\left|\begin{vmatrix}3&8&1\\-4&2&1\\5&1&1\end{vmatrix}\right|$$
  2. Expand the determinant along the first row.
    $$\Delta=\frac{1}{2}|3(2-1)-8(-4-5)+1(-4-10)|$$
  3. Simplify and take the absolute value.
    $$\Delta=\frac{1}{2}|3+72-14|=\frac{61}{2}$$
Answer: The area is \(\frac{61}{2}\) square units.
⚠ Common mistake

Do not report a negative area: take the absolute value of the determinant. If the area is given, retain both determinant signs while solving.

Practice

Use determinants to find the line through \(A(1,3)\) and \(B(0,0)\). If \(D(k,0)\) makes the area of \(\triangle ABD\) equal to \(3\) square units, find \(k\).

Hint: Set the determinant for the variable point \((x,y)\) to \(0\) for the line; use the absolute value for the area.
Key step: \(\frac{1}{2}(y-3x)=0\) and \(\frac{1}{2}|-3k|=3\).
Solution: The line is \(y=3x\). Since \(|k|=2\), \(k=2\) or \(k=-2\).

Minors and Cofactors

Definition

Minor — The minor \(M_{ij}\) of element \(a_{ij}\) is the determinant formed by deleting row \(i\) and column \(j\).

Cofactors and determinant expansion

For an element \(a_{ij}\), its cofactor is \(A_{ij}=(-1)^{i+j}M_{ij}\). A determinant can be expanded along any one row or column by summing each entry times its corresponding cofactor.

Key properties

  • Order: In an order-\(n\) determinant, each minor is of order \(n-1\).
  • Sign pattern: Cofactor signs alternate, starting with \(+\) at the top left.
  • Expansion along row \(i\): Sum \(a_{ij}A_{ij}\) across that row.
  • Expansion along column \(j\): Sum \(a_{ij}A_{ij}\) down that column.
  • Different row or column: Entries from one row (or column) multiplied by cofactors from a different row (or column) sum to \(0\).
Cofactor signs for a \(3\times3\) determinant
Row 1Row 2Row 3
\(+\quad-\quad+\)\(-\quad+\quad-\)\(+\quad-\quad+\)
Cofactor and expansion formulas
$$A_{ij}=(-1)^{i+j}M_{ij},\qquad \det(A)=\sum_{j=1}^{n}a_{ij}A_{ij}\ \text{(fixed row }i\text{)},\qquad \det(A)=\sum_{i=1}^{n}a_{ij}A_{ij}\ \text{(fixed column }j\text{)}$$

For a fixed row or column, multiply each entry by its cofactor and add.

Zero sum using cofactors from a different row or column

For distinct rows \(r\) and \(s\), \(\sum_{j=1}^{n}a_{rj}A_{sj}=0\). For distinct columns \(r\) and \(s\), \(\sum_{i=1}^{n}a_{ir}A_{is}=0\).

Finding a minor and cofactor

  1. Locate \(a_{ij}\) and delete its row and column to get \(M_{ij}\).
    $$M_{ij}$$
  2. Evaluate the remaining determinant; for a \(2\times2\) determinant, subtract the cross-products.
    $$\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc$$
  3. Multiply the minor by \((-1)^{i+j}\) to obtain the cofactor.
    $$A_{ij}=(-1)^{i+j}M_{ij}$$
Worked example

Find the minors and cofactors of all elements of \(\begin{vmatrix}1&4\\-2&3\end{vmatrix}\).

  1. Find all minors and cofactors of the determinant.
    $$\Delta=\begin{vmatrix}1&4\\-2&3\end{vmatrix}$$
  2. Each minor is the single entry left after deleting the element’s row and column.
    $$M_{11}=3,\quad M_{12}=-2,\quad M_{21}=4,\quad M_{22}=1$$
  3. Apply the alternating cofactor signs.
    $$A_{11}=3,\quad A_{12}=-(-2)=2,\quad A_{21}=-4,\quad A_{22}=1$$
Answer: The minors are \(M_{11}=3\), \(M_{12}=-2\), \(M_{21}=4\), \(M_{22}=1\); the cofactors are \(A_{11}=3\), \(A_{12}=2\), \(A_{21}=-4\), \(A_{22}=1\).
⚠ Common mistake

Do not confuse a minor with a cofactor: the cofactor includes the sign factor \((-1)^{i+j}\).

