Math Problem Statement

(a)Calculatethedeterminantsofthefollowingmatrices, ifpossible(1p): A=   −2 1 3 1 0 3 0 1 3   (i) B= 3 4 2 −5 −6 −3 (ii) C=     2 3 6 −3 0 −2 −5 9 0 0 0 −4 0 0 0 1    . (iii) (b)Whichofthematricesfrompart(a)haveaninversematrix? (c)Forwhichsystemsofequationscorrespondingtothematricesinpart(a)isthesolution unique?

Solution

Part (a): Determinants of the matrices

(i) Matrix A:

-2 & 1 & 3 \\ 1 & 0 & 3 \\ 0 & 1 & 3 \end{bmatrix}$$ The determinant of a 3x3 matrix is calculated as: $$\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)$$ Where the matrix is: $$\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$$ For matrix A, we substitute the values: $$a = -2, b = 1, c = 3, d = 1, e = 0, f = 3, g = 0, h = 1, i = 3$$ Now we compute each part: $$ei - fh = (0)(3) - (3)(1) = 0 - 3 = -3$$ $$di - fg = (1)(3) - (3)(0) = 3 - 0 = 3$$ $$dh - eg = (1)(1) - (0)(0) = 1$$ Now, substitute these values back into the determinant formula: $$\det(A) = -2(-3) - 1(3) + 3(1) = 6 - 3 + 3 = 6$$ Thus, $$\det(A) = 6$$. #### (ii) Matrix B: $$B = \begin{bmatrix} 3 & 4 & 2 \\ -5 & -6 & -3 \end{bmatrix}$$ This is a $$2 \times 3$$ matrix, and determinants are only defined for square matrices. Therefore, $$\det(B)$$ is **not defined**. #### (iii) Matrix C: $$C = \begin{bmatrix} 2 & 3 & 6 & -3 \\ 0 & -2 & -5 & 9 \\ 0 & 0 & 0 & -4 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$ The determinant of a $$4 \times 4$$ matrix can be computed using cofactor expansion. Expanding along the first row (since the other rows have many zeros): $$\det(C) = 2 \cdot \det\begin{bmatrix} -2 & -5 & 9 \\ 0 & 0 & -4 \\ 0 & 0 & 1 \end{bmatrix}$$ The $$3 \times 3$$ submatrix is upper triangular, so its determinant is the product of the diagonal elements: $$\det\begin{bmatrix} -2 & -5 & 9 \\ 0 & 0 & -4 \\ 0 & 0 & 1 \end{bmatrix} = (-2) \cdot 0 \cdot 1 = 0$$ Thus, $$\det(C) = 2 \times 0 = 0$$. ### Part (b): Inverse of the matrices A matrix has an inverse if and only if its determinant is non-zero. - Matrix **A** has $$\det(A) = 6$$, which is non-zero. Therefore, matrix A **has an inverse**. - Matrix **B** is not square, so it **does not have an inverse**. - Matrix **C** has $$\det(C) = 0$$, so it **does not have an inverse**. ### Part (c): Systems with a unique solution A system of equations corresponding to a matrix has a unique solution if the matrix is square and has a non-zero determinant. - For matrix **A** ($$\det(A) = 6$$), the system has a **unique solution**. - Matrix **B** is not square, so it does not represent a system with a unique solution. - For matrix **C** ($$\det(C) = 0$$), the system does not have a unique solution; it may have no solutions or infinitely many solutions. --- Do you have any questions or need further details on any part of the explanation? Here are 5 related questions you might find interesting: 1. How do you find the determinant of a 4x4 matrix using cofactor expansion? 2. What conditions must a matrix satisfy for it to have an inverse? 3. Why is the determinant important in solving systems of linear equations? 4. Can a non-square matrix ever have an inverse? 5. How does the rank of a matrix relate to the existence of a unique solution? **Tip:** When working with larger matrices, simplifying calculations using properties like triangularity can save time.

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Math Problem Analysis

Mathematical Concepts

Matrix Determinants
Inverse Matrices
Systems of Linear Equations

Formulas

Determinant of a 3x3 matrix: det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)
Determinant of a 4x4 matrix using cofactor expansion

Theorems

A matrix has an inverse if and only if its determinant is non-zero
A system of equations has a unique solution if the corresponding matrix is square and has a non-zero determinant

Suitable Grade Level

Grades 11-12 or College Level

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