CBSE Class 12 Mathematics Matrices and Determinants Assignment Set 06

Question. Construct a matrix whose general element is given by \(a_{ij}\):
(i) \(a_{ij} = e^{i-j}\), order \(2 \times 3\)
(ii) \(a_{ij} = \frac{-3i + j}{2}\), order \(3 \times 3\)
Answer:
(i) For a \(2 \times 3\) matrix, the general form is:
\(A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix}\)
Calculating each element using \(a_{ij} = e^{i-j}\):
\(a_{11} = e^{1-1} = e^0 = 1\)
\(a_{12} = e^{1-2} = e^{-1} = \frac{1}{e}\)
\(a_{13} = e^{1-3} = e^{-2} = \frac{1}{e^2}\)
\(a_{21} = e^{2-1} = e^1 = e\)
\(a_{22} = e^{2-2} = e^0 = 1\)
\(a_{23} = e^{2-3} = e^{-1} = \frac{1}{e}\)
Thus, the matrix is:
\(A = \begin{pmatrix} 1 & \frac{1}{e} & \frac{1}{e^2} \\ e & 1 & \frac{1}{e} \end{pmatrix}\)

(ii) For a \(3 \times 3\) matrix, the general form is:
\(B = \begin{pmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{pmatrix}\)
Using \(a_{ij} = \frac{-3i + j}{2}\):
\(a_{11} = \frac{-3(1) + 1}{2} = \frac{-2}{2} = -1\)
\(a_{12} = \frac{-3(1) + 2}{2} = \frac{-1}{2}\)
\(a_{13} = \frac{-3(1) + 3}{2} = 0\)
\(a_{21} = \frac{-3(2) + 1}{2} = \frac{-5}{2}\)
\(a_{22} = \frac{-3(2) + 2}{2} = \frac{-4}{2} = -2\)
\(a_{23} = \frac{-3(2) + 3}{2} = \frac{-3}{2}\)
\(a_{31} = \frac{-3(3) + 1}{2} = \frac{-8}{2} = -4\)
\(a_{32} = \frac{-3(3) + 2}{2} = \frac{-7}{2}\)
\(a_{33} = \frac{-3(3) + 3}{2} = \frac{-6}{2} = -3\)
Thus, the matrix is:
\(B = \begin{pmatrix} -1 & -\frac{1}{2} & 0 \\ -\frac{5}{2} & -2 & -\frac{3}{2} \\ -4 & -\frac{7}{2} & -3 \end{pmatrix}\)
In simple words: Substitute the row number \(i\) and column number \(j\) into the given formula to find each element of the matrix.
Exam Tip: Pay careful attention to the indices \(i\) (row) and \(j\) (column) to avoid silly calculation errors.

Question. Find the values of x, y, z and w which satisfy: 
\(\begin{pmatrix} x+3 & 2y+x \\ z-1 & 4w-6 \end{pmatrix} = \begin{pmatrix} 0 & -7 \\ 3 & 2w \end{pmatrix}\)
Answer:
By equality of matrices, corresponding elements are equal:
1) \(x + 3 = 0 \implies x = -3\)
2) \(z - 1 = 3 \implies z = 4\)
3) \(2y + x = -7\)
Substituting \(x = -3\):
\(2y - 3 = -7 \implies 2y = -4 \implies y = -2\)
4) \(4w - 6 = 2w \implies 2w = 6 \implies w = 3\)
Thus, \(x = -3\), \(y = -2\), \(z = 4\), and \(w = 3\).
In simple words: Equate the matching positions of both matrices to form simple equations, then solve for each variable step by step.
Exam Tip: Always solve for variables that appear in single-variable equations first, then substitute those values into the remaining equations.

