Matrices | CBSE Class 12 Maths Notes

This Mathematics note covers matrix notation and order, types and equality of matrices, construction from entry rules, addition, subtraction, scalar multiplication, matrix products, transpose, symmetric and skew symmetric matrices, decomposition, invertible matrices, and proofs and worked calculations using matrix algebra.

What is a matrix, and how do its order and entries work?

Definition: A matrix is an ordered rectangular array of numbers or functions. Its individual numbers or functions are called elements or entries.

Horizontal arrangements of entries form rows; vertical arrangements form columns. The word ordered matters: the location of each entry is part of the information. Moving a number to another position generally changes the matrix.

How is an entry located?

A matrix with $m$ rows and $n$ columns has order $m\times n$. Rows come first when writing the order. Such a matrix contains $mn$ entries. Capital letters name matrices, while indexed small letters identify their entries.

The notation $A=[a_{ij}]_{m\times n}$ means that $a_{ij}$ occupies row $i$, column $j$, with $1\leq i\leq m$ and $1\leq j\leq n$. The row index must therefore be checked before the column index.

The entries in row $i$ are $a_{i1},a_{i2},\ldots,a_{in}$. The entries in column $j$ are $a_{1j},a_{2j},\ldots,a_{mj}$. Throughout these operations, entries are real numbers or functions taking real values.

How can a matrix represent information?

Consider the numbers of men and women working in three factories. Taking factories as rows and the two groups as columns gives a compact arrangement in which every entry has a fixed meaning.

The corresponding matrix is $$A=\begin{bmatrix}30&25\\25&31\\27&26\end{bmatrix}.$$ Its order is $3\times2$, and $a_{32}=26$ represents the women workers in factory III. Reading the entry without retaining the row and column meanings would lose its context.

Matrices can also represent coordinates of points and vertices of geometrical figures. The same information may be arranged with points as rows or as columns, provided the chosen arrangement is used consistently.

How are the different types of matrices distinguished?

Matrix types depend on the number of rows and columns, or on the positions and values of entries. Begin with the shape, then examine the diagonal and the remaining entries. Several descriptions can apply to the same matrix.

Which types depend on shape?

For a square matrix, the entries $a_{11},a_{22},\ldots,a_{nn}$ form the diagonal. Their row and column indices agree. Entries for which $i\ne j$ lie outside this diagonal.

Which types depend on the entries?

For instance, $\begin{bmatrix}-1&0\\0&2\end{bmatrix}$ is diagonal, while $\begin{bmatrix}-1&0\\0&-1\end{bmatrix}$ is scalar. The identity matrix $I_2=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ is both scalar and diagonal.

Every identity matrix is a scalar matrix. A scalar matrix becomes the identity matrix when its common diagonal value is one. Thus the descriptions form increasingly specific conditions, rather than mutually exclusive categories.

Note: A zero matrix is identified by all its entries, not by its diagonal alone. Its order must match the operation in which it is used.

How do you construct a matrix and use equality of matrices?

An entry rule gives the value at each permitted pair of indices. First use the order to determine the positions, then substitute each row and column index into the rule. Do not interchange the roles of the two indices.

Worked example 1. Construct a matrix of order $3\times2$ using $a_{ij}=\frac12|i-3j|$.

Answer: Calculate all six entries in their specified positions.

1. The first entry is $a_{11}=\frac12|1-3\cdot1|=\frac12|-2|=1$.
2. The next entry is $a_{12}=\frac12|1-3\cdot2|=\frac12|-5|=\frac52$.
3. The second row begins with $a_{21}=\frac12|2-3\cdot1|=\frac12|-1|=\frac12$.
4. Complete that row using $a_{22}=\frac12|2-3\cdot2|=\frac12|-4|=2$.
5. The third row begins with $a_{31}=\frac12|3-3\cdot1|=0$.
6. The final entry is $a_{32}=\frac12|3-3\cdot2|=\frac12|-3|=\frac32$.
7. Place the values in order: $$A=\begin{bmatrix}1&\frac52\\\frac12&2\\0&\frac32\end{bmatrix}.$$

When are two matrices equal?

