Question 1

Which one of the following matrices has an inverse which is not listed?

Accepted Answer

E

Question 2

Matrices, A, B and C are given by   If D = A(2B + 3C)find d23.&#10;A)- 56&#10;B)19&#10;C)- 109&#10;D)217&#10;E)324

Accepted Answer

To find $D = A(2B + 3C)$, we first need to calculate $2B + 3C$, then multiply the result by $A$.Given matrices:$B = \left[\begin{array}{lrr}1 & 5 & 0 \-8 & 4 & 7 \2 & 3 & -9\end{array}\right]$and$C = \left[\begin{array}{lrr}3 & -2 & -7 \-4 & 5 & 1 \3 & 0 & 6\end{array}\right]$Calculate $2B + 3C$:$2B = \left[\begin{array}{lrr}2 & 10 & 0 \-16 & 8 & 14 \4 & 6 & -18\end{array}\right]$$3C = \left[\begin{array}{lrr}9 & -6 & -21 \-12 & 15 & 3 \9 & 0 & 18\end{array}\right]$Adding $2B$ and $3C$:$\left[\begin{array}{lrr}11 & 4 & -21 \-28 & 23 & 17 \13 & 6 & 0\end{array}\right]$Now, multiply this result by $A$:$A = \left[\begin{array}{lrr}3 & -2 & 4 \6 & 1 & 0 \-5 & 9 & 5\end{array}\right]$$A(2B + 3C) = \left[\begin{array}{lrr}3 & -2 & 4 \6 & 1 & 0 \-5 & 9 & 5\end{array}\right] \left[\begin{array}{lrr}11 & 4 & -21 \-28 & 23 & 17 \13 & 6 & 0\end{array}\right]$Multiplying the matrices:$\left[\begin{array}{lrr}(3*11 + -2*(-28) + 4*13) & (3*4 + -2*23 + 4*6) & (3*(-21) + -2*17 + 4*0) \(6*11 + 1*(-28) + 0*13) & (6*4 + 1*23 + 0*6) & (6*(-21) + 1*17 + 0*0) \(-5*11 + 9*(-28) + 5*13) & (-5*4 + 9*23 + 5*6) & (-5*(-21) + 9*17 + 5*0)\end{array}\right]$Calculating the entries:$\left[\begin{array}{lrr}(33 + 56 + 52) & (12 - 46 + 24) & (-63 - 34 + 0) \(66 - 28 + 0) & (24 + 23 + 0) & (-126 + 17 + 0) \(-55 - 252 + 65) & (-20 + 207 + 30) & (105 + 153 + 0)\end{array}\right]$$\left[\begin{array}{lrr}141 & -10 & -97 \38 & 47 & -109 \-242 & 217 & 258\end{array}\right]$The determinant $d₂₃$ refers to the determinant of the resulting matrix $D$, but the question seems to ask for a specific entry rather than calculating a determinant. If we interpret the question as asking for the entry in the third row and second column of $D$, the answer is $217$.

Question 3

For the commodity market C = aY + b and I = cr + d&#10;For the money market&#10;MS = MS* and MD = k1Y + k2r + k3&#10;If both markets are in equilibrium, find the matrix A such that Ax = b where

Accepted Answer

The matrix A is the coefficient matrix of the system of equations that results from setting the commodity market and money market equations equal to zero. The matrix A is:$$\left[\begin{array}{ccc}k_{2} & & k_{1} \-c & 1 & -a\end{array}ight]$$

Question 4

One unit of I1 uses 0.2 units of I1 and 0.6 units of I2. One unit of I2 uses 0.4 units of I1 and 0.2 units of I2. Find the associated Leontief inverse.

Accepted Answer

The answer of One unit of I1 uses 0.2 units...

Question 5

Find the value of a if the following matrix is singular

Accepted Answer

A matrix is singular if its determinant is 0. The determinant of the given matrix is $(-4) \cdot a - (2) \cdot (-6) = -4a + 12$. Setting this equal to 0 gives $-4a + 12 = 0$, solving for $a$ gives $a = 3$.

Question 6

Find the determinant of the matrix  &#10;A)364&#10;B)340&#10;C)100&#10;D)76&#10;E)14

Accepted Answer

We can use cofactor expansion along the first row to compute the determinant: $$\begin{aligned} \det\left(\begin{array}{rrr} 5 & -2 & 3 \ 4 & -1 & -5 \ 6 & 7 & 9 \end{array}\right) &= 5 \det\left(\begin{array}{rr} -1 & -5 \ 7 & 9 \end{array}\right) - (-2) \det\left(\begin{array}{rr} 4 & -5 \ 6 & 9 \end{array}\right) + 3\det\left(\begin{array}{rr} 4 & -1 \ 6 & 7 \end{array}\right) \ &= 5( (-1)(9) - (-5)(7)) - (-2)((4)(9)-(5)(6)) + 3((4)(7)-(6)(-1)) \ &= 364. \end{aligned}$$Therefore, the answer is $\boxed{\textbf{(A)}\ 364}$.

Question 7

If A, B and C are matrices with orders 3&#215;3, 2&#215;3 and 4&#215;2 respectively, how many of the following matrix calculations are possible? 4B, A + B, 3BT + C, AB, BTA, (CB)T, CBA&#10;A)2&#10;B)4&#10;C)1&#10;D)3&#10;E)0

Accepted Answer

To perform matrix multiplication, the number of columns of the first matrix must be equal to the number of rows of the second matrix. 1) 4B is possible as we can multiply a matrix by a scalar. 2) A+B is possible as matrices with the same order can be added. 3) 3B^T+C is not possible as the orders of the matrices do not match for addition. 4) AB is possible as the number of columns in matrix A (3) is equal to the number of rows in matrix B (2). 5) B^TA is not possible as the number of columns in matrix B (3) does not match the number of rows in matrix A (3) for matrix multiplication. 6) (CB)^T is not possible as the number of columns in matrix C (2) does not match the number of rows in matrix B (3) for matrix multiplication. 7) CBA is possible as the number of columns in matrix C (2) is equal to the number of rows in matrix B (3) and the number of columns in matrix B (3) is equal to the number of rows in matrix A (3) for matrix multiplication.

Question 8

Find the cofactor, A23, of the matrix  &#10;A)23&#10;B)-23&#10;C)0&#10;D)-7&#10;E)7

Accepted Answer

A23 is the cofactor of the element in the 2nd row and 3rd column. The element in the 2nd row and 3rd column is -9. The cofactor of an element is found by multiplying the element by -1 and then taking the determinant of the submatrix formed by deleting the row and column of the element. The submatrix formed by deleting the 2nd row and 3rd column is:$$\left[ \begin{array} { r r } 5 & - 2 \4 & - 3 \end{array} ight]$$The determinant of this submatrix is:$$(5)(-3) - (-2)(4) = -15 + 8 = -7$$Therefore, the cofactor of the element in the 2nd row and 3rd column is:$$(-1)(-7) = 7$$

Deck 7: Matrices