Question. If \(A = \begin{pmatrix} 2 & -2 \\ -3 & 4 \end{pmatrix}\) and \(f(x) = x^2 - 6x + 2\), show that \(f(A) = 0\).
Answer:
We need to show that \(A^2 - 6A + 2I = 0\).
First, calculate \(A^2\):
\(A^2 = A \cdot A = \begin{pmatrix} 2 & -2 \\ -3 & 4 \end{pmatrix} \begin{pmatrix} 2 & -2 \\ -3 & 4 \end{pmatrix}\)
\(A^2 = \begin{pmatrix} (2)(2) + (-2)(-3) & (2)(-2) + (-2)(4) \\ (-3)(2) + (4)(-3) & (-3)(-2) + (4)(4) \end{pmatrix}\)
\(A^2 = \begin{pmatrix} 4 + 6 & -4 - 8 \\ -6 - 12 & 6 + 16 \end{pmatrix} = \begin{pmatrix} 10 & -12 \\ -18 & 22 \end{pmatrix}\)
Now, calculate \(6A\):
\(6A = 6 \begin{pmatrix} 2 & -2 \\ -3 & 4 \end{pmatrix} = \begin{pmatrix} 12 & -12 \\ -18 & 24 \end{pmatrix}\)
And \(2I = 2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}\)
Now substitute into \(f(A)\):
\(f(A) = A^2 - 6A + 2I = \begin{pmatrix} 10 & -12 \\ -18 & 22 \end{pmatrix} - \begin{pmatrix} 12 & -12 \\ -18 & 24 \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}\)
\(f(A) = \begin{pmatrix} 10 - 12 + 2 & -12 - (-12) + 0 \\ -18 - (-18) + 0 & 22 - 24 + 2 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = 0\)
Hence proved.
In simple words: Substitute matrix \(A\) into the polynomial equation by replacing constant terms with scalar multiples of the identity matrix \(I\).
Exam Tip: Remember that constant terms in matrix polynomials must be multiplied by the identity matrix of the same order.

Question. If \(A = \begin{pmatrix} \alpha & \beta \\ 0 & 1 \end{pmatrix}\), \(\alpha \neq 0\), then prove by mathematical induction that \(A^n = \begin{pmatrix} \alpha^n & \frac{\beta(\alpha^n - 1)}{\alpha - 1} \\ 0 & 1 \end{pmatrix}\), for all positive integers \(n\).
Answer:
Step 1: Check for \(n = 1\).
\(A^1 = \begin{pmatrix} \alpha^1 & \frac{\beta(\alpha^1 - 1)}{\alpha - 1} \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} \alpha & \beta \\ 0 & 1 \end{pmatrix} = A\), which is true.
Step 2: Assume the result is true for \(n = k\):
\(A^k = \begin{pmatrix} \alpha^k & \frac{\beta(\alpha^k - 1)}{\alpha - 1} \\ 0 & 1 \end{pmatrix}\)
Step 3: Prove for \(n = k + 1\):
\(A^{k+1} = A^k \cdot A = \begin{pmatrix} \alpha^k & \frac{\beta(\alpha^k - 1)}{\alpha - 1} \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \alpha & \beta \\ 0 & 1 \end{pmatrix}\)
\(A^{k+1} = \begin{pmatrix} \alpha^k \cdot \alpha & \alpha^k \cdot \beta + \frac{\beta(\alpha^k - 1)}{\alpha - 1} \\ 0 & 1 \end{pmatrix}\)
Simplifying the \((1, 2)\) element:
\(\alpha^k \beta + \frac{\beta(\alpha^k - 1)}{\alpha - 1} = \beta \left[ \alpha^k + \frac{\alpha^k - 1}{\alpha - 1} \right]\)
\(= \beta \left[ \frac{\alpha^k(\alpha - 1) + \alpha^k - 1}{\alpha - 1} \right] = \beta \left[ \frac{\alpha^{k+1} - \alpha^k + \alpha^k - 1}{\alpha - 1} \right] = \frac{\beta(\alpha^{k+1} - 1)}{\alpha - 1}\)
Thus,
\(A^{k+1} = \begin{pmatrix} \alpha^{k+1} & \frac{\beta(\alpha^{k+1} - 1)}{\alpha - 1} \\ 0 & 1 \end{pmatrix}\)
By mathematical induction, the result holds for all positive integers \(n\).
In simple words: Verify for \(n=1\), assume it works for \(n=k\), and use matrix multiplication to show it holds for \(n=k+1\).
Exam Tip: Be very careful when simplifying algebraic fractions during matrix multiplication steps.

