Question

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

Answer 1

Concept:
To convert a word problem into a matrix equation, we first assign variables to the unknown quantities and then form linear equations based on the given conditions. Let: 
• Length of the wooden cuboid box = \(x\) 
• Breadth of the wooden cuboid box = \(y\) 
• Height of the wooden cuboid box = \(z\) The system of linear equations can be represented in the matrix form: \[ AX=B \] where \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constant matrix. 

Step 1: Form the first equation.
Given statement: 
"The sum of its length and breadth is 3 cm more than its height." This can be written as: \[ x+y=z+3 \] Rearranging, \[ x+y-z=3 \] 

Step 2: Form the second equation.
Given statement: 
"Twice its length, thrice its breadth, and its height add up to 10 cm." Therefore, \[ 2x+3y+z=10 \] 

Step 3: Form the third equation.
Given statement: 
"Its breadth added to 7 times its height is 1 cm less than 3 times its length." Therefore, \[ y+7z=3x-1 \] Rearranging, \[ -3x+y+7z=-1 \] 

Step 4: Convert the equations into matrix form.
The three equations are: \[ x+y-z=3 \] \[ 2x+3y+z=10 \] \[ -3x+y+7z=-1 \] Separating the coefficients, variables, and constants: \[ A= \begin{bmatrix} 1 & 1 & -1\\ 2 & 3 & 1\\ -3 & 1 & 7 \end{bmatrix} \] \[ X= \begin{bmatrix} x\\ y\\ z \end{bmatrix} \] \[ B= \begin{bmatrix} 3\\ 10\\ -1 \end{bmatrix} \] Hence, the required matrix equation is: \[ \begin{bmatrix} 1 & 1 & -1\\ 2 & 3 & 1\\ -3 & 1 & 7 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} 3\\ 10\\ -1 \end{bmatrix} \] Therefore, \[ \boxed{AX=B} \] which matches **Option (A)**.

Answer 2

Concept:
The inverse of a square matrix \(A\) exists only when the matrix is non-singular. A matrix is non-singular if its determinant is non-zero: \[ |A| \neq 0 \] If \[ |A|=0, \] then the matrix is singular and \(A^{-1}\) does not exist. Therefore, we need to calculate the determinant of matrix \(A\). 

Step 1: Write the determinant of matrix \(A\).
Given, \[ A= \begin{bmatrix} 1 & 1 & -1\\ 2 & 3 & 1\\ -3 & 1 & 7 \end{bmatrix} \] Expanding the determinant along the first row, \[ |A|= 1 \begin{vmatrix} 3 & 1\\ 1 & 7 \end{vmatrix} - 1 \begin{vmatrix} 2 & 1\\ -3 & 7 \end{vmatrix} + (-1) \begin{vmatrix} 2 & 3\\ -3 & 1 \end{vmatrix} \] 

Step 2: Calculate the \(2 \times 2\) determinants.
First minor: \[ \begin{vmatrix} 3 & 1\\ 1 & 7 \end{vmatrix} = (3)(7)-(1)(1) = 21-1 = 20 \] 
Second minor: \[ \begin{vmatrix} 2 & 1\\ -3 & 7 \end{vmatrix} = (2)(7)-(1)(-3) \] \[ =14+3=17 \] 
Third minor: \[ \begin{vmatrix} 2 & 3\\ -3 & 1 \end{vmatrix} = (2)(1)-(3)(-3) \] \[ =2+9=11 \] 

Step 3: Substitute the values.
\[ |A|=1(20)-1(17)+(-1)(11) \] \[ |A|=20-17-11 \] \[ |A|=-8 \] 

Step 4: Determine whether \(A^{-1}\) exists.
Since, \[ |A|=-8\neq0 \] the matrix \(A\) is non-singular. Therefore, the inverse of matrix \(A\) exists: \[ \boxed{A^{-1}\text{ exists}} \]

Answer 3

Concept:
The inverse of a square matrix can be found using the adjugate method: \[ A^{-1}=\frac{1}{|A|}\text{adj}(A) \] where \(\text{adj}(A)\) is the adjugate matrix of \(A\). The adjugate matrix is the transpose of the cofactor matrix: \[ \text{adj}(A)=C^T \] The cofactor of an element is given by: \[ C_{ij}=(-1)^{i+j}M_{ij} \] where \(M_{ij}\) is the minor determinant obtained by deleting the \(i\)-th row and \(j\)-th column. 

