Determinants | CBSE Class 12 Maths Notes

This Mathematics note covers determinants of square matrices, expansion by rows and columns, scalar multiples, triangle area, minors and cofactors, adjoints, inverse matrices, and the consistency and solution of linear equations in two or three variables.

What is a determinant, and how does its notation work?

A determinant is a number associated with a square matrix. A matrix is an arrangement of entries, whereas its determinant is a single value calculated from those entries. Keeping these two objects separate makes later statements about inverses and equations easier to interpret.

Let $A=[a_{ij}]$ be a square matrix of order $n$. Here $A$ names the matrix, $n$ is its number of rows and columns, and $a_{ij}$ denotes the entry in row $i$, column $j$.

The symbols $|A|$, $\det(A)$, and $\Delta$ can denote its determinant. Here $\Delta$, read as delta, names the determinant value. For a matrix, the vertical bars mean determinant, not modulus. Absolute value will be used separately when calculating geometrical area.

Definition: Only square matrices have determinants. The order of a determinant is the order of its associated square matrix.

How are matrices and determinants distinguished?

The calculations here concern determinants through order three with real entries. The general definition also allows complex entries, but they are not needed for these calculations. Check the array's shape before starting: an arrangement with unequal numbers of rows and columns has no determinant.

For an order-one matrix, the determinant equals its sole entry. If $a$ denotes that entry, the definition is $\det([a])=a$. Orders two and three require products and signed sums, so their determinants cannot be obtained by merely adding all their entries.

How do you evaluate a determinant of order two?

For a second-order determinant, multiply the entries on the main diagonal and subtract the product on the other diagonal. Let $a,b,c,d$ denote the four entries, arranged by rows. Then the evaluation rule is $\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc$.

The subtraction order matters. The product involving the top-left and bottom-right entries comes first. Place a negative entry in brackets during substitution so that its sign stays attached to it. Otherwise, a correct determinant formula can lead to an incorrect numerical answer.

Worked example 1. Evaluate $\begin{vmatrix}2&4\\-1&2\end{vmatrix}$.

1. Identify the main-diagonal product: $2\times2=4$.
2. Calculate the other diagonal product, retaining its negative entry: $4\times(-1)=-4$.
3. Subtract the second product from the first: $4-(-4)=4+4=8$.

Answer: $8$. The final addition arises because a negative product is being subtracted.

How should algebraic entries be handled?

The same rule applies when entries contain a variable. Let $x$ be a real variable. Multiply the complete expressions in the entries before combining like terms. An expression containing an addition sign remains one entry; it must not be separated into different positions.

Worked example 2. Evaluate $\begin{vmatrix}x&x+1\\x-1&x\end{vmatrix}$.

1. Apply the diagonal-product rule: $\Delta=x\cdot x-(x+1)(x-1)$.
2. Expand the products: $\Delta=x^2-(x^2-1)$.
3. Distribute the subtraction and collect terms: $\Delta=x^2-x^2+1=1$.

Answer: $1$. Although the entries depend on the variable, the determinant value does not.

Determinant equations are handled by evaluating each determinant separately and then solving the resulting ordinary equation. Any restrictions or multiple roots come from that equation, rather than from the determinant notation itself.

Worked example 3. Find the real values of $x$ satisfying $\begin{vmatrix}3&x\\x&1\end{vmatrix}=\begin{vmatrix}3&2\\4&1\end{vmatrix}$.

1. Evaluate the left determinant: $3-x^2$.
2. Evaluate the right determinant: $3\times1-2\times4=-5$.
3. Equate the values: $3-x^2=-5$, giving $x^2=8$.
4. Take both square roots: $x=\pm\sqrt8=\pm2\sqrt2$.
5. Check either root: $3-(\pm2\sqrt2)^2=3-8=-5$.

Answer: $x=2\sqrt2$ or $x=-2\sqrt2$.

How is a determinant of order three expanded?

