By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
A matrix is a rectangular array of numbers arranged in rows and columns; e.g., a 2×3 matrix has 2 rows and 3 columns: [ \begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ \end{bmatrix} ]
Two matrices are equal only if they have the same order and corresponding elements are equal; e.g., [ \begin{bmatrix} a & b \ c & d \end{bmatrix} = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix} \Rightarrow a=1, b=2, c=3, d=4 ]
Matrix addition is commutative: ( A + B = B + A ); e.g., if ( A = \begin{bmatrix} 1 & 2 \end{bmatrix}, B = \begin{bmatrix} 3 & 4 \end{bmatrix} ), then ( A+B = B+A = \begin{bmatrix} 4 & 6 \end{bmatrix} )
Matrix multiplication is associative: ( (AB)C = A(BC) ), but not commutative; e.g., for ( A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}, B = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} ), ( AB \ne BA )
The identity matrix ( I_n ) of order ( n ) satisfies ( AI = IA = A ); e.g., ( I_2 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} )
A square matrix ( A ) is invertible if there exists a matrix ( B ) such that ( AB = BA = I ); the inverse is denoted ( A^{-1} )
The determinant of a 2×2 matrix ( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} ) is ( |A| = ad - bc ); e.g., ( A = \begin{bmatrix} 2 & 3 \ 1 & 4 \end{bmatrix} \Rightarrow |A| = (2)(4) - (3)(1) = 8 - 3 = 5 )
A matrix is singular if its determinant is zero; non-singular if determinant-0; only non-singular matrices are invertible
The inverse of a 2×2 matrix ( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} ) is ( A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} ), provided ( ad-bc \ne 0 )
For a 3×3 matrix, the determinant can be expanded using cofactors along any row or column; e.g., expand along R1: [ \begin{vmatrix} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{vmatrix} = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} ]
The adjoint of a square matrix is the transpose of the cofactor matrix; ( A^{-1} = \frac{1}{|A|} \text{adj}(A) ), if ( |A| \ne 0 )
If ( A ) is a 3×3 matrix and ( |A| = 4 ), then ( |\text{adj}(A)| = |A|^{n-1} = 4^{2} = 16 ) (where ( n=3 ))
For square matrices of order ( n ), ( |kA| = k^n |A| ); e.g., if ( |A| = 3 ) and ( n=3 ), then ( |2A| = 2^3 \times 3 = 24 )
If ( A ) is invertible, then ( (A^{-1})^{-1} = = A )
If ( A ) and ( B ) are invertible matrices of same order, then ( (AB)^{-1} = B^{-1}A^{-1} )
Cramer’s Rule: For system ( a_1x + b_1y = c_1 ), ( a_2x + b_2y = c_2 ), ( x = \frac{D_1}{D}, y = \frac{D_2}{D} ), where ( D = \begin{vmatrix} a_1 & b_1 \ a_2 & b_2 \end{vmatrix}, D_1 = \begin{vmatrix} c_1 & b_1 \ c_2 & b_2 \end{vmatrix}, D_2 = \begin{vmatrix} a_1 & c_1 \ a_2 & c_2 \end{vmatrix} ), and ( D \ne 0 )
A system of equations has a unique solution if ( D \ne 0 ); infinite solutions or no solution if ( D = 0 )
The determinant of a diagonal matrix is the product of its diagonal elements; e.g., ( \begin{vmatrix} 2 & 0 \ 0 & 3 \end{vmatrix} = 6 )
The determinant of a triangular matrix (upper or lower) is the product of diagonal entries
If any two rows or columns of a determinant are identical, the value is zero
Intermediate — Requires conceptual clarity on operations and conditions for invertibility, but most formulas are direct from NCERT Class 12, Chapter 4.
Trap: Assuming matrix multiplication is commutative and writing ( AB = BA ) in simplifications. Avoid: Always verify order; remember ( AB \ne BA ) in general.
Trap: Finding inverse using formula without checking if determinant is zero. Avoid: First compute ( |A| ); if ( |A| = 0 ), inverse does not exist.
Trap: Using Cramer’s Rule when determinant ( D = 0 ), leading to incorrect conclusion of unique solution. Avoid: Cramer’s Rule is applicable only if ( D \ne 0 ); otherwise, check consistency separately.
Q1. If ( A = \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix} ), what is ( |A| )? A. 5 B. 6 C. 7 D. 8
Answer: C Explanation: ( |A| = (3)(2) - (1)(-1) = 6 + 1 = 7 ) Why others fail: Option B (6) ignores the negative sign in ( -1 \times -1 )
Q2. Which of the following is the identity matrix of order 2? A. ( \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} ) B. ( \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix} ) C. ( \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} ) D. ( \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} )
Answer: C Explanation: Identity matrix has 1s on diagonal and 0s elsewhere Why others fail: Option A is a permutation matrix, often mistaken for identity
Q3. If ( A ) is a 3×3 matrix with ( |A| = 5 ), what is ( |A^{-1}| )? A. 5 B. 1/5 C. 25 D. -5
Answer: B Explanation: ( |A^{-1}| = 1/|A| = 1/5 ) Why others fail: Option C confuses with ( |\text{adj}(A)| = |A|^{n-1} = 25 )
Q4. For the system ( 2x + 3y = 7 ), ( 4x + 6y = 14 ), what is the value of determinant ( D )? A. 0 B. 12 C. 14 D. 21
Answer: A Explanation: ( D = \begin{vmatrix} 2 & 3 \ 4 & 6 \end{vmatrix} = (2)(6) - (3)(4) = 12 - 12 = 0 ) Why others fail: Students compute ( c_1b_2 - c_2b_1 ) instead of coefficient determinant
Q5. If ( A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix} ), what is the first element of ( \text{adj}(A) )? A. 4 B. -4 C. -3 D. 3
Answer: A Explanation: Cofactor of ( a_{11} = 1 ) is ( C_{11} = (-1)^{1+1} \cdot 4 = 4 ); adjoint is transpose of cofactor matrix Why others fail: Option B (-4) arises from sign error or confusing with ( a_{22} ) cofactor
Join 4M+ learners. Unlock unlimited quizzes, wrong-answer tracking, flashcards + reminders, study guides, and 1-on-1 challenges.