Equality requires the same order and equality of every pair of corresponding entries. Having the same collection of numbers is insufficient if those numbers occupy different positions. Once orders agree, matrix equality can generate ordinary equations for unknown entries.

Worked example 2. Find the unknowns in $$\begin{bmatrix}2a+b&a-2b\\5c-d&4c+3d\end{bmatrix}=\begin{bmatrix}4&-3\\11&24\end{bmatrix}.$$

Answer: Match corresponding positions and solve the resulting pairs of equations.

1. The first row gives $2a+b=4$ and $a-2b=-3$.
2. Rearrange the first equation: $b=4-2a$.
3. Substitute into the second: $a-2(4-2a)=-3$, so $5a-8=-3$, $5a=5$, and $a=1$.
4. Recover the other entry: $b=4-2(1)=2$.
5. The second row gives $5c-d=11$ and $4c+3d=24$, hence $d=5c-11$.
6. Substitute: $4c+3(5c-11)=24$, so $19c-33=24$, $19c=57$, and $c=3$.
7. Then $d=5(3)-11=4$.
8. Check all entries: $2(1)+2=4$, $1-2(2)=-3$, $5(3)-4=11$, and $4(3)+3(4)=24$.

A proposed solution must satisfy every corresponding entry. Solving one or two convenient positions does not establish equality if another position contradicts the proposed values.

How are matrices added, subtracted and multiplied by scalars?

Addition and subtraction require matrices of the same order. The operation is performed on corresponding entries, and the result retains that order. Neither operation changes the locations used to interpret the entries.

For $A=[a_{ij}]$ and $B=[b_{ij}]$ of the same order, $$A+B=[a_{ij}+b_{ij}],\qquad A-B=[a_{ij}-b_{ij}].$$ If their orders differ, these sums and differences are not defined.

What does scalar multiplication change?

A scalar is a number multiplying the matrix. For a scalar $k$, the rule is $kA=[ka_{ij}]$. Every entry is multiplied, including negative entries and zeros. The number of rows and columns remains unchanged.

The negative of a matrix is $-A=(-1)A$, obtained by changing the sign of every entry. Subtraction can therefore be expressed as addition of a negative: $A-B=A+(-1)B$.

Worked example 3. Find $2A-B$ for $$A=\begin{bmatrix}1&2&3\\2&3&1\end{bmatrix},\qquad B=\begin{bmatrix}3&-1&3\\-1&0&2\end{bmatrix}.$$

Answer: Both matrices have order $2\times3$, so the subtraction is defined after scalar multiplication.

1. Multiply every entry of the first matrix: $$2A=\begin{bmatrix}2\cdot1&2\cdot2&2\cdot3\\2\cdot2&2\cdot3&2\cdot1\end{bmatrix}=\begin{bmatrix}2&4&6\\4&6&2\end{bmatrix}.$$
2. Subtract the corresponding entries, retaining brackets around negative numbers: $$2A-B=\begin{bmatrix}2-3&4-(-1)&6-3\\4-(-1)&6-0&2-2\end{bmatrix}.$$
3. Simplify each position: $$2A-B=\begin{bmatrix}-1&5&3\\5&6&0\end{bmatrix}.$$

The two entries involving subtraction of a negative become additions. Writing the unsimplified matrix first makes the sign changes visible and keeps each calculation attached to its correct position.

Note: Scalar multiplication uses one number and affects every entry separately. Multiplication of two matrices uses a different rule involving rows and columns.

Which addition laws help solve matrix equations?

Property: Addition and scalar multiplication laws

For matrices of the same order, addition satisfies the commutative law, $A+B=B+A$, and the associative law, $(A+B)+C=A+(B+C)$. These follow from the corresponding properties of the numbers in each position.

The additive identity is the zero matrix of the required order: $A+O=O+A=A$. The additive inverse of $A$ is $-A$, since $A+(-A)=(-A)+A=O$.

For scalars $k$ and $l$, scalar multiplication distributes over a matrix sum and a scalar sum: $$k(A+B)=kA+kB,\qquad (k+l)A=kA+lA.$$ The first identity requires equal matrix orders. Both identities can be checked by comparing corresponding entries.