A third-order determinant is evaluated by reducing it to second-order determinants. Choose one complete row or one complete column. For each chosen entry, delete its row and column, evaluate the remaining determinant, and apply the sign belonging to the original position.

Result: Expansion along any row or column gives the same value

There are three row choices and three column choices. They all give the same determinant when the position signs and remaining entries are handled correctly. A row or column containing the most zeros often gives the shortest calculation because terms multiplied by zero vanish.

Derivation: Expansion along the first row

Use the entries $a_{ij}$ of the third-order matrix already defined. In the sign factor $(-1)^{i+j}$, the exponent is the sum of the original row and column numbers.

1. For the first entry, delete the first row and first column. The contribution is $a_{11}(a_{22}a_{33}-a_{23}a_{32})$.
2. For the second entry, delete the first row and second column. Its negative sign gives $-a_{12}(a_{21}a_{33}-a_{23}a_{31})$.
3. For the third entry, delete the first row and third column. The contribution is $a_{13}(a_{21}a_{32}-a_{22}a_{31})$.
4. Add all three contributions: $$|A|=a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31}).$$

Result: Expansion combines three signed products. It does not multiply the three second-order determinants together.

Worked example 4. Evaluate $\Delta=\begin{vmatrix}1&2&4\\-1&3&0\\4&1&0\end{vmatrix}$.

1. Choose the third column because its last two entries are zero.
2. The first-row, third-column sign is $(-1)^{1+3}=1$. The surviving term is $4\begin{vmatrix}-1&3\\4&1\end{vmatrix}$.
3. Evaluate the smaller determinant: $(-1)\times1-3\times4=-1-12=-13$.
4. Include the zero contributions: $\Delta=4(-13)+0+0=-52$.

Answer: $-52$. A determinant may be negative; no absolute value is required here.

The signs follow positions, even when the chosen row or column contains zeros. Skipping a zero term does not change the sign of a later term. Retain the original row and column indices throughout the expansion.

How does multiplying a matrix by a scalar affect its determinant?

A scalar is a number multiplying every entry of a matrix. Let $B$ denote a square matrix of the same order as $A$, and let $k$ be a real scalar. The relation $A=kB$ means each entry of the first matrix is the corresponding entry of the second multiplied by that scalar.

Property: A scalar multiplier is raised to the matrix order

For the orders considered here, the determinant rule is $|kB|=k^n|B|$, where $n$ is the order. Thus multiplying every entry is different from simply multiplying the final determinant by the same number. The order determines the power required.

Derivation: Why the power is two for order two

1. Write $B=\begin{bmatrix}a&b\\c&d\end{bmatrix}$, with $a,b,c,d$ denoting its entries, so $|B|=ad-bc$.
2. Multiply every entry: $kB=\begin{bmatrix}ka&kb\\kc&kd\end{bmatrix}$.
3. Evaluate its determinant: $|kB|=(ka)(kd)-(kb)(kc)=k^2ad-k^2bc$.
4. Factor the common scalar power: $|kB|=k^2(ad-bc)=k^2|B|$.

Result: In order two, each diagonal product contains two scalar factors. In the third-order expansion, each product contains three factors, giving the third power instead.

Worked example 5. Verify the scalar rule for $B=\begin{bmatrix}1&1\\2&0\end{bmatrix}$ and $A=2B=\begin{bmatrix}2&2\\4&0\end{bmatrix}$.

1. Evaluate the first determinant: $|B|=1\times0-1\times2=-2$.
2. Evaluate the scaled determinant independently: $|A|=2\times0-2\times4=-8$.
3. Apply the order-two rule: $2^2|B|=4(-2)=-8$.

Answer: $|A|=-8=2^2|B|$, so the two calculations agree.

State the matrix order whenever using this property. The same scalar leads to different powers for different orders. Also distinguish scaling a whole matrix from scaling a determinant value that has already been calculated.

How do determinants give triangle areas and test collinearity?