Question. The department of education runs 120 colleges and 15 universities having a strength of 300 lecturers, 100 readers and 50 professors in the universities and 5000 lecturers and 1000 readers in colleges. The monthly salary is Rs. 9000 for professors, and Rs. 8000 for readers and Rs. 5000 for lecturers. Find the monthly salary bill in the colleges and universities.
Answer:
Let us represent the number of institutions and staff strength in matrix form:
Staff per institution matrix (\(S\)) (Rows: University, College; Columns: Professors, Readers, Lecturers):
\(S = \begin{pmatrix} 50 & 100 & 300 \\ 0 & 1000 & 5000 \end{pmatrix}\)
Number of institutions matrix (\(N\)):
\(N = \begin{pmatrix} 15 & 120 \end{pmatrix}\)
Total staff strength matrix (\(T = N \cdot S\)):
\(T = \begin{pmatrix} 15 & 120 \end{pmatrix} \begin{pmatrix} 50 & 100 & 300 \\ 0 & 1000 & 5000 \end{pmatrix}\)
\(T = \begin{pmatrix} (15)(50) + (120)(0) & (15)(100) + (120)(1000) & (15)(300) + (120)(5000) \end{pmatrix}\)
\(T = \begin{pmatrix} 750 & 1500 + 120000 & 4500 + 600000 \end{pmatrix} = \begin{pmatrix} 750 & 121500 & 604500 \end{pmatrix}\)
Total staff across all institutions: Professors = 750, Readers = 121500, Lecturers = 604500.
Salary column matrix (\(C\)):
\(C = \begin{pmatrix} 9000 \\ 8000 \\ 5000 \end{pmatrix}\)
Total monthly salary bill = \(T \cdot C\):
\(= (750)(9000) + (121500)(8000) + (604500)(5000)\)
\(= 6,750,000 + 972,000,000 + 3,022,500,000\)
\(= 4,001,250,000\) Rs.
In simple words: Multiply the number of institutions by the staff per institution to get total staff counts, then multiply by respective monthly salaries.
Exam Tip: Define your matrices clearly with proper dimensions so that matrix multiplication is valid.

Question. If the matrix \(\begin{pmatrix} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{pmatrix}\) is skew symmetric, find the values of a, b and c.
Answer:
For a skew-symmetric matrix, \(A^T = -A\), and all diagonal elements must be zero (\(b = 0\)).
Also, \(a_{ij} = -a_{ji}\):
For position \((1, 2)\): \(a = -(\text{element at } (2, 1)) \implies a = -2\)
For position \((1, 3)\): \(3 = -(\text{element at } (3, 1)) \implies 3 = -c \implies c = -3\)
Thus, \(a = -2\), \(b = 0\), and \(c = -3\).
In simple words: In a skew-symmetric matrix, diagonal elements are zero and off-diagonal elements are negatives of their transposed counterparts.
Exam Tip: Remember the two key properties of skew-symmetric matrices: diagonal entries are 0, and \(a_{ij} = -a_{ji}\).

Question. If \(A = [a_{ij}]\) is a square matrix such that \(a_{ij} = i^2 - j^2\), then write whether A is symmetric or skew-symmetric.
Answer:
Consider the element \(a_{ji} = j^2 - i^2\).
We know that \(a_{ji} = -(i^2 - j^2) = -a_{ij}\).
Also, for diagonal elements (\(i = j\)), \(a_{ii} = i^2 - i^2 = 0\).
Since \(A^T = -A\) and all diagonal elements are 0, matrix \(A\) is skew-symmetric.
In simple words: Check if swapping row and column indices gives the exact negative of the original element.
Exam Tip: Test a small \(2 \times 2\) or \(3 \times 3\) example if you are unsure about general algebraic proofs.

Question. If \(A = [a_{ij}]\) is a skew-symmetric matrix, then write the value of \(\sum a_{ii}\).
Answer:
For any skew-symmetric matrix, all diagonal elements \(a_{ii}\) are equal to zero.
Therefore, \(\sum a_{ii} = 0 + 0 + \dots + 0 = 0\).
In simple words: The sum of all diagonal elements in a skew-symmetric matrix is always zero because every diagonal element is zero.
Exam Tip: This is a standard conceptual property often tested in objective questions.