Practice

Find the minor and cofactor of the element \(6\) in \(\begin{vmatrix}1&2&3\\4&5&6\\7&8&9\end{vmatrix}\).

Hint: The element \(6\) is in row \(2\), column \(3\); delete that row and column.
Key step: \(M_{23}=\begin{vmatrix}1&2\\7&8\end{vmatrix}\) and the cofactor sign is \((-1)^5\).
Solution: \(M_{23}=8-14=-6\) and \(A_{23}=(-1)^5(-6)=6\).
Practice

For \(\Delta=\begin{vmatrix}2&-3&5\\6&0&4\\1&5&-7\end{vmatrix}\), find all minors and cofactors, then verify \(a_{11}A_{31}+a_{12}A_{32}+a_{13}A_{33}=0\).

Hint: Delete the matching row and column for each minor; use \((-1)^{i+j}\) to find each cofactor.
Key step: The third-row cofactors are \(A_{31}=-12\), \(A_{32}=22\), and \(A_{33}=18\).
Solution: Minors by row: \((-20,-46,30)\), \((-4,-19,13)\), \((-12,-22,18)\). Cofactors by row: \((-20,46,30)\), \((4,-19,-13)\), \((-12,22,18)\). The required sum is \(2(-12)+(-3)(22)+5(18)=0\).

Quick recap

  • Triangle area is half the absolute value of the coordinate determinant.
  • Three collinear points have determinant \(0\) and form a zero-area triangle.
  • A minor \(M_{ij}\) is formed by deleting row \(i\) and column \(j\).
  • A cofactor is \(A_{ij}=(-1)^{i+j}M_{ij}\).
  • Expand a determinant along any one row or column using matching cofactors.
  • Entries from one row multiplied by cofactors from a different row sum to \(0\).
Lecture 3

Adjoint And Inverse Of A Matrix

Reference video · Lecture 3Adjoint And Inverse Of A Matrix

Adjoint of a Matrix

Adjoint of a 2 × 2 Matrix
$$\text{If } A=\begin{bmatrix}a&b\\c&d\end{bmatrix},\quad \operatorname{adj}A=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$

For a matrix of order \(2\), interchange the diagonal entries and change the signs of the off-diagonal entries.

For a 2 × 2 matrix, swap the diagonal entries and change the signs of the other two entries.
For a 2 × 2 matrix, swap the diagonal entries and change the signs of the other two entries.
Worked example

Find \(\operatorname{adj} A\) for \(A=\begin{bmatrix}2&3\\1&4\end{bmatrix}\).

  1. Find the adjoint by using the 2 × 2 rule.
    $$A=\begin{bmatrix}2&3\\1&4\end{bmatrix}$$
  2. Swap the diagonal entries and negate the off-diagonal entries.
    $$\operatorname{adj}A=\begin{bmatrix}4&-3\\-1&2\end{bmatrix}$$
Answer: \(\operatorname{adj} A=\begin{bmatrix}4&-3\\-1&2\end{bmatrix}\)

Adjoint Identity and Determinant Properties

Adjoint Identity

For a square matrix \(A\) of order \(n\), \(A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I\).

$$A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I$$
Determinant of the Adjoint

For a square matrix \(A\) of order \(n\), \(|\operatorname{adj}A|=|A|^{n-1}\).

$$|\operatorname{adj}A|=|A|^{n-1}$$
Definition

Singular matrix — A square matrix is singular if its determinant is zero.

Definition

Nonsingular matrix — A square matrix is nonsingular if its determinant is nonzero.

Singular vs Nonsingular
PointSingularNonsingular
Determinant\(|A|=0\)\(|A|\ne 0\)
InverseDoes not existExists

Examples

  • Singular: \(\begin{bmatrix}1&2\\4&8\end{bmatrix}\) has determinant \(0\).
  • Nonsingular: \(\begin{bmatrix}1&2\\3&4\end{bmatrix}\) has determinant \(-2\), so it is nonsingular.

Inverse of a Matrix

Invertibility Criterion

A square matrix \(A\) is invertible if and only if \(|A|\ne0\).

$$A^{-1}=\frac{1}{|A|}\operatorname{adj}A,\quad |A|\ne0$$

Find an Inverse Using the Adjoint

  1. Calculate the determinant \(|A|\).
  2. If \(|A|=0\), the inverse does not exist; otherwise find the cofactors and form \(\operatorname{adj}A\).
  3. Use the inverse formula.
    $$A^{-1}=\frac{1}{|A|}\operatorname{adj}A$$
Worked example

For \(A=\begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}\), verify \(A(\operatorname{adj}A)=|A|I\) and find \(A^{-1}\).