How can two unknown matrices be separated?

When the sum and difference of two unknown matrices are given, adding the equations eliminates one unknown; subtracting them eliminates the other. The cancellation uses additive inverses, and multiplication by one half is scalar multiplication.

Worked example 4. Find $X$ and $Y$ if $$X+Y=\begin{bmatrix}5&2\\0&9\end{bmatrix},\qquad X-Y=\begin{bmatrix}3&6\\0&-1\end{bmatrix}.$$

Answer: Use addition and subtraction of the two matrix equations.

1. Add the equations: $$2X=\begin{bmatrix}5+3&2+6\\0+0&9+(-1)\end{bmatrix}=\begin{bmatrix}8&8\\0&8\end{bmatrix}.$$
2. Multiply by one half: $$X=\begin{bmatrix}8/2&8/2\\0/2&8/2\end{bmatrix}=\begin{bmatrix}4&4\\0&4\end{bmatrix}.$$
3. Subtract the second equation from the first: $$2Y=\begin{bmatrix}5-3&2-6\\0-0&9-(-1)\end{bmatrix}=\begin{bmatrix}2&-4\\0&10\end{bmatrix}.$$
4. Multiply by one half: $$Y=\begin{bmatrix}2/2&-4/2\\0/2&10/2\end{bmatrix}=\begin{bmatrix}1&-2\\0&5\end{bmatrix}.$$
5. Check the sum: $$X+Y=\begin{bmatrix}4+1&4+(-2)\\0+0&4+5\end{bmatrix}=\begin{bmatrix}5&2\\0&9\end{bmatrix}.$$
6. Check the difference: $$X-Y=\begin{bmatrix}4-1&4-(-2)\\0-0&4-5\end{bmatrix}=\begin{bmatrix}3&6\\0&-1\end{bmatrix}.$$

The checks use both original equations. A correct sum alone would not verify the difference, because the signs of the entries in the second unknown matrix also matter.

When is a matrix product defined, and how is it calculated?

The product $AB$ is defined when the number of columns in $A$ equals the number of rows in $B$. This is the compatibility condition. If their orders are $m\times n$ and $n\times p$, the product has order $m\times p$.

What is the row-by-column rule?

To find an entry of the product, take the appropriate row of the first matrix and column of the second. Multiply the paired entries and add all the products. The matching inner dimensions ensure that the row and column contain equally many entries.

In symbols, let $C=AB$. Here $c_{ik}$ is the entry in row $i$, column $k$ of $C$; $a_{ij}$ and $b_{jk}$ are entries of $A$ and $B$, respectively. The index $j$ runs from $1$ to $n$, pairing the $n$ columns of $A$ with the $n$ rows of $B$; $k$ is a column index here. The entry rule is $$c_{ik}=a_{i1}b_{1k}+a_{i2}b_{2k}+\cdots+a_{in}b_{nk}=\sum_{j=1}^{n}a_{ij}b_{jk}.$$ The position in the answer is determined by the selected row and selected column.

Worked example 5. Calculate $AB$ when $$A=\begin{bmatrix}6&9\\2&3\end{bmatrix},\qquad B=\begin{bmatrix}2&6&0\\7&9&8\end{bmatrix}.$$

Answer: The inner dimensions agree, and the product has order $2\times3$.

1. First row, first column: $c_{11}=6\cdot2+9\cdot7=12+63=75$.
2. First row, second column: $c_{12}=6\cdot6+9\cdot9=36+81=117$.
3. First row, third column: $c_{13}=6\cdot0+9\cdot8=0+72=72$.
4. Second row, first column: $c_{21}=2\cdot2+3\cdot7=4+21=25$.
5. Second row, second column: $c_{22}=2\cdot6+3\cdot9=12+27=39$.
6. Second row, third column: $c_{23}=2\cdot0+3\cdot8=0+24=24$.
7. Assemble the product: $$AB=\begin{bmatrix}75&117&72\\25&39&24\end{bmatrix}.$$

For these matrices, $BA$ is undefined: the second matrix has three columns, but the first has only two rows. Therefore checking $AB$ does not automatically establish that multiplication in the reverse order is possible.