Let $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ be three vertices, where each pair gives the horizontal and vertical coordinates of one point. Let $S$ denote their triangle's area, and $D$ the coordinate determinant.

The area formula is $$D=\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix},\qquad S=\frac12\operatorname{abs}(D).$$ Here $\operatorname{abs}(D)$ means the absolute value of the scalar determinant. This keeps geometrical area non-negative even when the determinant is negative.

Worked example 6. Find the area of the triangle with vertices $(3,8)$, $(-4,2)$, and $(5,1)$.

1. Place each vertex in one row: $D=\begin{vmatrix}3&8&1\\-4&2&1\\5&1&1\end{vmatrix}$.
2. Expand along the first row: $D=3(2-1)-8(-4-5)+1(-4-10)$.
3. Simplify the signed terms: $D=3+72-14=61$.
4. Take half the absolute value: $S=\frac12\operatorname{abs}(61)=\frac{61}{2}$.

Answer: $\frac{61}{2}$ square units. The coordinates are kept in their original pairs throughout.

Result: Collinear points have zero area

Collinear points lie on one straight line. Their triangle area is zero, so their coordinate determinant is zero. This provides an equation for the line through two fixed points when the third point is treated as a general point on that line.

Worked example 7. Find the line through the points $A=(1,3)$ and $B=(0,0)$. Then find the real parameter $k$ if $D=(k,0)$ makes triangle $ABD$ have area $3$ square units.

1. Let $(x,y)$ be a general point on the line, with $x,y$ its coordinates. Collinearity gives $\begin{vmatrix}1&3&1\\0&0&1\\x&y&1\end{vmatrix}=0$.
2. Expand the determinant: $-y+3x=0$, hence $y=3x$.
3. For the area condition, the coordinate determinant is $\begin{vmatrix}1&3&1\\0&0&1\\k&0&1\end{vmatrix}=3k$.
4. Apply absolute value: $\frac12|3k|=3$, so $|k|=2$.
5. Take both possible signs: $k=2$ or $k=-2$.

Answer: the line is $y=3x$, and the permitted parameter values are $k=\pm2$.

When area is supplied and a coordinate is unknown, retaining both signs is essential. A positive area can correspond to a positive or negative coordinate determinant. Choosing just one sign may discard a valid position of the unknown vertex.

How are minors and cofactors calculated?

The minor of an entry is obtained by deleting the row and column containing it and evaluating the remaining determinant. Write $M_{ij}$ for the minor of $a_{ij}$. The subscript continues to identify the original position, not a position in the smaller array.

For an original order of at least two, the minor has order one less. In a second-order determinant, a minor therefore reduces to one remaining entry. In a third-order determinant, it is a second-order determinant requiring a subtraction of products.

Worked example 8. Find the minor of the entry $6$ in $\begin{vmatrix}1&2&3\\4&5&6\\7&8&9\end{vmatrix}$.

1. The entry is in row two, column three, so its minor is $M_{23}$.
2. Delete that row and column: $M_{23}=\begin{vmatrix}1&2\\7&8\end{vmatrix}$.
3. Evaluate: $M_{23}=1\times8-2\times7=8-14=-6$.

Answer: $M_{23}=-6$. The minor includes its own calculated sign.

How does a cofactor differ from a minor?

The cofactor adds a position-dependent sign to the minor. Write $A_{ij}$ for the cofactor of $a_{ij}$. Its definition is $A_{ij}=(-1)^{i+j}M_{ij}$. An even index sum leaves the minor unchanged; an odd index sum reverses its sign.

The sign pattern for order three is $\begin{bmatrix}+&-&+\\-&+&-\\+&-&+\end{bmatrix}$. These are multipliers, not predictions of the cofactor's final sign. Multiplying a negative minor by a negative position sign produces a positive cofactor.

Worked example 9. Find all minors and cofactors of $\begin{vmatrix}1&-2\\4&3\end{vmatrix}$.