  1. Expand the determinant along the first row.
    $$|A|=1(16-9)-3(4-3)+3(3-4)=1$$
  2. The cofactor matrix transposed gives the adjoint.
    $$\operatorname{adj}A=\begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}$$
  3. Since \(|A|=1\), the inverse equals the adjoint.
    $$A^{-1}=\frac{1}{1}\operatorname{adj}A=\begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}$$
  4. The adjoint identity verifies the product.
    $$A(\operatorname{adj}A)=I=|A|I$$
Answer: \(A^{-1}=\begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}\).

Inverse of a Product

Inverse of a Product
$$(AB)^{-1}=B^{-1}A^{-1}$$

For invertible matrices \(A\) and \(B\) of the same order:

Worked example

Verify \((AB)^{-1}=B^{-1}A^{-1}\) for \(A=\begin{bmatrix}2&3\\1&-4\end{bmatrix}\) and \(B=\begin{bmatrix}1&-2\\-1&3\end{bmatrix}\).

  1. Multiply the matrices and check the product is invertible.
    $$AB=\begin{bmatrix}-1&5\\5&-14\end{bmatrix},\quad |AB|=-11$$
  2. Find the inverse of the product using the 2 × 2 adjoint formula.
    $$(AB)^{-1}=\frac{1}{-11}\begin{bmatrix}-14&-5\\-5&-1\end{bmatrix}=\frac{1}{11}\begin{bmatrix}14&5\\5&1\end{bmatrix}$$
  3. Calculate each inverse, then multiply them in reverse order.
    $$B^{-1}A^{-1}=\frac{1}{11}\begin{bmatrix}14&5\\5&1\end{bmatrix}$$
  4. The two results agree.
    $$(AB)^{-1}=B^{-1}A^{-1}$$
Answer: \((AB)^{-1}=B^{-1}A^{-1}=\frac{1}{11}\begin{bmatrix}14&5\\5&1\end{bmatrix}\).

Finding an Inverse from a Matrix Equation

Worked example

Show that \(A^2-4A+I=O\) for \(A=\begin{bmatrix}2&3\\1&2\end{bmatrix}\), then find \(A^{-1}\).

  1. Square the matrix and confirm the given identity.
    $$A=\begin{bmatrix}2&3\\1&2\end{bmatrix},\quad A^2=\begin{bmatrix}7&12\\4&7\end{bmatrix},\quad A^2-4A+I=O$$
  2. Rearrange the equation and postmultiply by \(A^{-1}\).
    $$A^2-4A=-I\quad\Longrightarrow\quad A-4I=-A^{-1}$$
  3. Solve for the inverse.
    $$A^{-1}=4I-A=\begin{bmatrix}2&-3\\-1&2\end{bmatrix}$$
Answer: \(A^{-1}=\begin{bmatrix}2&-3\\-1&2\end{bmatrix}\).
⚠ Common mistake

Do not reverse the wrong way: \((AB)^{-1}=B^{-1}A^{-1}\), not \(A^{-1}B^{-1}\). Also, never use the adjoint inverse formula when \(|A|=0\).

Practice

Find \(\operatorname{adj}A\) for \(A=\begin{bmatrix}3&-2\\5&1\end{bmatrix}\).

Hint: Swap the diagonal entries and negate the off-diagonal entries.
Key step: \(\operatorname{adj}A=\begin{bmatrix}1&2\\-5&3\end{bmatrix}\).
Solution: \(\operatorname{adj}A=\begin{bmatrix}1&2\\-5&3\end{bmatrix}\).
Practice

Find the inverse of \(A=\begin{bmatrix}4&1\\2&1\end{bmatrix}\), if it exists.

Hint: Compute \(|A|\) first, then apply the adjoint formula.
Key step: \(|A|=2\) and \(\operatorname{adj}A=\begin{bmatrix}1&-1\\-2&4\end{bmatrix}\).
Solution: \(A^{-1}=\frac{1}{2}\begin{bmatrix}1&-1\\-2&4\end{bmatrix}\).
Practice

If \(A\) and \(B\) are invertible, state and verify the inverse formula for their product.