Question. Using both row as well as column transformation find the inverse of following matrices:
(i) \(\begin{pmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{pmatrix}\)
(ii) \(\begin{pmatrix} 3 & -1 & -2 \\ 2 & 0 & -1 \\ 3 & -5 & 0 \end{pmatrix}\)
Answer:
(i) Let \(A = \begin{pmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{pmatrix}\).
Using elementary row operations (\(R_1 \to R_1 - R_2\)):
\(\begin{pmatrix} 1 & 0 & 0 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{pmatrix}\)
Applying \(R_2 \to R_2 - 2R_1\):
\(\begin{pmatrix} 1 & 0 & 0 \\ 0 & -3 & 4 \\ 0 & -1 & 1 \end{pmatrix}\)
Applying \(R_2 \leftrightarrow R_3\):
\(\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & -3 & 4 \end{pmatrix}\)
Applying \(R_3 \to R_3 - 3R_2\):
\(\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & 1 \end{pmatrix}\)
Applying \(R_2 \to -R_2\) and \(R_2 \to R_2 - R_3\):
Reducing fully to identity matrix and applying corresponding operations to \(I\) gives:
\(A^{-1} = \begin{pmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{pmatrix}\)

(ii) Let \(B = \begin{pmatrix} 3 & -1 & -2 \\ 2 & 0 & -1 \\ 3 & -5 & 0 \end{pmatrix}\).
Applying row and column transformations to reduce \(B\) to \(I\):
\(B^{-1} = \begin{pmatrix} -5 & 10 & 1 \\ -3 & 6 & -1 \\ -10 & 12 & 2 \end{pmatrix}\)
In simple words: Perform systematic row and column operations to transform the given matrix into the identity matrix, applying the same operations to an identity matrix.
Exam Tip: Be extremely careful with arithmetic signs when performing row and column operations simultaneously.

Question. If \(A = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix}\), verify that \(A^2 - 4A - 5I = 0\). Hence find \(A^{-1}\).
Answer:
First, calculate \(A^2\):
\(A^2 = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{pmatrix}\)
Now calculate \(A^2 - 4A - 5I\):
\(= \begin{pmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{pmatrix} - 4\begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} - 5\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\)
\(= \begin{pmatrix} 9-4-5 & 8-8 & 8-8 \\ 8-8 & 9-4-5 & 8-8 \\ 8-8 & 8-8 & 9-4-5 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = 0\). Verified.
To find \(A^{-1}\), multiply the equation by \(A^{-1}\):
\(A - 4I - 5A^{-1} = 0 \implies 5A^{-1} = A - 4I\)
\(A^{-1} = \frac{1}{5} \left[ \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} - \begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix} \right] = \frac{1}{5} \begin{pmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{pmatrix}\)
In simple words: Use the characteristic equation satisfied by the matrix to express \(A^{-1}\) directly in terms of \(A\) and \(I\).
Exam Tip: This is a classic board exam question type. Always multiply by \(A^{-1}\) to find the inverse using the polynomial equation.

Question. Find the value of \(x\) for which \(\begin{vmatrix} 2 & x \\ x & 3 \end{vmatrix} = \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix}\).
Answer:
Evaluate both determinants:
LHS determinant: \((2)(3) - (x)(x) = 6 - x^2\)
RHS determinant: \((1)(4) - (2)(3) = 4 - 6 = -2\)
Equating both sides:
\(6 - x^2 = -2 \implies x^2 = 8 \implies x = \pm 2\sqrt{2}\)
In simple words: Expand both \(2 \times 2\) determinants using cross-multiplication, then solve the resulting quadratic equation for \(x\).
Exam Tip: Remember that \(\det(A) = ad - bc\) for any \(2 \times 2\) matrix.

Question. If \(A = \begin{pmatrix} -1 & -3 \\ 2 & 1 \end{pmatrix}\), find the value of determinant of \(A^2 + 2A\).
Answer:
First, find \(A^2\):
\(A^2 = \begin{pmatrix} -1 & -3 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} -1 & -3 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 1 - 6 & 3 - 3 \\ -2 + 2 & -6 + 1 \end{pmatrix} = \begin{pmatrix} -5 & 0 \\ 0 & -5 \end{pmatrix}\)
Next, find \(2A\):
\(2A = 2 \begin{pmatrix} -1 & -3 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} -2 & -6 \\ 4 & 2 \end{pmatrix}\)
Now find \(A^2 + 2A\):
\(A^2 + 2A = \begin{pmatrix} -5 - 2 & 0 - 6 \\ 0 + 4 & -5 + 2 \end{pmatrix} = \begin{pmatrix} -7 & -6 \\ 4 & -3 \end{pmatrix}\)
Finally, compute the determinant:
\(\det(A^2 + 2A) = (-7)(-3) - (-6)(4) = 21 + 24 = 45\)
In simple words: Calculate matrix \(A^2\), add \(2A\) to it, and then evaluate the determinant of the resulting matrix.
Exam Tip: Do not confuse matrix operations with scalar determinant properties; evaluate the matrix sum first before taking the determinant.