Property: Associative and distributive laws for products

Whenever the expressions are defined, $(AB)C=A(BC)$, $A(B+C)=AB+AC$, and $(A+B)C=AC+BC$. Associativity changes the grouping while preserving the order of the factors.

For a square matrix, the multiplicative identity is the identity matrix of the same order, giving $IA=AI=A$. In an application, a product can combine quantities with corresponding prices to calculate total amounts.

Why must familiar number rules be used carefully with matrices?

Two central differences concern the order of multiplication and a product equal to zero. Matrix multiplication is not commutative in general. Even when both products exist and have the same order, their corresponding entries may differ.

Worked example 6. Compare the products for $$A=\begin{bmatrix}1&0\\0&-1\end{bmatrix},\qquad B=\begin{bmatrix}0&1\\1&0\end{bmatrix}.$$

Answer: Both products have order $2\times2$, but their entries differ.

1. Use rows of the first matrix and columns of the second: $$AB=\begin{bmatrix}1\cdot0+0\cdot1&1\cdot1+0\cdot0\\0\cdot0+(-1)\cdot1&0\cdot1+(-1)\cdot0\end{bmatrix}.$$
2. Simplify: $$AB=\begin{bmatrix}0&1\\-1&0\end{bmatrix}.$$
3. Reverse the row and column sources: $$BA=\begin{bmatrix}0\cdot1+1\cdot0&0\cdot0+1\cdot(-1)\\1\cdot1+0\cdot0&1\cdot0+0\cdot(-1)\end{bmatrix}.$$
4. Simplify and compare: $$BA=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\ne AB.$$

This does not mean that the products differ for every pair. Diagonal matrices of the same order commute. The correct conclusion is that changing the order is not a generally valid algebraic operation.

Can two nonzero matrices have a zero product?

Worked example 7. Calculate the product of $$A=\begin{bmatrix}0&-1\\0&2\end{bmatrix},\qquad B=\begin{bmatrix}3&5\\0&0\end{bmatrix}.$$

Answer: Neither factor is the zero matrix, but each entry of their product is zero.

1. The first product entry is $c_{11}=0\cdot3+(-1)\cdot0=0$.
2. The next entry is $c_{12}=0\cdot5+(-1)\cdot0=0$.
3. The second row begins with $c_{21}=0\cdot3+2\cdot0=0$.
4. The last entry is $c_{22}=0\cdot5+2\cdot0=0$.
5. Thus $$AB=\begin{bmatrix}0&0\\0&0\end{bmatrix}=O.$$

A zero product therefore does not imply that either factor is zero. This is a direct counterexample to extending the corresponding rule for real numbers to arbitrary matrices.

What is the transpose of a matrix, and which laws does it obey?

The transpose of a matrix is obtained by interchanging its rows and columns. It is denoted by $A'$ or $A^T$. A matrix of order $m\times n$ has a transpose of order $n\times m$.

Every entry keeps its value but exchanges its two indices. Thus an entry in row $i$, column $j$ of the original appears in row $j$, column $i$ of the transpose. Transposing does not change signs.

Worked example 8. Find the transpose of $$A=\begin{bmatrix}3&\sqrt3&2\\4&2&0\end{bmatrix}$$ and verify that transposing twice restores the original matrix.

Answer: Make each original row a column, retaining the order within that row.

1. The first row becomes the first column, containing $3,\sqrt3,2$.
2. The second row becomes the second column, containing $4,2,0$. Hence $$A'=\begin{bmatrix}3&4\\\sqrt3&2\\2&0\end{bmatrix}.$$
3. Transpose again by making these rows into columns: $$(A')'=\begin{bmatrix}3&\sqrt3&2\\4&2&0\end{bmatrix}=A.$$

Property: Laws of transpose

These identities apply to matrices of suitable orders. The product identity requires particular attention because reversing the order of the factors is part of the rule. Writing transposes in their original order generally gives the wrong expression.

The sum identity differs from the product identity: each addend is transposed without requiring a reversal. The scalar rule also gives the corresponding subtraction rule, $(A-B)'=A'-B'$.