Hint: Multiply \(AB\) by the proposed inverse and use associativity.
Key step: \((AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=I\).
Solution: \((AB)^{-1}=B^{-1}A^{-1}\), since \((AB)(B^{-1}A^{-1})=I\) and \((B^{-1}A^{-1})(AB)=I\).

Quick recap

  • The adjoint is the transpose of the cofactor matrix; for order \(2\), swap diagonal entries and negate the others.
  • \(A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I\).
  • A matrix is singular when \(|A|=0\) and nonsingular when \(|A|\ne0\).
  • A square matrix is invertible exactly when it is nonsingular; then \(A^{-1}=\frac{1}{|A|}\operatorname{adj}A\).
  • For matrices of the same order, \(|AB|=|A||B|\) and \(|\operatorname{adj}A|=|A|^{n-1}\) for order \(n\).
  • The inverse of a product reverses the order: \((AB)^{-1}=B^{-1}A^{-1}\).
Lecture 4

Applications of Determinants and Matrices

Reference video · Lecture 4Applications of Determinants and Matrices

Consistency and the matrix method

Matrix method and uniqueness

Write a linear system as \(AX=B\), where \(A\) is the coefficient matrix, \(X\) is the variable vector and \(B\) is the constants vector. If \(|A|\ne 0\), then \(A^{-1}\) exists and the system has the unique solution \(X=A^{-1}B\).

Singular coefficient matrix

If \(|A|=0\), then \(A^{-1}\) does not exist. For a square singular matrix, if \((\operatorname{adj}A)B\ne O\), the system has no solution; if \((\operatorname{adj}A)B=O\), it may have infinitely many solutions or no solution.

Inverse of a matrix and solution
$$A^{-1}=\frac{1}{|A|}\operatorname{adj}A,\qquad X=A^{-1}B\quad (|A|\ne 0)$$

Use these formulas when the coefficient matrix is nonsingular.

Solve a system using the inverse matrix

  1. Arrange the equations as \(AX=B\).
    $$AX=B$$
  2. Find \(|A|\). If it is nonzero, calculate \(A^{-1}=\frac{1}{|A|}\operatorname{adj}A\).
    $$A^{-1}=\frac{1}{|A|}\operatorname{adj}A$$
  3. Calculate \(X=A^{-1}B\) and read off the variable values.
    $$X=A^{-1}B$$
Worked example

Solve \(3x-2y+3z=8\), \(2x+y-z=1\) and \(4x-3y+2z=4\) using the matrix method.

  1. Solve the system using the matrix method.
    $$\begin{aligned}3x-2y+3z&=8\\2x+y-z&=1\\4x-3y+2z&=4\end{aligned}$$
  2. Form \(AX=B\) and check that the coefficient matrix is nonsingular.
    $$A=\begin{bmatrix}3&-2&3\\2&1&-1\\4&-3&2\end{bmatrix},\quad B=\begin{bmatrix}8\\1\\4\end{bmatrix},\quad |A|=-17\ne0$$
  3. Calculate the inverse using the adjugate.
    $$A^{-1}=\frac{1}{-17}\begin{bmatrix}-1&-5&-1\\-8&-6&9\\-10&1&7\end{bmatrix}$$
  4. Multiply by \(B\) to obtain the variable vector.
    $$X=A^{-1}B=\frac{1}{-17}\begin{bmatrix}-17\\-34\\-51\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}$$
Answer: \(x=1,\ y=2,\ z=3\)
⚠ Common mistake

Do not use \(A^{-1}\) when \(|A|=0\); the inverse exists only for a nonsingular matrix. Also check the sign of \(|A|\) when dividing the adjugate.

Practice

Solve \(2x+5y=1\) and \(3x+2y=7\) using the inverse-matrix method.

Hint: Use the coefficient matrix \(A=\begin{bmatrix}2&5\\3&2\end{bmatrix}\).
Key step: Check \(|A|=4-15=-11\), then calculate \(A^{-1}B\).
Solution: \(A^{-1}=\frac{1}{-11}\begin{bmatrix}2&-5\\-3&2\end{bmatrix}\), so \(\begin{bmatrix}x\\y\end{bmatrix}=A^{-1}\begin{bmatrix}1\\7\end{bmatrix}=\begin{bmatrix}3\\-1\end{bmatrix}\). Thus \(x=3,\ y=-1\).
Practice

The sum of three numbers is \(6\). Three times the third number plus the second is \(11\), and the sum of the first and third is twice the second. Find the numbers using the matrix method.