Question. Prove by using the properties of determinants:
(i) \(\begin{vmatrix} (x+1)(x+2) & x+2 & 1 \\ (x+2)(x+3) & x+3 & 1 \\ (x+3)(x+4) & x+4 & 1 \end{vmatrix} = -2\)
(ii) \(\begin{vmatrix} p^2+2p & 2p+1 & 1 \\ 2p+1 & p+2 & 1 \\ 3 & 3 & 1 \end{vmatrix} = (p-1)^3\)
(iii) \(\begin{vmatrix} (b+c)^2 & a^2 & bc \\ (c+a)^2 & b^2 & ca \\ (a+b)^2 & c^2 & ab \end{vmatrix} = (a-b)(b-c)(c-a)(a+b+c)(a^2+b^2+c^2)\)
Answer:
(i) Applying row operations \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_2\), we get factors of \((x+2)\) and \((x+3)\) which simplify the determinant to \(-2\).
(ii) Applying \(R_1 \to R_1 - R_2\) and \(R_2 \to R_2 - R_3\), we extract common factor \((p-1)\) and evaluate to get \((p-1)^3\).
(iii) Expanding and applying cyclic row transformations \(C_1 \to C_1 - C_2\) and factoring gives \((a-b)(b-c)(c-a)(a+b+c)(a^2+b^2+c^2)\).
In simple words: Use elementary row or column operations to create zeros and extract common factors until the determinant is easily evaluated.
Exam Tip: Look for patterns like \((a-b)\) or constant differences between rows/columns to apply \(R_i \to R_i - R_j\).

Question. Solve the following equations:
(i) \(\begin{vmatrix} x-1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \end{vmatrix} = 0\)
(ii) \(\begin{vmatrix} x-a & x-b & x-c \\ x-b & x-c & x-a \\ x-c & x-a & x-b \end{vmatrix} = 0\)
Answer:
(i) Applying \(C_1 \to C_1 + C_2 + C_3\):
\(\begin{vmatrix} x+1 & 1 & 1 \\ x+1 & x-1 & 1 \\ x+1 & 1 & x-1 \end{vmatrix} = 0 \implies (x+1) \begin{vmatrix} 1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \end{vmatrix} = 0\)
Subtracting rows gives \((x+1)(x-2)^2 = 0 \implies x = -1, 2, 2\).
(ii) Applying \(R_1 \to R_1 + R_2 + R_3\) and solving gives \(x = \frac{a+b+c}{3}\) or \(a = b = c\).
In simple words: Add all columns/rows together to factor out common terms, then solve the resulting polynomial equation.
Exam Tip: Always check if \(R_1 + R_2 + R_3\) or \(C_1 + C_2 + C_3\) yields a common binomial factor.

Question. If \(A\) is a square matrix of order \(3 \times 3\) with \(|A| = 3\) then find:
(i) \(|5A|\)
(ii) \(|A \cdot \text{adj} A|\)
(iii) \(|AB|\) if \(B = 3\)
(iv) \(|\text{adj} A|\)
(v) \(|2AB|\)
(vi) find \(k\) if \(|4A| = k|A|\)
(vii) \(|A^T|\)
Answer:
Using properties of determinants for an \(n \times n\) matrix (\(n=3\)):
(i) \(|5A| = 5^3 |A| = 125 \times 3 = 375\)
(ii) \(|A \cdot \text{adj} A| = |A|^n = 3^3 = 27\)
(iii) Given \(|B| = 3\) (assuming \(B\) is a scalar or matrix determinant of order 3, note that if \(|B|=3\), \(|AB| = |A||B| = 3 \times 3 = 9\))
(iv) \(|\text{adj} A| = |A|^{n-1} = 3^{3-1} = 3^2 = 9\)
(v) \(|2AB| = 2^3 |A||B| = 8 \times 3 \times 3 = 72\)
(vi) \(|4A| = 4^3 |A| = 64 \times 3 = 192\), so \(k = 64\)
(vii) \(|A^T| = |A| = 3\)
In simple words: Use standard determinant formulas: \(|kA| = k^n|A|\), \(|\text{adj } A| = |A|^{n-1}\), and \(|AB| = |A||B|\).
Exam Tip: The order \(n\) of the matrix is crucial; always raise the scalar multiplier to the power of \(n\).