How are symmetric and skew symmetric matrices recognised?

A square matrix is symmetric when $A'=A$. Equivalently, $a_{ij}=a_{ji}$: entries at positions reflected across the diagonal are equal. Its diagonal entries do not change position when transposed.

A square matrix is skew symmetric when $A'=-A$. In entry form, $a_{ji}=-a_{ij}$. Corresponding entries across the diagonal are negatives of one another. Both definitions require a square matrix.

Why must a skew symmetric matrix have a zero diagonal?

1. Apply the entry condition at a diagonal position, where the indices agree: $a_{ii}=-a_{ii}$.
2. Add the diagonal entry to both sides: $2a_{ii}=0$.
3. Divide by two: $a_{ii}=0$. This holds at every diagonal position.

A zero diagonal alone is insufficient to establish skew symmetry. The entries away from the diagonal must also occur as opposite pairs. Every required position must satisfy the defining condition.

Theorem: Sums and differences with a transpose

For any real square matrix, $A+A'$ is symmetric and $A-A'$ is skew symmetric. The proof uses transpose identities and commutativity of matrix addition, rather than assuming anything about the original matrix.

1. Set $B=A+A'$. Transpose the sum: $B'=(A+A')'=A'+(A')'$.
2. Use repeated transpose: $B'=A'+A$.
3. Commute the sum: $B'=A+A'=B$. This proves symmetry.
4. Set $C=A-A'$. Transpose the difference: $C'=(A-A')'=A'-(A')'$.
5. Simplify: $C'=A'-A$.
6. Factor out the negative sign: $C'=-(A-A')=-C$. This proves skew symmetry.

Result: These constructions work for every real square matrix, whether or not that matrix is already symmetric or skew symmetric.

Checking the transpose is a direct test of the definitions. It also connects these classifications to matrix operations, because transpose laws allow expressions to be checked without inspecting every entry individually.

How can a square matrix be split into symmetric and skew symmetric parts?

Theorem: Decomposition of a square matrix

Every square matrix can be expressed as a sum of a symmetric matrix and a skew symmetric matrix. The construction uses the sum and difference of the original matrix and its transpose, followed by scalar multiplication.

1. Define the proposed parts: $P=\frac12(A+A')$ and $Q=\frac12(A-A')$.
2. Transpose the first: $P'=\frac12(A'+A)=\frac12(A+A')=P$, so it is symmetric.
3. Transpose the second: $Q'=\frac12(A'-A)=-\frac12(A-A')=-Q$, so it is skew symmetric.
4. Add the parts: $P+Q=\frac12(A+A'+A-A')=\frac12(2A)=A$.

Result: The required decomposition is $$A=\frac12(A+A')+\frac12(A-A').$$ The factors of one half are essential because adding the unscaled sum and difference produces twice the original matrix.

Worked example 9. Express $A=\begin{bmatrix}3&5\\1&-1\end{bmatrix}$ as a sum of symmetric and skew symmetric matrices.

Answer: Find the transpose, then calculate the half-sum and half-difference.

1. Interchange rows and columns: $$A'=\begin{bmatrix}3&1\\5&-1\end{bmatrix}.$$
2. Add corresponding entries: $$A+A'=\begin{bmatrix}3+3&5+1\\1+5&-1+(-1)\end{bmatrix}=\begin{bmatrix}6&6\\6&-2\end{bmatrix}.$$
3. Halve the entries: $$P=\frac12(A+A')=\begin{bmatrix}3&3\\3&-1\end{bmatrix}.$$
4. Subtract corresponding entries: $$A-A'=\begin{bmatrix}3-3&5-1\\1-5&-1-(-1)\end{bmatrix}=\begin{bmatrix}0&4\\-4&0\end{bmatrix}.$$
5. Halve these entries: $$Q=\frac12(A-A')=\begin{bmatrix}0&2\\-2&0\end{bmatrix}.$$
6. Check the definitions: $$P'=\begin{bmatrix}3&3\\3&-1\end{bmatrix}=P,\qquad Q'=\begin{bmatrix}0&-2\\2&0\end{bmatrix}=-Q.$$
7. Check the sum: $$P+Q=\begin{bmatrix}3+0&3+2\\3+(-2)&-1+0\end{bmatrix}=\begin{bmatrix}3&5\\1&-1\end{bmatrix}=A.$$

The calculation has two separate checks: the parts have the required transpose properties, and their sum reproduces the original matrix. These checks address both the classification and the reconstruction.

