Question

Write the equations representing the various dimensions and express them as the matrix equation \(AX = B\).

Answer 1

Concept:
Let the dimensions of the cuboid be \(x\) (length), \(y\) (breadth), and \(z\) (height) in cm.
A system of linear equations can be represented in matrix form as: \[ AX=B \] where \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constant matrix. 
Step 1: Form the algebraic equations
The sum of length and breadth is \(3\) cm more than the height: \[ x+y=z+3 \] Therefore, \[ x+y-z=3 \] Twice the length, thrice the breadth, and the height add up to \(10\) cm: \[ 2x+3y+z=10 \] The breadth added to \(7\) times the height is \(1\) cm less than \(3\) times the length: \[ y+7z=3x-1 \] Therefore, \[ 3x-y-7z=1 \] 
Step 2: Construct the matrix equation
From the coefficients of \(x\), \(y\), and \(z\), the coefficient matrix is: \[ A= \begin{bmatrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 3 & -1 & -7 \end{bmatrix} \] The variable matrix is: \[ X= \begin{bmatrix} x \\ y \\ z \end{bmatrix} \] The constant matrix is: \[ B= \begin{bmatrix} 3 \\ 10 \\ 1 \end{bmatrix} \] 
Step 3: Write in the form \(AX=B\)
Therefore: \[ \begin{bmatrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 3 & -1 & -7 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ 10 \\ 1 \end{bmatrix} \] 
Final Answer:
The required matrix equation is: \[ \boxed{ \begin{bmatrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 3 & -1 & -7 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ 10 \\ 1 \end{bmatrix} } \]

Answer 2

Concept:
A square matrix \(A\) is invertible if and only if it is non-singular.
A matrix is non-singular when: \[ |A|\neq 0 \] 
Step 1: Write the matrix \(A\)
The given coefficient matrix is: \[ A= \begin{bmatrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 3 & -1 & -7 \end{bmatrix} \] 
Step 2: Calculate the determinant of \(A\)
Expanding along the first row: \[ |A|= 1\begin{vmatrix} 3 & 1 \\ -1 & -7 \end{vmatrix} - 1\begin{vmatrix} 2 & 1 \\ 3 & -7 \end{vmatrix} + (-1)\begin{vmatrix} 2 & 3 \\ 3 & -1 \end{vmatrix} \] \[ |A| = 1[3(-7)-1(-1)] - 1[2(-7)-1(3)] - [2(-1)-3(3)] \] \[ |A| = (-21+1)-(-14-3)-(-2-9) \] \[ |A|=-20+17+11 \] \[ |A|=8 \] 
Step 3: Check the existence of \(A^{-1}\)
Since: \[ |A|=8\neq 0 \] the matrix \(A\) is non-singular.
Hence, \(A\) is invertible and \(A^{-1}\) exists. 
Final Answer:
\[ \boxed{|A|=8\neq 0} \] Therefore, \(\boxed{A^{-1}\text{ exists}}\).

Answer 3

Concept:
The inverse of a matrix is given by: \[ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A) \] where \(\operatorname{adj}(A)\) is the transpose of the cofactor matrix. 
Step 1: Calculate the cofactors
The cofactors of matrix \(A\) are: \[ C_{11}=-20,\quad C_{12}=17,\quad C_{13}=-11 \] \[ C_{21}=8,\quad C_{22}=-4,\quad C_{23}=4 \] \[ C_{31}=4,\quad C_{32}=-3,\quad C_{33}=1 \] Therefore, the cofactor matrix is: \[ C= \begin{bmatrix} -20 & 17 & -11 \\ 8 & -4 & 4 \\ 4 & -3 & 1 \end{bmatrix} \] 
Step 2: Find the adjoint of \(A\)
The adjoint is the transpose of the cofactor matrix: \[ \operatorname{adj}(A)=C^T \] \[ \operatorname{adj}(A)= \begin{bmatrix} -20 & 8 & 4 \\ 17 & -4 & -3 \\ -11 & 4 & 1 \end{bmatrix} \] 
Step 3: Calculate \(A^{-1}\)
Given: \[ |A|=8 \] Using: \[ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A) \] we get: \[ A^{-1} = \frac{1}{8} \begin{bmatrix} -20 & 8 & 4 \\ 17 & -4 & -3 \\ -11 & 4 & 1 \end{bmatrix} \] 
Final Answer:
Therefore, \[ \boxed{ A^{-1} = \frac{1}{8} \begin{bmatrix} -20 & 8 & 4 \\ 17 & -4 & -3 \\ -11 & 4 & 1 \end{bmatrix} } \]

Answer 4

Concept:
\(A^2\) represents the matrix product \(A\cdot A\).
The \(3\times3\) identity matrix is: \[ I= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] 
Step 1: Calculate \(A^2\)
Given: \[ A= \begin{bmatrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 3 & -1 & -7 \end{bmatrix} \] Therefore: \[ A^2= \begin{bmatrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 3 & -1 & -7 \end{bmatrix} \begin{bmatrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 3 & -1 & -7 \end{bmatrix} \] Multiplying the corresponding rows and columns: \[ A^2= \begin{bmatrix} 0 & 5 & 7 \\ 11 & 10 & -6 \\ -20 & 7 & 45 \end{bmatrix} \] 
Step 2: Calculate \(7I\)
\[ 7I= \begin{bmatrix} 7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7 \end{bmatrix} \] 
Step 3: Calculate \(A^2+7I\)
\[ A^2+7I= \begin{bmatrix} 0 & 5 & 7 \\ 11 & 10 & -6 \\ -20 & 7 & 45 \end{bmatrix} + \begin{bmatrix} 7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7 \end{bmatrix} \] \[ A^2+7I= \begin{bmatrix} 7 & 5 & 7 \\ 11 & 17 & -6 \\ -20 & 7 & 52 \end{bmatrix} \] 
Final Answer:
Therefore, \[ \boxed{ A^2+7I= \begin{bmatrix} 7 & 5 & 7 \\ 11 & 17 & -6 \\ -20 & 7 & 52 \end{bmatrix} } \]