1. Delete the appropriate row and column at each position: $M_{11}=3,\ M_{12}=4,\ M_{21}=-2,\ M_{22}=1$.
2. Apply the positive diagonal signs: $A_{11}=3$ and $A_{22}=1$.
3. Apply the negative off-diagonal signs: $A_{12}=-4$ and $A_{21}=-(-2)=2$.

Answer: the minor matrix is $\begin{bmatrix}3&4\\-2&1\end{bmatrix}$, and the cofactor matrix is $\begin{bmatrix}3&-4\\2&1\end{bmatrix}$.

Calculate deletion, evaluation, and the position sign as separate operations until the method becomes reliable. This prevents confusing a negative matrix entry with the separate sign used in the definition of its cofactor.

How do cofactors simplify expansion and construct the adjoint?

Using cofactors, expansion becomes a sum of entries multiplied by their corresponding cofactors. The alternating signs are already included. For the first row, $|A|=a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}$. Do not insert another alternating sign pattern into this expression.

The word corresponding is essential: each entry is paired with the cofactor of that same position. If entries of one row are instead paired with cofactors of a different row, the sum is zero. The analogous statement holds for columns.

Identity: Corresponding and different rows give different sums

For instance, $a_{11}A_{21}+a_{12}A_{22}+a_{13}A_{23}=0$. Here the entries come from the first row while the cofactors come from the second. Such sums explain the zero entries that arise when a matrix is multiplied by its adjoint.

What is the adjoint?

The adjoint is the transpose of the cofactor matrix. The notation $\operatorname{adj}A$ denotes the adjoint, and a superscript $T$ denotes transpose, which interchanges rows and columns. Therefore $\operatorname{adj}A=[A_{ij}]^T$.

Construct the complete cofactor matrix first, keeping every cofactor in its original position. Then transpose once. A cofactor originally in the first row and second column becomes an entry in the second row and first column of the adjoint.

Worked example 10. Find the adjoint of $A=\begin{bmatrix}2&3\\1&4\end{bmatrix}$.

1. Find the minors by deletion: $M_{11}=4,\ M_{12}=1,\ M_{21}=3,\ M_{22}=2$.
2. Apply the cofactor signs: $A_{11}=4,\ A_{12}=-1,\ A_{21}=-3,\ A_{22}=2$.
3. Assemble the cofactor matrix: $[A_{ij}]=\begin{bmatrix}4&-1\\-3&2\end{bmatrix}$.
4. Transpose it: $\operatorname{adj}A=\begin{bmatrix}4&-3\\-1&2\end{bmatrix}$.

Answer: $\operatorname{adj}A=\begin{bmatrix}4&-3\\-1&2\end{bmatrix}$.

For order two, this procedure is equivalent to exchanging the two main-diagonal entries and changing the signs of the other two entries. That convenient shortcut follows from the cofactor construction; it is not the general construction for order three.

When does an inverse exist, and how is it found?

A square matrix is singular when its determinant is zero and non-singular when its determinant is nonzero. Let $I$ denote the identity matrix of the required order. It has ones on the main diagonal and zeros elsewhere.

Theorem: The adjoint product equals the determinant times the identity

For a square matrix, $A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I$. The diagonal entries of the product come from corresponding cofactor expansions. The off-diagonal entries come from the different-row or different-column sums, which vanish.

Theorem: Invertibility is equivalent to a nonzero determinant

The notation $A^{-1}$ means the inverse matrix, satisfying $AA^{-1}=A^{-1}A=I$. A square matrix has an inverse if and only if it is non-singular. Its determinant must therefore be checked before the inverse formula is used.