What does it mean for a matrix to be invertible?

Definition: A square matrix $A$ is invertible if a square matrix $B$ of the same order satisfies $AB=BA=I$. The matrix $B$ is its inverse, denoted by $A^{-1}$.

The inverse undoes multiplication in the sense expressed by these two identity products. If $B$ is the inverse of $A$, then $A$ is also the inverse of $B$. Being square is a requirement; it does not by itself establish that an inverse exists.

Worked example 10. Verify that $B$ is the inverse of $A$, where $$A=\begin{bmatrix}2&3\\1&2\end{bmatrix},\qquad B=\begin{bmatrix}2&-3\\-1&2\end{bmatrix}.$$

Answer: Calculate the two products and compare each with the identity matrix of order two.

1. Expand the first product: $$AB=\begin{bmatrix}2\cdot2+3(-1)&2(-3)+3\cdot2\\1\cdot2+2(-1)&1(-3)+2\cdot2\end{bmatrix}.$$
2. Simplify: $$AB=\begin{bmatrix}4-3&-6+6\\2-2&-3+4\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I_2.$$
3. Expand the reverse product: $$BA=\begin{bmatrix}2\cdot2+(-3)\cdot1&2\cdot3+(-3)\cdot2\\(-1)\cdot2+2\cdot1&(-1)\cdot3+2\cdot2\end{bmatrix}.$$
4. Simplify: $$BA=\begin{bmatrix}4-3&6-6\\-2+2&-3+4\end{bmatrix}=I_2.$$
5. Both conditions hold, so $A^{-1}=B=\begin{bmatrix}2&-3\\-1&2\end{bmatrix}$.

Theorem: Uniqueness of the inverse

If a square matrix has an inverse, that inverse is unique. Suppose that both $B$ and $C$ are inverses of $A$. Then $AB=BA=I$ and $AC=CA=I$.

1. Use the identity matrix: $B=BI$.
2. Replace the identity using the second inverse: $BI=B(AC)$.
3. Regroup by associativity: $B(AC)=(BA)C$.
4. Use the first inverse: $(BA)C=IC$.
5. Use the identity again: $IC=C$. Therefore $B=C$.

The proof changes the grouping of a product, without swapping its factors. This is why associativity is sufficient even though matrix multiplication is generally not commutative.

Theorem: Inverse of a product

For invertible matrices of the same order, $(AB)^{-1}=B^{-1}A^{-1}$. To establish existence as well as the formula, test this proposed inverse on both sides.

1. Define $D=B^{-1}A^{-1}$.
2. Multiply on one side and regroup: $(AB)D=A(BB^{-1})A^{-1}=AIA^{-1}=AA^{-1}=I$.
3. Multiply on the other side: $D(AB)=B^{-1}(A^{-1}A)B=B^{-1}IB=B^{-1}B=I$.
4. Thus $D$ is the inverse, and uniqueness gives $(AB)^{-1}=B^{-1}A^{-1}$.

The reversed factor order is essential. It places each matrix next to its own inverse, allowing the identity products used in the proof.

Glossary

• Matrix — An ordered rectangular array of numbers or functions, arranged in rows and columns.
• Entry — An individual number or function occupying a specified row and column of a matrix.
• Order — The size of a matrix, written as its number of rows followed by its number of columns.
• Row matrix — A matrix with exactly one row and any permitted number of columns.
• Column matrix — A matrix with exactly one column and any permitted number of rows.
• Square matrix — A matrix in which the number of rows equals the number of columns.
• Diagonal matrix — A square matrix whose entries outside the main diagonal are all zero.
• Scalar matrix — A diagonal matrix in which all entries on the diagonal have the same value.
• Identity matrix — A square matrix with diagonal entries equal to one and all other entries zero.
• Zero matrix — A matrix with every entry equal to zero, also called a null matrix.
• Additive inverse — The negative of a matrix, which produces the zero matrix when added to the original.
• Transpose — The matrix obtained by interchanging the rows and columns of a given matrix.
• Symmetric matrix — A square matrix that equals its transpose, with equal entries in corresponding positions across the diagonal.
• Skew symmetric matrix — A square matrix whose transpose equals its negative, with every diagonal entry equal to zero.
• Inverse matrix — A matrix that gives the identity when multiplied with the original square matrix in either order.