Step 1: Find the cofactor matrix.
Given, \[ A= \begin{bmatrix} 1 & 1 & -1\\ 2 & 3 & 1\\ -3 & 1 & 7 \end{bmatrix} \] The cofactors are: \[ C_{11} = \begin{vmatrix} 3 & 1\\ 1 & 7 \end{vmatrix} = 21-1=20 \] \[ C_{12} = - \begin{vmatrix} 2 & 1\\ -3 & 7 \end{vmatrix} = -(14+3)=-17 \] \[ C_{13} = \begin{vmatrix} 2 & 3\\ -3 & 1 \end{vmatrix} = 2+9=11 \] \[ C_{21} = - \begin{vmatrix} 1 & -1\\ 1 & 7 \end{vmatrix} = -(7+1)=-8 \] \[ C_{22} = \begin{vmatrix} 1 & -1\\ -3 & 7 \end{vmatrix} = 7-3=4 \] \[ C_{23} = - \begin{vmatrix} 1 & 1\\ -3 & 1 \end{vmatrix} = -(1+3)=-4 \] \[ C_{31} = \begin{vmatrix} 1 & -1\\ 3 & 1 \end{vmatrix} = 1+3=4 \] \[ C_{32} = - \begin{vmatrix} 1 & -1\\ 2 & 1 \end{vmatrix} = -(1+2)=-3 \] \[ C_{33} = \begin{vmatrix} 1 & 1\\ 2 & 3 \end{vmatrix} = 3-2=1 \] Therefore, the cofactor matrix is: \[ C= \begin{bmatrix} 20 & -17 & 11\\ -8 & 4 & -4\\ 4 & -3 & 1 \end{bmatrix} \] 

Step 2: Find the adjugate matrix.
The adjugate matrix is the transpose of the cofactor matrix: \[ \text{adj}(A)=C^T \] Hence, \[ \text{adj}(A)= \begin{bmatrix} 20 & -8 & 4\\ -17 & 4 & -3\\ 11 & -4 & 1 \end{bmatrix} \] 

Step 3: Calculate \(A^{-1}\).
From the previous calculation, \[ |A|=-8 \] Using the inverse formula: \[ A^{-1} = \frac{1}{-8} \begin{bmatrix} 20 & -8 & 4\\ -17 & 4 & -3\\ 11 & -4 & 1 \end{bmatrix} \] Therefore, \[ \boxed{ A^{-1} = -\frac{1}{8} \begin{bmatrix} 20 & -8 & 4\\ -17 & 4 & -3\\ 11 & -4 & 1 \end{bmatrix} } \] Hence, the inverse of matrix \(A\) is obtained using the adjugate method and matches **Option (A)**.

Answer 4

Concept:
To evaluate the matrix expression \(A^2+7I\), we first calculate the square of matrix \(A\) by performing matrix multiplication: \[ A^2=A\cdot A \] Then, we calculate \(7I\), where \(I\) is the \(3\times3\) identity matrix, and finally add the two matrices element-wise. 

Step 1: Calculate \(A^2\).
Given, \[ A= \begin{bmatrix} 1 & 1 & -1\\ 2 & 3 & 1\\ -3 & 1 & 7 \end{bmatrix} \] Therefore, \[ A^2= \begin{bmatrix} 1 & 1 & -1\\ 2 & 3 & 1\\ -3 & 1 & 7 \end{bmatrix} \begin{bmatrix} 1 & 1 & -1\\ 2 & 3 & 1\\ -3 & 1 & 7 \end{bmatrix} \] Calculating each element: 
First row: \[ (A^2)_{11}=(1)(1)+(1)(2)+(-1)(-3)=1+2+3=6 \] \[ (A^2)_{12}=(1)(1)+(1)(3)+(-1)(1)=1+3-1=3 \] \[ (A^2)_{13}=(1)(-1)+(1)(1)+(-1)(7)=-1+1-7=-7 \] 
Second row: \[ (A^2)_{21}=(2)(1)+(3)(2)+(1)(-3)=2+6-3=5 \] \[ (A^2)_{22}=(2)(1)+(3)(3)+(1)(1)=2+9+1=12 \] \[ (A^2)_{23}=(2)(-1)+(3)(1)+(1)(7)=-2+3+7=8 \] 
Third row: \[ (A^2)_{31}=(-3)(1)+(1)(2)+(7)(-3)=-3+2-21=-22 \] \[ (A^2)_{32}=(-3)(1)+(1)(3)+(7)(1)=-3+3+7=7 \] \[ (A^2)_{33}=(-3)(-1)+(1)(1)+(7)(7)=3+1+49=53 \] Hence, \[ A^2= \begin{bmatrix} 6 & 3 & -7\\ 5 & 12 & 8\\ -22 & 7 & 53 \end{bmatrix} \] 

Step 2: Calculate \(7I\).
The identity matrix is: \[ I= \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} \] Therefore, \[ 7I= \begin{bmatrix} 7&0&0\\ 0&7&0\\ 0&0&7 \end{bmatrix} \] 

Step 3: Add \(A^2\) and \(7I\).
\[ A^2+7I= \begin{bmatrix} 6&3&-7\\ 5&12&8\\ -22&7&53 \end{bmatrix} + \begin{bmatrix} 7&0&0\\ 0&7&0\\ 0&0&7 \end{bmatrix} \] Adding corresponding elements: \[ A^2+7I= \begin{bmatrix} 6+7&3+0&-7+0\\ 5+0&12+7&8+0\\ -22+0&7+0&53+7 \end{bmatrix} \] \[ = \begin{bmatrix} 13&3&-7\\ 5&19&8\\ -22&7&60 \end{bmatrix} \] Therefore, \[ \boxed{ A^2+7I= \begin{bmatrix} 13&3&-7\\ 5&19&8\\ -22&7&60 \end{bmatrix} } \]