Hint: Let the numbers be \(x,y,z\) in order and translate each condition into an equation.
Key step: The equations are \(x+y+z=6\), \(y+3z=11\) and \(x-2y+z=0\); the coefficient determinant is \(9\).
Solution: \(A=\begin{bmatrix}1&1&1\\0&1&3\\1&-2&1\end{bmatrix}\) and \(B=\begin{bmatrix}6\\11\\0\end{bmatrix}\). Since \(|A|=9\), the system has a unique solution; \(A^{-1}B=\begin{bmatrix}1\\2\\3\end{bmatrix}\). The numbers are \(1,2,3\).

Quick recap

  • A system is consistent if it has at least one solution; otherwise it is inconsistent.
  • This chapter's discussion is restricted to systems having unique solutions.
  • Write a linear system in the form \(AX=B\).
  • If \(|A|\ne0\), then \(X=A^{-1}B\) gives the unique solution.
  • For a nonsingular matrix, \(A^{-1}=\frac{1}{|A|}\operatorname{adj}A\).
  • If \(|A|=0\), the inverse method cannot be used; the system may have no solution or infinitely many solutions.
Lecture 5

Miscellaneous Examples

Reference video · Lecture 5Miscellaneous Examples

Determinants, Adjugates and Systems of Equations

Determinants of Orders 2 and 3
$$\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc,\qquad \begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}=a_1\begin{vmatrix}b_2&c_2\\b_3&c_3\end{vmatrix}-b_1\begin{vmatrix}a_2&c_2\\a_3&c_3\end{vmatrix}+c_1\begin{vmatrix}a_2&b_2\\a_3&b_3\end{vmatrix}$$

For a third-order determinant, the expansion shown is along the first row.

Cofactor Expansion and Adjugate Properties

A determinant equals the sum of the products of entries in any row or column with their corresponding cofactors. For example, expansion along the first row gives \(|A|=a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}\). The sum of entries of one row multiplied by the corresponding cofactors of a different row is zero, such as \(a_{11}A_{21}+a_{12}A_{22}+a_{13}A_{23}=0\). The adjugate is the transpose of the cofactor matrix, and \(A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I\). A matrix is singular when \(|A|=0\) and non-singular when \(|A|\ne 0\).

Triangle Area from Coordinates
$$\Delta=\frac{1}{2}\left|\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}\right|$$

Use the absolute value of the determinant; the third row contains the third vertex coordinates.

Inverse Matrix and Matrix Equation
$$A^{-1}=\frac{1}{|A|}\operatorname{adj}A\quad (|A|\ne 0),\qquad AX=B\Longrightarrow X=A^{-1}B$$

A square matrix has an inverse exactly when it is non-singular. If \(AB=BA=I\), then \(B=A^{-1}\) and \((A^{-1})^{-1}=A\).

Solutions of a Square Matrix System

For \(AX=B\), a non-singular coefficient matrix gives exactly one solution. If \(|A|=0\), then \((\operatorname{adj}A)B\ne 0\) implies no solution; if \((\operatorname{adj}A)B=0\), the system may or may not be consistent. A system is consistent if at least one solution exists, and inconsistent if none exists.

Solve a Linear System Using the Inverse

  1. Write the system as \(AX=B\), where \(A\) contains the coefficients and \(X\) contains the variables.
    $$AX=B$$
  2. Check that \(|A|\ne 0\) so the inverse exists.
    $$|A|\ne 0$$
  3. Find \(A^{-1}\) and calculate \(X=A^{-1}B\).
    $$X=A^{-1}B$$
  4. Read the values of the variables from the resulting vector.
Worked example

Solve \(x-y+2z=1\), \(2y-3z=1\), and \(3x-2y+4z=2\) using the inverse-matrix method.

  1. Write the system in matrix form.
    $$\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\1\\2\end{bmatrix}$$
  2. The coefficient determinant is non-zero, so the inverse exists.
    $$|A|=-1$$
  3. Use the inverse matrix and verify that it gives the identity when multiplied by \(A\).
    $$A^{-1}=\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix},\qquad AA^{-1}=I$$
  4. Multiply the inverse by the constants vector.
    $$\begin{bmatrix}x\\y\\z\end{bmatrix}=A^{-1}B=\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}\begin{bmatrix}1\\1\\2\end{bmatrix}=\begin{bmatrix}0\\5\\3\end{bmatrix}$$
Answer: \(x=0,\ y=5,\ z=3\).
⚠ Common mistake

Do not use \(A^{-1}=\frac{1}{|A|}\operatorname{adj}A\) when \(|A|=0\); a singular matrix has no inverse. For coordinate-area questions, use the absolute value of the determinant.