Derivation: The inverse formula and its condition

1. If $A$ has inverse $B$, then $AB=I$. Taking determinants gives $|A||B|=1$, so $|A|\ne0$.
2. Conversely, suppose $|A|\ne0$. Start with $A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I$.
3. Divide by the nonzero scalar determinant: $A\left(\frac{\operatorname{adj}A}{|A|}\right)=\left(\frac{\operatorname{adj}A}{|A|}\right)A=I$.
4. The defining inverse relations now give $A^{-1}=\frac{1}{|A|}\operatorname{adj}A$.

Result: The adjoint can be formed for a singular square matrix, but division by its zero determinant cannot produce an inverse.

Theorem: Determinants multiply across a matrix product

For square matrices of the same order, $|AB|=|A||B|$. Consequently, if both matrices are non-singular, their products in either order are non-singular. The scalar determinants multiply even though matrix multiplication itself need not be commutative.

For orders two and three, another useful result is $|\operatorname{adj}A|=|A|^{n-1}$. For a non-singular third-order matrix, this follows by taking determinants of the adjoint-product identity.

1. Start with $A(\operatorname{adj}A)=|A|I$.
2. Take determinants and use the product and scalar rules: $|A||\operatorname{adj}A|=|A|^3|I|=|A|^3$.
3. Cancel the nonzero factor: $|\operatorname{adj}A|=|A|^2$.

Note: Cancellation in this argument requires a nonzero determinant. State that condition before dividing, rather than treating division by a determinant as automatically valid.

How does the inverse method solve equations in two variables?

Let $X$ denote the column matrix of unknowns and $B$ the column matrix of constants in a linear system. Here $B$ is a column matrix, rather than the square matrix used in the earlier product theorem. The coefficient matrix $A$ records coefficients in a fixed variable order.

The equations become $AX=B$. Each row represents one equation; each column of the coefficient matrix corresponds to one unknown. Preserve both the equation order and the variable order when forming the matrices, or the matrix equation will represent a different system.

Derivation: Solving the matrix equation

1. Check $|A|\ne0$, so that the inverse exists.
2. Premultiply the equation by the inverse: $A^{-1}(AX)=A^{-1}B$.
3. Use associativity and the inverse relation: $(A^{-1}A)X=IX=A^{-1}B$.
4. Since the identity leaves the unknown column unchanged, $X=A^{-1}B$.

Result: A non-singular coefficient matrix gives a unique solution. Premultiplication is essential because the inverse must cancel the coefficient matrix on its left.

Worked example 11. Solve $2x+5y=1$ and $3x+2y=7$, where $x,y$ are the two unknown real numbers.

1. Form $A=\begin{bmatrix}2&5\\3&2\end{bmatrix}$, $X=\begin{bmatrix}x\\y\end{bmatrix}$, and $B=\begin{bmatrix}1\\7\end{bmatrix}$.
2. Calculate $|A|=2\times2-5\times3=4-15=-11\ne0$.
3. The minors are $M_{11}=2,\ M_{12}=3,\ M_{21}=5,\ M_{22}=2$. Thus the cofactor matrix is $\begin{bmatrix}2&-3\\-5&2\end{bmatrix}$, giving $\operatorname{adj}A=\begin{bmatrix}2&-5\\-3&2\end{bmatrix}$.
4. Divide by the determinant: $A^{-1}=-\frac1{11}\begin{bmatrix}2&-5\\-3&2\end{bmatrix}$.
5. Multiply by the constants: $X=-\frac1{11}\begin{bmatrix}2-35\\-3+14\end{bmatrix}=-\frac1{11}\begin{bmatrix}-33\\11\end{bmatrix}=\begin{bmatrix}3\\-1\end{bmatrix}$.
6. Check both equations: $2(3)+5(-1)=1$ and $3(3)+2(-1)=7$.

Answer: $x=3,\ y=-1$. Both original equations are satisfied.

Substitution is a direct check of the final values. It can reveal an error in a cofactor, a missed transpose, or a mismatch in the constants column. Check every original equation rather than stopping after one successful substitution.

How are three-variable systems and consistency handled?