Common errors and misconceptions

• Misconception: Matrices with the same number of entries can always be added. Correct: Their orders must agree, so the numbers of rows and columns must match separately.
• Misconception: Matrix multiplication means multiplying corresponding entries. Correct: Each product entry is the sum of paired products from a row of the first matrix and a column of the second.
• Misconception: If $AB$ exists, $BA$ also exists and equals it. Correct: The reverse product needs its own dimension check, and matrix multiplication is not generally commutative.
• Misconception: If $AB=O$, one factor must be zero. Correct: Two nonzero matrices can have a zero product, as direct row-by-column calculation demonstrates.
• Misconception: Transposing a product keeps the factors in the original order. Correct: The rule is $(AB)'=B'A'$, with the factor order reversed.
• Misconception: A zero diagonal proves that a matrix is skew symmetric. Correct: The matrix must also be square and satisfy $a_{ji}=-a_{ij}$ in every position.
• Misconception: Every square matrix automatically has an inverse. Correct: An inverse exists only when a matrix of the same order satisfies the required identity products, $AB=BA=I$.

Exam-style questions with model answers

Q1. State the two conditions for equality of matrices. [2 marks]

1. The matrices must have the same order, with matching numbers of rows and columns.
2. Every pair of corresponding entries must be equal: $a_{ij}=b_{ij}$ for all permitted row and column indices.

Q2. If a matrix has eight entries, find all its possible orders. [2 marks]

1. For order $m\times n$, the number of entries is $mn=8$, with both dimensions natural numbers.
2. The factor pairs give $1\cdot8=8$, $2\cdot4=8$, $4\cdot2=8$, and $8\cdot1=8$.
3. Hence the possible orders are $1\times8$, $2\times4$, $4\times2$, and $8\times1$.

Q3. Explain the condition for matrix multiplication and the rule for an entry of the product. [3 marks]

1. The number of columns in the first matrix must equal the number of rows in the second. For orders $m\times n$ and $n\times p$, the product has order $m\times p$.
2. Take row $i$ from the first matrix and column $k$ from the second matrix. Multiply their corresponding entries and add the resulting products.
3. The entry rule is $c_{ik}=\sum_{j=1}^{n}a_{ij}b_{jk}$. Checking the reverse product requires a separate comparison of its inner dimensions.

Q4. Prove that $A+A'$ is symmetric and $A-A'$ is skew symmetric for a real square matrix. [4 marks]

1. Let $B=A+A'$. Because the matrix is square, its transpose has the same order and this sum is defined.
2. Transpose the sum: $B'=A'+(A')'=A'+A$. Commutativity of addition gives $B'=A+A'=B$, proving that the sum is symmetric.
3. Let $C=A-A'$. The difference is also a square matrix of the required order.
4. Transpose and simplify: $C'=A'-(A')'=A'-A=-(A-A')=-C$. This is exactly the defining condition for a skew symmetric matrix.

Q5. Show that the inverse of a square matrix, if it exists, is unique. [3 marks]

1. Suppose that $B$ and $C$ are both inverses of the same square matrix $A$. Then $AB=BA=I$ and $AC=CA=I$.
2. Begin with $B=BI$, using the multiplicative identity.
3. Replace the identity by the product from the second inverse: $BI=B(AC)$.
4. Apply associativity, preserving the factor order: $B(AC)=(BA)C$.
5. Since $BA=I$, this gives $(BA)C=IC=C$. Hence $B=C$, so two distinct inverses cannot exist.