Practice

Find \(\begin{vmatrix}1&2&0\\0&1&1\\2&0&1\end{vmatrix}\).

Hint: Expand along the first row.
Key step: \(1(1)-2(-2)+0=5\).
Solution: The determinant is \(5\).
Practice

Find the area of the triangle with vertices \((0,0)\), \((4,0)\) and \((1,3)\).

Hint: Use the coordinate determinant area formula.
Key step: \(\Delta=\frac{1}{2}|12|\).
Solution: The area is \(6\) square units.
Practice

Classify the solutions of \(x+y=1\) and \(2x+2y=2\).

Hint: Check whether the second equation is a multiple of the first.
Key step: The second equation is twice the first, so both represent the same line.
Solution: The system is consistent and has infinitely many solutions: \(x+y=1\).

Quick recap

  • Expand a determinant using entries and their matching cofactors.
  • The cofactor is \(A_{ij}=(-1)^{i+j}M_{ij}\); the adjugate is the transpose of the cofactor matrix.
  • A matrix is invertible exactly when its determinant is non-zero.
  • For \(AX=B\) with \(|A|\ne 0\), the unique solution is \(X=A^{-1}B\).
  • For triangle vertices, the area is one-half the absolute value of the coordinate determinant.
  • When \(|A|=0\), use the adjugate consistency test carefully: a zero result does not alone guarantee consistency.

Formula sheet

Determinant of a Two-Equation Coefficient Matrix
$$\det(A)=a_1b_2-a_2b_1$$
Order One
$$\det([a])=a$$
Order Two
$$\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}=a_{11}a_{22}-a_{12}a_{21}$$
Order-three Determinant: Six-term Form
$$a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}-a_{12}a_{21}a_{33}-a_{11}a_{23}a_{32}$$
Scaling a Determinant
$$|A|=k^n|B|$$
Triangle area
$$\Delta=\frac{1}{2}\left|\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}\right|=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$$
Cofactor and expansion formulas
$$A_{ij}=(-1)^{i+j}M_{ij},\qquad \det(A)=\sum_{j=1}^{n}a_{ij}A_{ij}\ \text{(fixed row }i\text{)},\qquad \det(A)=\sum_{i=1}^{n}a_{ij}A_{ij}\ \text{(fixed column }j\text{)}$$
Adjoint of a 2 × 2 Matrix
$$\text{If } A=\begin{bmatrix}a&b\\c&d\end{bmatrix},\quad \operatorname{adj}A=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$
Inverse of a Product
$$(AB)^{-1}=B^{-1}A^{-1}$$
Inverse of a matrix and solution
$$A^{-1}=\frac{1}{|A|}\operatorname{adj}A,\qquad X=A^{-1}B\quad (|A|\ne 0)$$
Determinants of Orders 2 and 3
$$\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc,\qquad \begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}=a_1\begin{vmatrix}b_2&c_2\\b_3&c_3\end{vmatrix}-b_1\begin{vmatrix}a_2&c_2\\a_3&c_3\end{vmatrix}+c_1\begin{vmatrix}a_2&b_2\\a_3&b_3\end{vmatrix}$$
Triangle Area from Coordinates
$$\Delta=\frac{1}{2}\left|\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}\right|$$
Inverse Matrix and Matrix Equation
$$A^{-1}=\frac{1}{|A|}\operatorname{adj}A\quad (|A|\ne 0),\qquad AX=B\Longrightarrow X=A^{-1}B$$

Mathematics — ALL CHAPTERS

1

Relations And Functions

2

Inverse Trigonometric Functions

3

Matrices

4

DETERMINANTS 

5

Continuity And Differentiability

6

Application Of Derivatives

7

Integrals

8

APPLICATION OF INTEGRALS

9

DIFFERENTIAL EQUATIONS

10

VECTOR ALGEBRA

11

Three Dimensional Geometry

12

Linear Programming

13

Probability

14

Proofs In Mathematics

15

Mathematical Modelling