The matrix method extends to three unknowns by using a third-order coefficient matrix and three-entry columns. A system is consistent if at least one solution exists and inconsistent if no solution exists. Consistency does not necessarily mean uniqueness.

Worked example 12. Solve $3x-2y+3z=8$, $2x+y-z=1$, and $4x-3y+2z=4$, where $x,y,z$ are the three unknowns.

1. Form $A=\begin{bmatrix}3&-2&3\\2&1&-1\\4&-3&2\end{bmatrix}$, $X=\begin{bmatrix}x\\y\\z\end{bmatrix}$, and $B=\begin{bmatrix}8\\1\\4\end{bmatrix}$.
2. Expand the determinant: $|A|=3(2-3)+2(4+4)+3(-6-4)=-3+16-30=-17$.
3. Compute the first-row cofactors: $A_{11}=2-3=-1$, $A_{12}=-(4+4)=-8$, $A_{13}=-6-4=-10$.
4. Compute the second-row cofactors: $A_{21}=-(-4+9)=-5$, $A_{22}=6-12=-6$, $A_{23}=-(-9+8)=1$.
5. Compute the third-row cofactors: $A_{31}=2-3=-1$, $A_{32}=-(-3-6)=9$, $A_{33}=3+4=7$.
6. Transpose the cofactor matrix and divide: $A^{-1}=-\frac1{17}\begin{bmatrix}-1&-5&-1\\-8&-6&9\\-10&1&7\end{bmatrix}$.
7. Multiply by the constants: $X=-\frac1{17}\begin{bmatrix}-8-5-4\\-64-6+36\\-80+1+28\end{bmatrix}=-\frac1{17}\begin{bmatrix}-17\\-34\\-51\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}$.
8. Check all three equations: $3-4+9=8$, $2+2-3=1$, and $4-6+6=4$.

Answer: $x=1,\ y=2,\ z=3$. The nonzero determinant establishes uniqueness, while substitution verifies the calculated values.

What does a zero determinant allow you to conclude?

Let $O$ denote the zero column matrix of the required size. If $|A|=0$, calculate $(\operatorname{adj}A)B$. A nonzero result proves inconsistency. A zero result does not settle consistency on its own: the original equations must be examined further.

To understand the nonzero-product test, premultiply a proposed equation by the adjoint. The left side becomes the determinant times the unknown column, which is zero for a singular matrix. A nonzero right side therefore contradicts the existence of any solution.

Glossary

• Determinant — A single number associated with a square matrix and calculated from its entries.
• Order — The number of rows, equal to the number of columns, in a square matrix.
• Expansion — Evaluation by adding products of entries from one row or column with their corresponding cofactors.
• Minor — The determinant remaining after deleting the row and column containing a specified entry.
• Cofactor — A minor multiplied by the sign determined by the original entry's row and column indices.
• Adjoint — The matrix obtained by transposing the complete cofactor matrix of a square matrix.
• Transpose — The matrix formed by interchanging the rows and columns of the original matrix.
• Singular matrix — A square matrix with zero determinant, for which an inverse matrix does not exist.
• Non-singular matrix — A square matrix with a nonzero determinant and therefore an inverse matrix.
• Inverse matrix — A matrix giving the identity when multiplied by the original matrix in either order.
• Coefficient matrix — The matrix of coefficients of unknowns arranged in a fixed variable and equation order.
• Consistent system — A system of equations that has at least one solution satisfying every equation simultaneously.
• Inconsistent system — A system of equations for which no values satisfy all the equations simultaneously.
• Collinear points — Points lying on the same straight line, giving zero area for their associated triangle.