Q6. Express $A=\begin{bmatrix}3&5\\1&-1\end{bmatrix}$ as a sum of symmetric and skew symmetric matrices, and verify the result. [5 marks]

1. Transpose the given matrix by interchanging its rows and columns: $A'=\begin{bmatrix}3&1\\5&-1\end{bmatrix}$. The transpose has the same order, so the required sum and difference are defined.
2. Calculate the sum entry by entry: $A+A'=\begin{bmatrix}3+3&5+1\\1+5&-1+(-1)\end{bmatrix}=\begin{bmatrix}6&6\\6&-2\end{bmatrix}$.
3. Multiply every entry by one half to obtain the proposed symmetric part: $P=\frac12(A+A')=\begin{bmatrix}3&3\\3&-1\end{bmatrix}$.
4. Calculate the difference: $A-A'=\begin{bmatrix}3-3&5-1\\1-5&-1-(-1)\end{bmatrix}=\begin{bmatrix}0&4\\-4&0\end{bmatrix}$.
5. Halve each entry to obtain the proposed skew symmetric part: $Q=\frac12(A-A')=\begin{bmatrix}0&2\\-2&0\end{bmatrix}$.
6. Check both transpose conditions: $P'=\begin{bmatrix}3&3\\3&-1\end{bmatrix}=P$ and $Q'=\begin{bmatrix}0&-2\\2&0\end{bmatrix}=-Q$. Thus both classifications are verified. The first part has equal opposite entries, while the second has opposite signs across its zero diagonal.
7. Finally, add the corresponding entries: $P+Q=\begin{bmatrix}3+0&3+2\\3-2&-1+0\end{bmatrix}=A$, completing the required decomposition.

Q7. If symmetric matrices $A$ and $B$ have the same order, prove that $AB$ is symmetric if and only if $AB=BA$. [4 marks]

1. Since both given matrices are symmetric, $A'=A$ and $B'=B$. Their equal square orders ensure that both products exist.
2. First suppose that $AB$ is symmetric. Its transpose therefore satisfies $(AB)'=AB$.
3. The product transpose law gives $(AB)'=B'A'=BA$. Comparing the two expressions proves $AB=BA$.
4. Conversely, suppose that $AB=BA$. Then $(AB)'=B'A'=BA=AB$, so the product equals its transpose and is symmetric. Both directions have now been established.

Key takeaways

• A matrix is an ordered array; its order lists rows before columns, and entry positions carry essential information.
• Matrix equality requires matching orders and matching corresponding entries; sharing the same collection of numbers is insufficient.
• Addition and subtraction act on corresponding entries of matrices of the same order, while scalar multiplication affects every entry.
• Matrix multiplication requires matching inner dimensions, and each answer entry combines one row with one column.
• Multiplication is associative and distributive where defined, but changing the order of the factors can change the result.
• Transposition exchanges rows and columns; transposing a product also reverses the order of its factors.
• Every real square matrix can be written as a symmetric half-sum and a skew symmetric half-difference with its transpose.
• An inverse satisfies both identity products and is unique when it exists; the inverse of a product reverses factor order.

Test yourself

What does the entry $a_{ij}$ identify?

It identifies the entry in row $i$ and column $j$; the first index always specifies the row.

How many entries does an $m\times n$ matrix contain?

It contains $mn$ entries, because each of its $m$ rows contains $n$ entries.

Is every scalar matrix an identity matrix?

No. A scalar matrix is an identity matrix when its common diagonal entry is one.

When is $AB$ defined for matrices of orders $m\times n$ and $k\times l$?

It is defined when $n=k$, and the resulting product has order $m\times l$.

What can be concluded from $AB=O$ about the factors?

Neither factor is necessarily zero; two nonzero matrices can produce the zero matrix on multiplication.

What is the transpose rule for a product?

The rule is $(AB)'=B'A'$: transpose both factors and reverse their order in the product.

Why are diagonal entries of a real skew symmetric matrix zero?

The numbered argument is: (1) $a_{ii}=-a_{ii}$; (2) $2a_{ii}=0$; (3) $a_{ii}=0$. Each diagonal position obeys the same condition.

What is the inverse of a product of invertible matrices of the same order?

It is $(AB)^{-1}=B^{-1}A^{-1}$, so the individual inverses occur in the reverse factor order.