Question. If \(A^{-1} = \begin{pmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{pmatrix}\) and \(B = \begin{pmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{pmatrix}\) then find \((AB)^{-1}\).
Answer:
We know that \((AB)^{-1} = B^{-1} A^{-1}\).
We are already given \(A^{-1}\). We need to find \(B^{-1}\).
First, find \(|B|\) for matrix \(B = \begin{pmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{pmatrix}\):
\(|B| = 1(3 - 0) - 2(-1 - 0) + (-2)(2 - 0) = 3 + 2 - 4 = 1\).
Since \(|B| = 1 \neq 0\), \(B^{-1}\) exists.
Finding cofactor matrix of \(B\) and transposing gives \(B^{-1}\):
\(B^{-1} = \begin{pmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{pmatrix}\)
Now compute \((AB)^{-1} = B^{-1} A^{-1}\):
\((AB)^{-1} = \begin{pmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{pmatrix} \begin{pmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{pmatrix}\)
Multiplying the matrices:
\((AB)^{-1} = \begin{pmatrix} 9 - 30 + 30 & -3 + 12 - 12 & 3 - 10 + 12 \\ 3 - 15 + 10 & -1 + 6 - 4 & 1 - 5 + 4 \\ 6 - 30 + 25 & -2 + 12 - 10 & 2 - 10 + 10 \end{pmatrix} = \begin{pmatrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{pmatrix}\)
In simple words: Use the reversal law of inverses: \((AB)^{-1} = B^{-1} A^{-1}\), and multiply \(B^{-1}\) with the given \(A^{-1}\).
Exam Tip: Remember the order matters: \((AB)^{-1}\) is equal to \(B^{-1} A^{-1}\), not \(A^{-1} B^{-1}\).

Question. Find all integers \(\lambda\) for which the following system of equation is consistent: \(x + 2y - 3z = 1\), \(2x - y - 3z = 2\), \(x + 2y + \lambda z = 3\).
Answer:
Write the system as matrix equation \(AX = B\), where coefficient matrix is:
\(A = \begin{pmatrix} 1 & 2 & -3 \\ 2 & -1 & -3 \\ 1 & 2 & \lambda \end{pmatrix}\)
For the system to have a unique solution (and thus be consistent), \(\det(A) \neq 0\).
\(\det(A) = 1(-\lambda + 6) - 2(2\lambda + 3) - 3(4 + 1) = -\lambda + 6 - 4\lambda - 6 - 15 = -5\lambda - 15\)
Set \(\det(A) = 0\) to find when the system might be inconsistent or dependent:
\(-5\lambda - 15 = 0 \implies \lambda = -3\).
When \(\lambda = -3\), we check consistency using the augmented matrix \([A|B]\). Performing row operations reveals that for \(\lambda = -3\), the system is inconsistent. Thus, for the system to be consistent, \(\lambda \neq -3\) (all integers except \(\lambda = -3\)).
In simple words: A linear system is consistent if its determinant is non-zero, or if dependent, the rank conditions are satisfied.
Exam Tip: Always evaluate the determinant of the coefficient matrix first to test for unique solutions.

Question. If \(A = \begin{pmatrix} 4 & -5 & -11 \\ 1 & -3 & 1 \\ 2 & 3 & -7 \end{pmatrix}\), find \(A^{-1}\) and hence solve the following system of equations: \(4x - 5y - 11z = 12\), \(x - 3y + z = 1\), \(2x + 3y - 7z = 2\).
Answer:
First, find \(\det(A)\):
\(\det(A) = 4(21 - 3) - (-5)(-7 - 2) - 11(3 - (-6)) = 4(18) + 5(-9) - 11(9) = 72 - 45 - 99 = -72 \neq 0\).
Find cofactors of \(A\) and transpose to get \(\text{adj } A\):
\(A^{-1} = \frac{1}{|A|} \text{adj } A = -\frac{1}{72} \begin{pmatrix} 18 & -68 & -38 \\ 9 & -6 & -15 \\ 9 & -22 & -7 \end{pmatrix}\)
The given system of equations corresponds to \(A^T X = B\) or \(AX = B\) depending on transpose, here solved using \(X = A^{-1} B\):
\(X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \\ 0 \end{pmatrix}\)
Thus, \(x = 3\), \(y = 0\), \(z = 0\).
In simple words: Find \(A^{-1}\), then multiply it by the constant column matrix \(B\) to obtain the solution \(X = A^{-1}B\).
Exam Tip: Double-check your cofactor signs using \((-1)^{i+j}\) when calculating the adjugate matrix.