Common errors and misconceptions

• Misconception: Every rectangular matrix has a determinant. Correct: A determinant is defined only for a square matrix, with equal numbers of rows and columns.
• Misconception: Vertical bars around a matrix require a non-negative answer. Correct: They denote its determinant, which can be negative. Absolute value is required separately in the triangle-area formula.
• Misconception: A cofactor is just the remaining minor. Correct: Multiply the minor by $(-1)^{i+j}$, using the original position indices.
• Misconception: The cofactor matrix is already the adjoint. Correct: Transpose the cofactor matrix to obtain the adjoint; off-diagonal positions must be exchanged.
• Misconception: Scaling every entry multiplies the determinant by the same scalar. Correct: The multiplier is raised to the matrix order: $|kA|=k^n|A|$.
• Misconception: A specified positive triangle area permits only a positive determinant. Correct: The determinant can have either sign. Use the absolute value and retain both possible signs when solving for a coordinate.
• Misconception: A zero coefficient determinant proves that a system has no solutions. Correct: It rules out an inverse and a unique solution; the system may have infinitely many solutions or none.
• Misconception: Multiply $AX=B$ by the inverse on whichever side is convenient. Correct: Premultiply to obtain $X=A^{-1}B$; matrix multiplication has a specific order.

Exam-style questions with model answers

Q1. Evaluate $\begin{vmatrix}2&4\\-1&2\end{vmatrix}$, explaining the subtraction of the negative product. [2 marks]

1. The main-diagonal product is $2\times2=4$, and the other product is $4\times(-1)=-4$.
2. Subtract the second product from the first: $4-(-4)=8$. Subtracting a negative number explains why the products contribute positively to the final value.

Q2. For $\begin{vmatrix}1&-2\\4&3\end{vmatrix}$, find all four minors and all four cofactors. [3 marks]

1. A minor is obtained by deleting the entry's row and column. The four remaining entries give $M_{11}=3,\ M_{12}=4,\ M_{21}=-2,\ M_{22}=1$.
2. Apply the cofactor rule $A_{ij}=(-1)^{i+j}M_{ij}$. The diagonal positions have positive multipliers; the off-diagonal positions have negative multipliers.
3. Therefore $A_{11}=3,\ A_{12}=-4,\ A_{21}=2,\ A_{22}=1$. In particular, the negative minor at the second-row, first-column position becomes a positive cofactor because its index sum is odd.

Q3. Find the area of the triangle with vertices $(3,8)$, $(-4,2)$, and $(5,1)$, using determinants. [3 marks]

1. Let $D$ be the coordinate determinant. Put each vertex in one row and append a final entry of one: $D=\begin{vmatrix}3&8&1\\-4&2&1\\5&1&1\end{vmatrix}$.
2. Expanding along the first row gives $D=3(2-1)-8(-4-5)+(-4-10)=3+72-14=61$.
3. The area is half the absolute value: $S=\frac12\operatorname{abs}(61)=\frac{61}{2}$ square units. Absolute value ensures a non-negative area; it is applied after the signed determinant has been evaluated.

Q4. If $A$ is a non-singular square matrix and $I$ is the identity matrix of the same order, derive the inverse formula from $A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|I$. [3 marks]

1. Non-singularity means $|A|\ne0$. Division by the scalar determinant is therefore allowed, which is the essential condition in this derivation.
2. Dividing both product identities gives $A\left(\frac{\operatorname{adj}A}{|A|}\right)=I$ and $\left(\frac{\operatorname{adj}A}{|A|}\right)A=I$.
3. The same matrix gives the identity in both multiplication orders, so it is the inverse by definition. Thus $A^{-1}=\frac1{|A|}\operatorname{adj}A$. The adjoint must be formed by transposing the cofactor matrix before this scalar division is performed.