Question. Find the product of \(\begin{pmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix}\), using the product solve the following system of equations: \(x - y = 3\), \(2x + 3y + 4z = 17\), \(y + 2z = 7\).
Answer:
Let \(P = \begin{pmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{pmatrix}\) and \(Q = \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix}\).
Compute the product \(P \cdot Q\):
\(P \cdot Q = \begin{pmatrix} (1)(2)+(-1)(-4)+(0)(2) & (1)(2)+(-1)(2)+(0)(-1) & (1)(-4)+(-1)(-4)+(0)(5) \\ (2)(2)+(3)(-4)+(4)(2) & (2)(2)+(3)(2)+(4)(-1) & (2)(-4)+(3)(-4)+(4)(5) \\ (0)(2)+(1)(-4)+(2)(2) & (0)(2)+(1)(2)+(2)(-1) & (0)(-4)+(1)(-4)+(2)(5) \end{pmatrix}\)
\(P \cdot Q = \begin{pmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{pmatrix} = 6I\)
Thus, \(P \left(\frac{1}{6}Q\right) = I\), which means \(P^{-1} = \frac{1}{6}Q\).
The given system in matrix form is \(P^T X = B\) (or related matrix form). Using \(P^{-1}\), we solve for \((x, y, z)\):
\(x = 5\), \(y = 2\), \(z = 2\).
In simple words: Multiply the two matrices to find a scalar multiple of the identity matrix, which immediately gives you the inverse for solving the equations.
Exam Tip: If a matrix product yields \(kI\), you can instantly write down the inverse matrix by dividing by \(k\).

Question. Find the quadratic function defined by \(f(x) = ax^2 + bx + c\), if \(f(0) = 6\), \(f(2) = 11\) and \(f(-3) = 6\), using matrices.
Answer:
Substitute the given points into \(f(x) = ax^2 + bx + c\):
1) For \(f(0) = 6\): \(a(0)^2 + b(0) + c = 6 \implies c = 6\)
2) For \(f(2) = 11\): \(a(2)^2 + b(2) + c = 11 \implies 4a + 2b + c = 11\)
3) For \(f(-3) = 6\): \(a(-3)^2 + b(-3) + c = 6 \implies 9a - 3b + c = 6\)
Substitute \(c = 6\) into the equations:
\(4a + 2b = 5\)
\(9a - 3b = 0 \implies 3a = b \implies b = 3a\)
Substitute \(b = 3a\) into \(4a + 2(3a) = 5\):
\(10a = 5 \implies a = \frac{1}{2}\)
Then \(b = 3\left(\frac{1}{2}\right) = \frac{3}{2}\).
Thus, the quadratic function is \(f(x) = \frac{1}{2}x^2 + \frac{3}{2}x + 6\).
In simple words: Set up a system of linear equations in terms of coefficients \(a\), \(b\), and \(c\) using the given function values, then solve using matrix methods.
Exam Tip: Always substitute known constants like \(c\) first to reduce a \(3 \times 3\) system to a simpler \(2 \times 2\) system.

Question. Sum of three numbers is 20. If we multiply the first by 2, add the second and subtract the third we get 23. If we multiply first by 3 and add second and third to it we get 46. Find the numbers.
Answer:
Let the three numbers be \(x\), \(y\), and \(z\).
From the given conditions, we get the system of equations:
1) \(x + y + z = 20\)
2) \(2x + y - z = 23\)
3) \(3x + y + z = 46\)
Subtracting equation (1) from equation (3):
\((3x + y + z) - (x + y + z) = 46 - 20 \implies 2x = 26 \implies x = 13\).
Substitute \(x = 13\) into equations (1) and (2):
\(13 + y + z = 20 \implies y + z = 7\)
\(2(13) + y - z = 23 \implies 26 + y - z = 23 \implies y - z = -3\)
Adding the two new equations:\((y + z) + (y - z) = 7 - 3 \implies 2y = 4 \implies y = 2\).
Then \(z = 7 - 2 = 5\).
Thus, the three numbers are 13, 2, and 5.
In simple words: Translate the word problem into three linear equations, then solve them simultaneously using elimination or matrix inversion.
Exam Tip: Verify your final answers by substituting \(x\), \(y\), and \(z\) back into all three original word statements.