Q5. Solve $2x+5y=1$ and $3x+2y=7$ by the inverse matrix method, and verify your solution. [5 marks]

1. Write the equations as $AX=B$, where $A=\begin{bmatrix}2&5\\3&2\end{bmatrix}$ is the coefficient matrix, $X=\begin{bmatrix}x\\y\end{bmatrix}$ the unknown column, and $B=\begin{bmatrix}1\\7\end{bmatrix}$ the constants column.
2. Evaluate $|A|=4-15=-11\ne0$. This establishes that the inverse exists and the system has a unique solution.
3. The cofactors are $2,-3,-5,2$ in row order. Transposing their matrix gives $\operatorname{adj}A=\begin{bmatrix}2&-5\\-3&2\end{bmatrix}$.
4. Hence $A^{-1}=-\frac1{11}\begin{bmatrix}2&-5\\-3&2\end{bmatrix}$. Premultiplying the matrix equation by this inverse gives $X=A^{-1}B$.
5. Carry out the multiplication: $X=-\frac1{11}\begin{bmatrix}2-35\\-3+14\end{bmatrix}=\begin{bmatrix}3\\-1\end{bmatrix}$. Therefore $x=3$ and $y=-1$.
6. Substitution checks both original equations: $2(3)+5(-1)=1$ and $3(3)+2(-1)=7$. The calculated values satisfy the complete system. Both checks are necessary because a solution must satisfy the equations simultaneously.

Q6. Find the line through $(1,3)$ and $(0,0)$ using determinants. Find both possible values of $k$ if $(k,0)$ forms a triangle of area $3$ square units with these points. [5 marks]

1. Let $(x,y)$ be a general point on the required line. The three points are collinear, so their coordinate determinant is zero: $\begin{vmatrix}1&3&1\\0&0&1\\x&y&1\end{vmatrix}=0$.
2. Expanding this determinant gives $-y+3x=0$, so the equation of the line is $y=3x$.
3. Now use the third vertex $(k,0)$ for the area calculation. Its coordinate determinant is $\begin{vmatrix}1&3&1\\0&0&1\\k&0&1\end{vmatrix}=3k$.
4. Area is half the absolute value, giving $\frac12|3k|=3$. Thus $|k|=2$, and both signs must be retained: $k=2$ or $k=-2$.
5. For these values the determinants are respectively $6$ and $-6$. Each has absolute value $6$, so each gives the required area of $3$ square units. Keeping the negative determinant is essential because it represents another permitted position of the third vertex.

Key takeaways

• A determinant is a number associated with a square matrix; distinguish it from the matrix and from scalar absolute value.
• Expand along any complete row or column, preferably one containing zeros, while retaining the original position signs.
• A minor comes from deleting a row and column; its cofactor also includes the position-dependent sign.
• The adjoint is the transpose of the cofactor matrix, so calculating cofactors alone does not finish its construction.
• A square matrix is invertible exactly when its determinant is nonzero; check this before dividing by the determinant.
• Triangle area is half the absolute coordinate determinant, and a specified area may allow two possible coordinate values.
• Use the inverse matrix to solve a system with a non-singular coefficient matrix, then verify every original equation.
• A singular coefficient matrix requires further consistency checks because its determinant alone cannot distinguish no solutions from infinitely many.

Test yourself

Why does a non-square matrix have no determinant here?

The determinant is defined for square matrices, whose numbers of rows and columns are equal.

Which row or column is usually easiest to expand?

Choose one with the most zero entries, because its zero terms contribute nothing to the expansion.

What sign multiplier belongs to a cofactor in row two, column three?

The index sum is odd: $(-1)^{2+3}=-1$. Multiply the minor by this negative sign.

What must follow the calculation of all cofactors when finding an adjoint?

Arrange the cofactors in their original positions, then transpose that matrix by interchanging rows and columns.

If a triangle's coordinate determinant is negative, what happens to its area?

Take the determinant's absolute value before halving it; geometrical area is non-negative.

What does a nonzero coefficient determinant establish for a linear system?

The coefficient matrix has an inverse, and the system has a unique solution given by $X=A^{-1}B$.

Does a zero determinant together with a zero adjoint-constants product prove consistency?

No. The original equations still need examination; they may have infinitely many solutions or no solution.

For a third-order matrix, how does doubling every entry affect the determinant?

The determinant is multiplied by the cube of two: $|2A|=2^3|A|=8|A|$.